Putt 1 travels flat and then shortly before the hole, it rises 6 inches.
Putt 2 has an rise of 6 inches right after you hit it, and then travels flat to the hole the rest of the way.
Putt 3 rises 6 inches gently from the ball to the hole at a constant angle.
Do you hit all three putts the same initial speed?
Follow up question : What if you have the same 20 foot putt but halfway to the hole the ground rises 8 inches, and then drops 2 inches and then flattens out the rest of the way to the hole. . .same speed?
ANSWER: v hole = √(v putt 2 -2gh ).
However, there is friction which will slow the ball further and the
frictional force on a slope (f=μmg cosθ ) will
be different than on level ground (f=μmg ); here μ
is the coefficient of friction (essentially, I believe, what is
measured by the "stimp meter"), m is the mass of the ball, and
g is the acceleration due to gravity. Since the frictional
force is smaller on the slope, you might think that less energy is lost
to friction on the slope. But, what matters is not the force but its
product with the distance s over which it acts, specifically
the work done by friction is W f =-fs where
the negative sign indicates that the friction takes energy away from the
ball. When the ball is moving forward along along the segment L 2 ,
the distance it travels is s=L 2 /cosθ
and so the work done is W f =fs=(μmg cosθ )(L 2 /cosθ )=μmgL 2 ,
exactly the same as if the slope was not there. The total work done by
friction, regardless of the path, is W f =μmgL.
The final velocity is then
v hole = √[v putt 2 -2g (h+μL )].
ANSWER:
ANSWER: a such that he moves a
distance y in a time t : y= ½at2
, so for this case a =2y /t 2 =2x3/(1.5)=2.7
m/s2 . Choosing the man plus pulley above him as the body
(assuming the mass of the pulley is negligible), Newton's second law is
2T-mg =ma or T =½m (g+a )=½x80x(9.8+2.7)=500
N where g =9.8 m/s2 is the acceleration due to
gravity. Finally sum forces for the weight Mg which has the
same magnitude of acceleration as the actor: Mg-T=Ma , so
M=T /(g-a )=500/(9.8-2.7)=70.4 kg. You may want to keep in
mind that the speed of the actor at the end of the 1.5 s is v=at =2.7x1.5=4.05
m/s; if the weight is stopped after the 1.5 s, he will continue moving
upwards until he has risen another h =½v 2 /g =0.84
m and then fall back.
table weight answer , makes me think maybe it could be.
ANSWER: S as the distance between the wheels, L the
distance from the axle to the tongue support, q the distance
from the ground to the center of mass (COM
© ) of the boat plus trailer, d
the distance of COM forward of the axle, and W the weight of
the boat plus trailer. The normal forces N i
are the forces up on each of the three points of contact; if an ideal
scale of zero thickness were placed under a wheel, the magnitude of the
appropriate normal force is what it would read and N 1 +N 2 +N 3 =W. I have also assumed that the
COM is centered laterally. The geometry and algebra are a bit tedious,
so I give only final answers. In each case I imagine lifting one point
of contact by the angle θ and then measuring the normal
force which I will call N' i . N' 3
is the easiest since it does not depend on θ , N' 3 =Wd /L .
The other two are, by symmetry, the same: N' 1 =N' 2 = ½W [1-(d /L )-(½q /S )tanθ ].
If you now compute the sum, N' 1 +N' 2 +N' 3 ≡W'=W [1-(q /S )tanθ ].
So your measurement of the weight will be wrong. As an example, suppose
that S =3 ft and q =5 ft and you lift a wheel h =4
in=1/3 ft off the ground; then tanθ =(1/3)/(2x3)=1/18 and
the measured weight will be W (1-(5/3)/18)=0.91W ,
nearly a 10% error. The culprit here is that the center of gravity is so
far off the ground. To get a more accurate measurement, block the
opposite wheel so that it is also h above the ground (the
tongue support does not need to be blocked); you only have to do this
once since N 1 =N 2 and N 3
does not depend on its height.
table weight example , wouldn't elevating one end of the table with a 2x4 and putting a scale beneath it shift (rotate) the table's weight (CoG) somewhat toward its other end and lessen the measured weight of the weighed end, and error be particularly noticeable if the span of the table is short and the scale is several inches high? If I could remember my trig I could pose this question more intelligently, but the extreme, if hypothetical, case would be if the table end were elevated all the way to vertical, where it would weigh nothing.
ANSWER: The Physicist's rare
errors! I must admit that I was thinking small angle here and that is
the only case where it is a good approximation to say that elevating one
end of the table does not change the normal force. And, again, I did not
really appreciate the importance of the fact that the COM is not
colinear with the other two forces until I started puzzling over your
boat trailer question. You can see an example of a case where this is
not an issue in an earlier question about a
boat dock . The correct
answer to the table weight
example would be W' =W [1-(q /S )tanθ ].
(I have changed the notation and assumptions from original problem just
to make the solution more concise. In particular, I have assumed that
the COM is a distance q above the floor and equidistant from
the ends and sides; I have neglected the weight w of the 2x4,
and ½W=W 1 =W 2 ; S
is the length of the table.) So, if q =3 ft, S =10 ft,
and h =1/3 ft (COM above floor), W'=W [1-(3/10)(1/3)/10]=0.91W .
ANSWER: R =6.37x106 m. An
object which moves in a circle of that radius with a speed v
requires a force pointing toward the center of the circle of F=mv 2 /R
where m is its mass. But your object would have its weight
W=mg as the only force acting on it where g =9.8 m/s2
is the acceleration due to gravity. Therefore, mg =mv 2 /R ;
solving for v , v = √(gR )=√(9.8x6.37x106 )=7901
m/s=17,674 mph.
Here is an even simpler way to figure it out. An object with speed v
moving in a circle of radius R has an acceleration of v 2 /R ,
the direction of which is toward the center of the circle. But an object
near the earth's surface is g , toward the center of the earth.
Therefore,
v = √(gR ).
ANSWER: very rough
calculation. I assumed that that the initial speed of the sedan was
u =20 mph and the initial speed v of the SUV was unknown. I
assumed that both cars slide after the collision. The coefficient of
kinetic friction μ for rubber on wet asphalt can be anything
from 0.25 to 0.75 so I did the calculation for both extremes. I first
estimated the speed V immediately after the collision for each
car using their masses M , the distance they slid S : ½MV 2 -μMgS= 0,
or V =μgS ).
Next I applied momentum conservation to the collision. SUV v-M sedan u=M SUV V SUV +M sedan V sedan
v [M SUV V SUV +M sedan (V sedan +u )]/M SUV .
First slope : 1.721 Meter high, 35 degrees, friction coefficient
0.14.
Second slope : 2.536 Meter high, 25 degrees, friction
coefficient
0.14.
Obviously the final speed will not be the total of both speeds, so how
do I add the speed of the first slope to the second slope to obtain
final velocity?
ANSWER:
ANSWER: m moving with speed v and at a height h above
some arbitrarily chosen h =0 level is E = ½mv 2 +mgh ;
g =9.8 m/s2 is the acceleration due to gravity. For a
given system the energy is conserved (the same everywhere) as long as
there is no agent adding or subtracting energy from the system. If there
were no friction in your slides, energy would be conserved. so, if you
chose h =0 at the bottom and started from the top with v =0,
the initial energy would be E 1 =mgh and the
final energy would be E 2 =½mv 2 ;
setting them equal and solving for v , v =√(2gh );
note that the mass does not matter. In your case, v =√(2x9.8x(1.721+2.536))=9.13
m/s.
Alas, there is friction and the
frictional force f for a mass m sliding on an incline
θ is f=μmg cosθ. The amount
which a force changes the energy if it acts over a distance s
is called the work done and the work done by friction is W f =-μmgs ·cosθ ;
the minus sign indicates that friction takes energy away from the
system. Now, instead of writing E 2 =E 1 ,
we write E 2 =E 1 +W —the
energy you end up with equals the energy you started with plus what you
added (negative for the case of friction).
Now we can address your specific
case. ½m v2 =m gh-μm gs1 cosθ 1 -μm gs 2 cosθ 2 =9.8x[(1.721+2.536)-0.14( 0 +6cos250 )]=31.6
m2 /s2 . Solving for the velocity, v =7.95
m/s.
The formula you seek if you are
doing a slide with two different slopes is v =√{2g [h-μ (s 1 cosθ 1 +s 2 cosθ 2 )]}.
T his
is the motor I am thinking of buying:
Rated power 0.01 W
ANSWER:
v you want the camera to move on the slope and the nature
of how the camera moves on the surface (in particular coefficient of
friction μ ). But the general solution will have those
in there and you can calculate with them. You will want to attach a
spool of some sort the the drive shaft of the motor to act as a reel to
pull up the string attached to the camera, say its radius is R .
The angular velocity of this motor is ω =20 rpmx(2π
rad/1 rev)x(1 min/60 s)=2.1 s-1 . The weight of the camera is
mg =(1.5 kg)x(9.8 m/s2 )=14.7 N. The angle of the
incline is stipulated to be 450 . The force necessary to pull
the camera up the ramp with constant speed may be shown to be F=mg (1+μ )/√2;
since the torque τ FR=τ = mg R (1+μ )/√2 .
In order for the camera to move at speed v , the radius of the
spool should be R=v / ω.
If the speed of the camera is v and the force is F , the
power being generated in the motor is P=Fv =mgv (1+μ )/√2.
Let's do an example. Suppose you want the camera to move with a speed of
v =1 cm/s=0.01 m/s and the friction is negligible, μ ≈0.
Then the spool has a radius R =0.01/2.1=0.0048 m=0.48 cm; the
required torque would be
τ =0.0048x14.7/√2=0.05
N·m; the required power would be P =14.7x0.01/√2=0.1
W. I am afraid that your proposed motor is inadequate.
ANSWER: 0 incline; that force was not even big
enough to get the car started! In fact a car is geared and the
assumption that the car's acceleration is uniform is incorrect and there
are therefore too many variables to be able to solve this problem.
ANSWER: change in energy
force" has no meaning in physics. If friction is neglected, the only
thing which matters is the vertical distance fallen and so the speed at
the bottom would be identically the same regardless of the shape of the
ramp. Refer to the figures on the left. Without friction, the speed at
the bottom is v = √(2gh ) where h is the
height. Including friction complicates things.
The ramp is easiest to do: the three forces on
the skater are his weight mg N
f . The magnitude of the frictional force
is f= μN where μ is the
coefficient of friction. Newton's equations are ΣF y =N-mg cosφ =0
and ΣF x =mg sinφ -μN =ma ;
the solutions are N=mg cosφ and a=g (sinφ -μ cosφ ). Now,
notice that the acceleration depends on the angle of the incline
which means that the speed at the bottom will also depend on the
angle, the steeper the faster. Therefore it is not really fair to
compare the flat ramp with the quarter pipe because the quarter pipe
will always have its horizontal distance equal to its vertical fall.
So the only fair comparison is for the flat ramp to have φ= 450 .
In that case, a=g (1-μ )/√2. I calculate that
the speed at the bottom is v flat_ramp =√(2gh (1-μ )).
For the quarter pipe, the situation is much harder to calculate
because the normal force gets larger as the skater falls thereby
increasing the friction. Newton's equations are ΣF x =mg cosθ -μN =ma
and ΣF y =N-mg sinθ =mv 2 /R
where a is the tangential acceleration and R
is the radius of the pipe. I have no idea how to solve these
because they are coupled, second-order, nonlinear differential
equations. I am guessing that there is not a closed-form
analytical solution and the equations would have to be solved
numerically. For this case, though, we are interested in cases
where the friction is low, μ<< 1. We can calculate the
velocity v as a function of θ for the
no-friction case, μ =0 and then use that v to
approximate the normal force as a function of theta and use that
to find the energy lost to friction: v (θ )≈√(2gR sinθ ),
N ≈3mg sinθ , f ≈3μmg sinθ.
Using these, I find v ¼pipe ≈√(2gh (1-3μ)).
So, to do a
numerical example, R=h =12 ft, μ= 0.1, g =32
ft/s2 , v flat_ramp =26.3 ft/s, v ¼pipe ≈23.2
ft/s. (v =27.7 ft/s for zero friction.)
ADDED
COMMENT:
His post : "Out of curiosity, I simulated the setup. With μ =0.001 and
m=R=g =1 I got E =0.99701, in agreement with your result.
μ =0.01 leads to
E =0.9704. Even with μ =0.1, the approximation is not bad:
E =0.732. μ >0.60. Tested with 100, 200 and 500 steps: The critical value is somewhere between 0.603 and 0.605. There is no obvious mathematical constant in that range."
Interpreting, his E is
the calculated value of the kinetic energy divided by mgh. For
this quantity, my approximation was [½mv 2 /(mgh )]=(1-3μ ).
Comparing the computed values with the approximated values: 0.99701≈0.997
for μ= 0.001, 0.9704≈0.97 for μ= 0.01, and
0.732≈0.7 for μ= 0.1. Thanks to
mfb for
his interest and calculations!
jet on the conveyor belt. I believe that your answer is incorrect.
If an aircraft was like a car, relying on the wheels to push the vehicle forward it would indeed, never gain any airspeed. However, the wheels are not driven, the accelerating
ANSWER:
Although we envision the wheels spinning and the conveyor moving
backwards, there is another possible scenario. The original question
states that the "…conveyor belt is designed to exactly match the
speed of the wheels, moving in the opposite direction." To my mind, this
can mean that the wheels are not moving and the conveyor is not moving.
The figure shows the forces on the plane in blue and the forces on the
conveyor in red. The plane is being pushed forward by the thrust
T . Suppose that the motor running the conveyor is
pulling with a force F =-T f =- T
-f = T .
Forces on both the plane and the conveyor are in equilibrium and nothing
moves. It is just the same as if the brakes were locked.
ANSWER: g
(acceleration due to gravity) can be considered constant, it will land
west of the point where you threw it by a distance d =(4 ωv 3 cosθ )/(3g 2 )
where v is the initial speed, ω is the angular
velocity of the earth, and θ is the latitude where you
threw it.
ANSWER: earlier answer .
ANSWER: s is given by v = √(2μgs )
where g =32 ft/s is the acceleration due to gravity and μ is
the coefficient of kinetic friction. Rubber on wet asphalt has a range
of possible μ =0.25-0.75 with a corresponding range of speeds
v =27-47 mph. Rubber on wet concrete has a range of possible μ =0.45-0.75
with a corresponding range of speeds v =37-47 mph.
ANSWER: W N f ΣF y =-W+N cosθ -f sinθ =0
and ΣF x =N sinθ -f cosθ =0.
The solutions are f=W sinθ and N=W cosθ. Now, if the car has some speed v , the car is still in
equilibrium in the y direction, but now has a centripetal acceleration
in the +x -direction of a=v 2 /R where R is
the radius of the track. So now, ΣF y =-W+N cosθ -f sinθ =0
and ΣF x =N sinθ -f cosθ =mv 2 /R.
Solving these equations, f=m (g sinθ-v 2 cosθ /R ).
Now, notice that if v 0 =√(gR tanθ ), f =0; at
this speed the car can go around the track even if it is perfectly
slippery. For speeds greater than v 0 , f will be
negative which means it must point down the slope. As the speed
increases beyond v 0 , the magnitude of the frictional
force will get bigger and bigger down the incline; but, there is a
maximum value which f can be, f max =μN
where μ is the coefficient of static friction. The
corresponding velocity may be
shown to be v max =√[gR (sinθ +μ cosθ )/(cosθ -μ sinθ )].
μ will result in the car
slipping, however. Whenever (cosθ -μ sinθ )]=0,
v max =∞ and the car will not slip no matter how
fast it goes. For any given μ there will be a minimum angle θ min = tan-1 (1/μ )
beyond which slipping will never occur; for example, for μ= 0.5,
θ min = tan-1 (2)=63.40 .
The graph shows the behavior of v max as a function
of angle for several values of μ.
If you want a way to understand
this qualitatively and you have worked with fictitious forces in
accelerated frames, you can look in the frame of the car and there will
be a centrifugal force pointing in the -x direction; this force
will have a magnitude of mv 2 /R and, if it is greater than N sinθ -f cosθ=N (sinθ -μ cosθ )
the car will move up the incline.
ANSWER: F =7/0.04=175 N. Ignoring all
friction, the
resulting acceleration would be a =175/85=2.1
m/s2 . If the final velocity is v =40 km/hr=11.1 m/s,
the time is t=v /a =5.4 s. Now, is friction really
negligible? First consider the frictional force f from the
wheels which could be approximated as f=μmg where μ
is a coefficeint kinetic friction and mg =85x9.8=833 N. A
reasonable approximation for μ would be μ ≈0.1;
with this μ the unpowered board would stop in about about 20
m if it had a speed of 2 m/s. The frictional force would then be about
83 N, not at all negligible. Redoing the calculations above for a net
force of 175-83=92 N, I find t =10.3 s. Finally we should talk
about air drag. The drag force D may be approximated (only in
SI units) as D≈ ¼Av 2 where
A is the cross-sectional area you present to the wind, let's say
about A ≈1 m2 ; so the biggest D gets
is about 30 N. Since most of the time spent is at lower speeds where
D is much smaller, I will not calculate the the time including
D (a much harder calculation because the force is not constant)
since all this is only a rough caclulation anyway; just think of 10 s as
a lower limit. You can, though, estimate the maximum speed if the torque
really did remain constant by finding the speed for which D =92
N: v max ≈√(4x92/1)≈19 m/s; this is
called the terminal velocity. For
constant torque, doubling it would result in a net force (again
neglecting D but including f ) of 267 N and a time of
3.5 s. If you know how the
torque varies with speed, you could do a similar calculation but it
would be trickier because the force and therefore the acceleration would
change with time.
ADDED
COMMENT: D for the two-wheel drive
case. I found that 11.1 m/s is reached in 11.5 s, only a very small
increase as I anticipated. The graph of v vs. t is
shown; you can see that the curve is nearly linear up to 11.1 m/s and
approaches 19.2 m/s at large times. The solutions are t =17.5·tanh-1 (v /19.2)
and v =19.2·tanh(t /19.2). Don't take any of this
too seriously because of the numerous approximations I have made; I
would guess there would be about a 30-40% uncertainty —probably
5-15 s would give the experimental time. If you get one of these, let me
know how long it actually takes.
ANSWER: P delivered by a torque τ at
angular velocity ω is given by P=τω .
In your case τ =7 N·m and ω=v /r= 11.1/0.04=278
s-1 , so P =1943 W, not kW.
ANSWER: earlier question about
submarines.
ANSWER:
description of how small an object the human eye can see:
"What is limited is the eye's resolution: how close two objects can become before they blur into one. At absolute best, humans can resolve two lines about 0.01 degrees apart: a 0.026mm gap, 15cm from your face. In practice, objects 0.04mm wide (the width of a fine human hair) are just distinguishable by good eyes, objects 0.02mm wide are not."
Suppose you had a door knob 10 cm in diameter. Scaling it down by 500
would make the size 0.02 cm=0.2 mm, easily seen by the standards above
but still very small, maybe the size of a period on this page. So maybe
the easiest thing to please your clients would be just to put a dot
there; I can understand how that might violate your aesthetics, though.
To answer your question, what would be the distance R at which
the angle subtended by 10 cm is the same as the angle subtended by 0.02
cm at a distance 15 cm from your eye? The angle, in radians, is
size/distance, so 0.02/15=10/R or R =7500 cm=75 m. To not be able to
distinguish the door knob at all, 0.002 cm, the distance would be 750 m
which would correspond to a scaling of 1:5000.
**The turntable is rotating about the vertical z-axis at a constant unspecified angular velocity.
**Radius and mass of the turntable and the ball are unspecified.
ANSWER: move in a circle
which is at rest in the laboratory frame. I will tell you at the outset
that this is a quite advanced classical mechanics problem and is well
beyond the purposes of this site. However, it was interesting enough to
me that I did a little research to try to get a feeling for how it can
be approached. I will not provide details, but they can be found in a
1979 article in American Journal of
Physics . I do not know your level, but if you are a high school
student, the math is probably beyond your grasp, all the vector
calculus. As you can see from the article, trying to write the solution
in Cartesian coordinates, as you have specified, is not the way to go.
Using the vectors r ω r ,
θ , z ).
COMMENT: Ask the Physicist . His argument
was "…the
question I brought to your attention does not exsist…" because he
could not find it in a book somewhere and therefore not classifiable as
homework; that, of course, is nonsense because this is a very standard
question. Nevertheless, since he is so avid to learn, I am going to
answer it!
ANSWER:
The question is kind of ambiguous because it refers to
constant speed; but if the car is to maintain constant speed down the
hill, there will have to be some drag friction. If a constant speed
v is maintained, the free body diagram is shown on the left. I will
choose a coordinate system with +y pointing along the direction
opposite the normal force and +x tangent to the circle and
pointing downhill. The car is inequilibrium in the x direction and has
an acceleration of magnitude v 2 /R in the
positive y direction. Newton's laws are ΣF x mg sinθ-f
and ΣFy =mv 2 /R=mg cosθ-N.
The solutions are simply obtained: f =mg sinθ
and N =m [g cosθ-v2 /R ].
An interesting result is that when θ =cos-1 (v2 /(gR )),
N =0 and the car will leave the road.
v
at the top of the hill; the car will therefore speed up as it goes down
the hill. Using energy conservation it is easy to show that its speed
when at angle
θ is v' 2 =v 2 +2gR (1-cosθ )
and so the centripetal acceleration is ay =v 2 /R +2g (1-cosθ ).
Notice that there will also be a tangential acceleration, ax ,
an unknown at this stage. Now the equations to solve are ΣF x max =mg sinθ
and ΣFy =m (v2 /R+ 2g (1-cosθ ))=mg cosθ-N.
The solutions are ax =g sinθ
and N =m (v2 /R+g (2- 3cosθ )).
Again, there will be an angle for which N =0 which is where the car would
leave the road, the determination of which I will leave to my AP
student.
ANSWER: a =7=17/0.1=170
m/s2 . The force necessary to hold the child is ma =8x170=1360
N. This corresponds to a weight 17 times greater than the weight of the
child. Although it might be argued that you might be able to exert such
a force for a tenth of a second, everything happens so fast in an
accident that the child would be ripped from your arms before you had a
chance to react.
ANSWER: k=f /x =2/0.01=200
N/m. The work you did to compress it was W = ½kx 2 =½x200x0.012 =0.01
J which is the energy which you put into the spring. So now the amount
of mass which would correspond to that amount of energy is m=W /c 2 =0.01/(3x108 )2 =1.1x10-19
kg. The mass of a dust particle is on the order of 10-8 kg,
so the mass change is about one ninety billionth of the mass of a
particle of dust. I think we can agree that this would be impossible
(not almost impossible) to observe.
ANSWER:
ANSWER: earlier answer .
ANSWER: W 1 =mMG /R 2
where M is the mass of the earth, R is the radius of
the earth, G is the universal gravitational constant, and m
is the mass of the point mass. This is usually written as W 1 =mg.
But, if the point mass is a distance H above the surface,
its weight is smaller, W 2 =mMG /(R+H )2 =mg /(1+ (H /R ))2 .
If you know integral calculus it is not hard to show that the weight of
a vertical uniform bar of length L and mass m is W 3 =mg /(1+ (L /R )).
In your example L =1000 mi is not small compared to R ≈4000
mi, so
W 3 =mg /(1.25)=0.8W 1 .
But, because L is so large, your horizontal bar does not have a
weight of mg either unless it bends to conform with the curved
surface of the earth. But, you can see from the figure that the
distances from the center of the earth are much smaller than for the
vertical bar, so it will surely have a larger weight, just not quite as
big as mg .
ADDED
THOUGHT: Actually, even if the bar conformed to the
surface of the earth, it would weigh less than mg because the
components along the length of the bar of the forces on each piece of
the rod would all cancel out. I think I will not calculate the exact
answer for the horizontal bar, just say that is slightly smaller than
mg .
ANSWER:
ANSWER: μ between the bottle
and the belt. In the figure, H is the position of the center of
mass ⊗ R is the radius of the base. When the
bottle touches the belt there will be a frictional force f
N 1 and N 2
acting on the leading and trailing edges of the base. Since the bottle
is accelerating, it is imperative to calculate the torques about the
center of mass of the bottle. The bottle is in equilibrium in the
vertical direction so N 1 +N 2 =mg .
The frictional force is f =μ (N 1 +N 2 )=μmg.
The torques will sum to zero if the bottle is not tipping, 0=fH+RN 1 -RN 2 .
Now, if the bottle is just about to tip over, N 1 =0
and so the torque equation becomes 0=fH -RN 2 =μmgH-Rmg
or μ max =R /H ; this is the
maximum that the coefficient of friction could without the bottle
tipping. The result does not depend on either m or v .
For example, if R =3 cm and H =6 cm, then μ ≤0.5.
ADDED
NOTE: N acting a distance d from the leading edge of the base of the
bottle. Then when d =2R the bottle would be about to
tip. You would get the same result as above.
ANSWER: E=mc 2 . The uncertainty principle is ΔE Δt ≈h /2π
where h/ 2π= 6.6x10-16 eV≈10-34
J·s. Since c =3x108 has no uncertainy, Δm ≈[h /(2πc 2 )]/Δt ≈10-51 /Δt
kg. Δt is the time over which you observe the mass. For a
stable particle or a stable nuclus, e.g ., Δm= 0 because Δt= ∞.
A neutron has a half life of 880 s, so the uncertainty in its mass is
approximately Δm ≈10-51 /880≈10-54
kg.
ANSWER:
ANSWER: earlier answer .
ANSWER: earlier answer .
ANSWER: earlier answer .
The carriage is shaped in a way that the scale is horizontal with respect to ground (that is is a special edge shaped carriage)
The carriage is regular so the the scale is not paralleled to the ground.
ANSWER: a=g sin θ down the
slope.
The figure
on the left
shows the forces on the man standing on the horizontal scale. There
is his own weight, mg N N . The contact between the man and the scale
cannot be frictionless or else the man would not accelerate along
with the carriage, so there is also a frictional force f ma=m g sinθ=-N sin θ+f cosθ+mg sin θ
and 0=N cosθ -f sinθ-mg sinθ .
Solving, I find N=mg cos2 θ and
f=mg sinθ cosθ .
The figure
on the right shows the man on the scale parallel to the incline. Now
no friction is needed and the scale, again, will read N but
N will be different. Newton's equations are 0=-mg cosθ+N
and ma=mg sinθ , just the same as the classic
object on the frictionless incline. So the scale reads N=mg cosθ.
The graph
compares the two cases.
ANSWER:
ANSWER: recent answer to get an
idea of how small the force exerted by light is.
ANSWER:
Q&A OF THE WEEK
2/26-3/4/2017
ANSWER: g may be written as g=MG /R 2
where M is the mass of the earth, G is the
gravitational constant, and R is the radius of the earth. For
some height h above the earth, g'=MG /(R+h )2 .
Therefore g' /g =R 2 /(R+h )2 =(1+(h /R ))-2 .
This is an exact expression and you can see that g' never
becomes exactly zero but continues decreasing forever as h gets
larger. (Note that your "formula" cannot be correct because g'
gets bigger as h increases.) However, for many applications you
want to know what g' is if h /R is very small;
one can do a binomial expansion of g' /g , g' /g= (1+(h /R ))-2 ≈1+(-2)(h/R)+…=1-2h /R .
So your formula was almost right, just need to change the + to -. But
this is only an approximation and will fail when h gets too
big. The graph above compares the exact (black) and approximate
solutions (red); as you can see, the approximation only works up to
about 0.1R . The fact that the approximation goes to zero at
h=R /2 has no meaning.
ANSWER:
ANSWER:
ANSWER:
ANSWER: —where the center of gravity is
and where the two people exert their forces. In the figure the weight
W F 1
and F 2 at distances at distances
d 1 and d 2 from the center of
gravity. The two people must hold up all the weight between them, so
F 1 +F 2 =W . The canoe is not
rotating about its center of gravity so the net torque must be zero, so
F 1 d 1 =F 2 d 2 .
Solving, F 1 =W [d 2 /(d 1 +d 2 )]
and F 2 =W [d 1 /(d 1 +d 2 )].
If d 1 =d 2 , each holds up half the
weight.
ANSWER: λ =550
nm=5.5x10-7 m (yellow); the corresponding frequency is f=c /λ =3x108 /5.5x10-7 =5.45x1014
Hz. The momentum of one photon is p=hf /c =6.6x10-34 x5.45x1014 /3x108 =1.2x10-27
kg·m/s. The
photon flux (photons per second per square meter) for photons of
this wavelength is about Φ =3x1021 s-1 ·m-2 .
So if I estimate your area to be about 1 m2 , there are about
3x1021 photons hitting you every second. The total momentum
imparted to you over that second (assuming that you absorb the photons)
is about Δp =3x1021 x1.2x10-27 =3.6x10-6
kg·m/s. The average force you feel is about F =Δp /Δt= 3.6x10-6
N; and this force is spread uniformly over your area. This is
really small; for example, if we take the typical weight of a one dust
particle to be about 10-8 N, the weight of about 360 dust
particles spread over your whole body is about what you "feel" from the
sunlight.
ANSWER:
ANSWER:
ANSWER: ¼ as large, but it would still be the same amount
of energy. You would not say that E=mc 2 was wrong if
we got two different numbers if we calculated it in Joules (using kg, m,
s) than in ergs (gm, cm, s) would you? E =(1 kg)(3x108
m/s)2 = 9x1016 J and E =(1000 g,)(3x1010
cm/s)2 = 9x1023 erg. In fact, 9x1023
erg=9x1016 J is a perfectly good equation.
ANSWER: —there
is no experiment you can perform to tell if you are at rest in empty
space or free-falling in a gravitational field. Another way to look at
it is that the acceleration due to gravity is independent of the mass
and so the cup and the fluid it contains both fall exactly the same.
ANSWER: c . Your statement that the
train rider would see the beams having is speed 0.00001c is
simply wrong. That is not to say that the beams are unchanged —they
would be Doppler-shifted to different wavelengths, but still move with
speed c .
See the faq page to help yourself to accept that this is true.
ANSWER:
there . You might also read an
earlier answer about relativistic velocity addition. You are right,
if light with frequency f is moving in the same direction as
you are, you will see a lower frequency light f' ,
red-shifted. The Doppler equation is f'=f √[(1-β )/(1+β )] where β=v /c.
Suppose we take your case where β= 0.9999999=10-7
so √[(1-β )/(1+β )]=2.24x10-4 and choose λ =550
nm (yellow light). Then f=c /λ =3x108 /5.5x10-7 =5.45x1014
Hz and f' =2.24x10-4 x5.45x1014 =1.23x1011
Hz=123 GHz. This frequency is in the short-wave radio range, nothing he could "…see
with his eyes…"! The Doppler effect has red-shifted the wave
way beyond visible red!
ANSWER: m may be written as
K=p 2 /(2m ) where p is the linear momentum.
The uncertainty principle states that Δp Δx>h
where h is Planck's constant and Δp and Δx
are the uncertainties in momentum and position x respectively.
This tells you that you cannot know either p or x
exactly, they will always be uncertain. For the uncertainty of K
to be zero the uncertainty of p would have to be zero. But, if
the uncertainty of p is zero, the uncertainty of x is Δx>h /0=∞.
You would lose all knowledge of the position of the particle.
ANSWER: m is
U=-GMm /r
where G =6.67x10-11 J·m/kg2 is the
gravitational constant, M =6x1024 kg is the mass of
the earth, r =3.85x108 m is the radius of the moon's
orbit, and m is the mass of the coin. So, U= -1.04x106 m
J. But, when the coin strikes the earth its energy will be E=-GMm /R+ ½mv 2
where v is its mass and R =6.4x106 m is the
radius of the earth. Ignoring air drag when the coin gets near earth
(which will not really be negligible), energy may be conserved: -1.04x106 m=-GMm/R+ ½mv 2 =(-6.25x107 +½v 2 )m.
Solving for v I find v =1.11x104 m/s. The
kinetic energy it would release would 6.15x107 m J.
ANSWER: —the sky is
also bright, just not as bright as the sun, and that light comes in
through the window also and illuminates other parts of the room from
which you could see the sky if you looked out. You will also note that
the wall of the room where the window is located is not as well
illuminated as the rest of the room.
ANSWER: W c from your
right hand. The position of your left hand I have denoted as being a
distance d from your right hand. For simplicity I have assumed
that you will exert a force L H V °, so
the horizontal and vertical components of L L /√2. The two equations for
translational equilibrium are ΣF vertical =0=-W +L /√2+V
and ΣF horizontal =0=-L /√2+H .
These may be simplified to H =L /√2 and V=W-H .
There are three unknowns and only two equations, so we need a third
equation; the sum of torques must also be zero. Summing torques about
the right hand, Στ=dL-cW /√2=0 or L =cW /(d √2).
Therefore, H =cW /(2d ) and V=W [1-cW /(2d )].
For example, suppose W =7 lb, d =3 ft, c =4 ft:
I find L =6.6 lb, H =4.7 lb, V =2.3 lb. The
force R R =√(V 2 +H 2 )=5.2
lb. The key to easy handling is to choose the saw with the smallest
c which will minimize the torque you must exert.
ANSWER:
ANSWER:
tokamak , an experimental fusion reactor. Nuclear fusion is the
energy which powers stars. Roughly, the idea is that if you get hydrogen
atoms hot enough (hundreds of millions of degrees) they will have enough
energy to fuse together to make helium, and that releases a great deal
of energy. The problem turned out to be much more challenging than
anticipated and there still is not a practicable fusion reactor which
will give you more energy out than you use to run it. The main problem
was the difficulty of keeping the hot plasma confined long enough to
have a sufficient number of fusion reactions. The current research is
centered on constructing a tokamak by an international collaboration
known as ITER ; the lab
is in France. There is an old cynical joke among physicists: "Fusion is
the energy fo the future and always will be!"
ANSWER: T would be proportional to the product of the
velocity v and the time t , T=cvt where c is
some constant. So, if you change v to v' but keep
t (measured) constant, the corrected thickness T' will be
T'=cv't. Take the ratio of the two equations and solve for the
thickness: T'=T (v /v' )=(5813/5920)T= 0.982T.
ANSWER:
ANSWER: U=mgh for potential energy, m =10,000 t=107
kg, h =100 m, g =9.8; U =9.8x109 J.
If it falls over a time t =10 min=600 s, the power generated
would be P=U /t =1.63x107 W=16.3 MW.
ANSWER: —energy,
linear momentum, angular momentum—are preserved but only if
changes to the definition of these quantities are changed. E.g .,
kinetic energy is no longer ½mv 2 and linear
momentum is no longer mv ; the new quantities, however, are
approximately their classical values if the velocity is very small
compared to the speed of light.
ANSWER:
ANSWER:
Lift it
from one end while the other end rests on the ground
In order for the rod to be
just barely lifted off the floor at one end, the torque due to
FW =1000,
9xF =4.5x1000=4500 or F =500 kg. The floor holds up the
other half of the weight.
Grab onto
one end and lift the whole rod keeping it horizontal.
It is
impossible to lift it horizontally using only a force at the end because
you could not exert enough torque to keep the rod from rotating due to
the torque caused by the weight. You might try to do it by using two
hands, one at the end pushing down and the other some distance d
from the end pushing up. In that case (see the figure above), summing
the torques about the end you find that F up =4500/d ;
for example, choosing d =20 cm=0.2 m, F up =22,500
kg. You can also find F down by summing all forces to
zero, F up -W-F down =0=22,500-1000-F down
or F down =21,500 kg.
ANSWER: run slowly . Do not mistake that
for saying that your clock appears
to run slowly —it might
appear to run faster or slower
than mine, but is running slowly. Furthermore, it has nothing to do with
mass increasing with speed. But how do you observe things? Your clock
appears to run just the same as if you were at rest; but wait, you are
at rest relative to yourself and you see me moving and therefore find
that my clock runs slowly! Now, the time that I see you getting to the
star is 4.3/0.7=6.14 years. But, you see it as a considerably shorter
time. How can that be if your clock is running normally and you clearly
see your speed as 0.7c ? The answer is length contraction—you
observe the distance to the star to be less than 4.3 light years by the
gamma factor, √(1-0.72 )=0.71, or 3.1 light years so the
time will be 3.1/0.7=4.43 years. If you could get going a speed 0.9c ,
you could get there in less than half a year by your clock but still
6.14 by mine. I have given you a couple of links to the
faq page , but you might want to peruse that page
for more relativity answers which might interest you, particularly the
twin paradox.
How much energy is contained in every stationary object on earth due to the rotation speed of the earth? I understand this this is only Potential energy and would require the object to be shielded from the earth's gravitational pull in order to release it but if a steel block weighing 100 kg were to no longer be held by gravity how much potential energy would it contain as it was flung from the earth rotational spin.
Also, once released from the earth's gravitational attraction, would you also need to factor in the speed of the earth's rotation around the sun to the calculations? The mass was travelling in two different directions (planetary rotational and orbital rotational) at extremely high speeds before the gravitational ties were cut?
And as a final factor, while trying to calculate this someone pointed out that the starting 100 kg piece of mass would instantly become a 0.KG piece of mass once it was no longer had any weight with respect to the Earth.
At that point, we pretty much tossed in the towel.
But, still feel that because we started with a 100 kg item that became weightless, that would not remove the potential energy it contained before gravity released it.
If afterward it were to crash into the moon or some other object in its path, wouldn't the amount of destruction still be equal to the energy it contained when it was "fired from the planet" much as a bullet fired from a gun?
ANSWER:
The most
important misconception should be dealt with first. Energy is
something which depends on your frame of reference. If a mass is on
the surface of the earth it has zero kinetic energy if you are also
on the surface of the earth. If you view that same mass from a frame
moving along with the earth in its orbit but not rotating like the
earth is, it has the kinetic energy you envision, that due to the
earth's rotation. Kinetic energy is given by K =½mv 2
where m is the mass and v is the velocity you see
it to have. The speed v of an object on the earth's surface
depends on the latitude λ , v=Rω cos(λ )
where R =6.4x106 m/s and ω =7.3x10-5
s-1 is the angular velocity of the earth. Let's just look
at the equator where v is largest, v=Rω= 6.4x106 x7.3x10-5 =470
m/s=1050 mph; you should note that this is not really that large.
The kinetic energy of 100 kg flung into space with this speed is
K =½x100x4702 =1.1x107 J.
By now you
should appreciate that it depends on who is looking at your
projectile. Someone who is at rest relative to the sun would need to
calculate the speed and therefore add in the velocity of the earth
around the sun, call it V . But, this would get complicated
because it would depend on time of day the projectile was launched.
If you are standing right on the orbit and the earth is coming
toward you, the speed would be V+v an noon and V-v
at midnight; at other times it would be somewhere between these two
extremes.
You have
this all wrong. Weight is the gravitational force which the earth
exerts on the mass; the weight does not "instantly" become zero when
it is launched, it gradually decreases as it gets farther away from
earth. But, that is beside the point because the kinetic energy
depends on the mass, not the weight, and the mass stays the same
regardless of whether there is any gravity or not.
Whenever
the projectile stikes something, like the moon in your question, all
that matters is what its velocity is with respect to the moon .
ANSWER: —his weight W F first law. But, what
about the reaction partner of F is not a force on the eagle
(which is why I did not even draw it in the force diagram of the eagle).
The picture on the right shows the eagle just as he is lifitng off. Now
there is an additional force L L+F>W , the eagle is no
longer in equilibrium and will accelerate upward. As soon as the talons
leave the branch, F
QUERY:
ANSWER:
N pressing the surfaces together. The force F to keep your
bundle moving with constant speed is given by F=μN where μ
is called the coefficient of friction and depends on what the surfaces
are. In your case, if the truck is level they are pressed together with
a force equal to the weight, 28 lb. For clean metal on wood, μ
can be anything between 0.2 and 0.6, so F will be somewhere
between 5.6 lb and 16.8 lb.
ANSWER:
here . Incidentally, if there were no gravity the only thing
slowing the bullet down is the air; it would slow down to the terminal
velocity and then continue moving in a straight line with that speed. In
the real world (with gravity), it would slow down and fall and
eventually be falling straight down with the terminal velocity. To think
that it would stop after 1700 miles with no gravity is absolutely wrong.
ANSWER:
earlier answer .
ANSWER:
ANSWER: M , and that the incoming speed is v ;
I will also assume that the time t which the collision lasts is the same
for either the two- or three-car case. If it is a two-car collision,
using momentum conservation Mv= 2Mu or u =½v ;
therefore you experience an acceleration of a =½v /t
and a force of ½mv /t
during the time t , where m is your mass. If it is a
three-car collision, using momentum conservation Mv= 3Mu or u =v /3
w here u is the speed of the three cars;
therefore you experience an acceleration of a =v /(3t )
and a force of mv /(3t ) during the time
t.
ANSWER:
ANSWER: —if
you were just standing on level ground, how far could you lean forward
without falling forward? Not very far! And it is even worse if you are
running because you have a forward momentum which also is tending to
topple you forward if you try to stop. All that I can think of that you
can do to avoid a fall is to try to slow enough that you can get
yourself more vertical so that your center of gravity is vertically
above your feet. If the hill is not too long, you can keep running until
you get to the bottom, but you will be speeding up the whole way.
ANSWER:
ANSWER: v=at and d = ½at 2
where, in this case, v =200 mph=89.4 m/s and d =¼
mile=402 m. The time can be eliminated using the first equation, t =89.4/a
and substituting into the second equation, 402=½a (89.4/a )2 =3996/a
or a =9.94 m/s2 . The acceleration due to gravity
is 9.8 m/s2 and so this is just about 1 g of acceleration,
a =1.01g . I figure that anyone who charges $650/hour
for his services can afford to send a little of that my way!
ANSWER: c . I doubt that you could fabricate any wheel which
could have a tangential speed 0.001c and not fly apart from the
stress of the centrifugal force. Think about a space ship in a vacuum!
ANSWER: F in the figure). You have not
given me enough information. You need to also know how much frictional
force (f in the figure) is on the bus as it moves. Assuming that the bus accelerates from rest with a uniform
acceleration a , F-f=ma where m is the mass of
the bus. Switching to SI units, m =13 ton=12,000 kg and 120
ft=37 m. Then, from kinematics, 37=½at 2 =½a (28)2 =392a ,
so a =0.094 m/s2 . So, F-f =0.094x12,000=1130
N=254 lb. One could make a reasonable guess about the friction by
assuming that it is mostly "rolling friction" of the tires, with a
coefficient of rolling friction of about 0.03, so f ≈0.03x12,000x9.8=3530
N=794 lb; the man would therefore need to exert a force of approximately
794+254=1048 lb.
ANSWER: currents
at times were sufficient to raise the already buoyant body(the ball)"? I
can guarantee that no thermal is going to cause a baseball to rise. It
could
keep the ball aloft longer, but can you imagine a
ball released at rest in one of these currents actually accelerating
upwards? What is most likely to cause unexpected behavior is spin
imparted to the ball when it leaves the bat. In particular, back topspin
can cause a ball to appear to rise because it is falling less than it
would with no spin. See the earlier answer on the
rising fastball .
QUESTION:
ANSWER: D =14d or d /D =2.
So, if d is greater than 2D , the whole system will
rotate about the rear wheels and topple off the bridge.
QUESTION:
ANSWER: 3 , so the volume of
the steel is about 100/8000=0.0125 m3 . If it is a sphere its
radius is about 0.144 m and its cross-sectional area is about A =0.0661
m2 . At sea level a pretty good approximation for the air drag
is F = ¼Av 2 (only for SI
units), so terminal velocity will be achieved when mg =¼Av 2 ,
or v =2√(mg /A )=62.6 m/s. Just an
estimate. Also, since it experiences less drag at higher altitudes
because of thinner air, I cannot guarantee that it would lose enough
speed from its orbital speed of about 8000 m/s to get to the
ground-level terminal velocity.
QUESTION:
ANSWER: —a
planet-sized object somehow gets inserted into an orbit which
periodically passes through the asteroid belt. The period of a typical
asteroid in a roughly circular orbit is about 3 years, so I imagine an
elliptical orbit for our rogue planet which has a semimajor axis about
equal to the radius of the asteroid belt as shown in the picture.
Kepler's first law then says that this orbit would also have a period of
about 3 years and therefore pass through the asteroid belt about once
every 1½ years; it would pass through the asteroid belt twice
each orbit spending only maybe a couple of months in the belt. So, we
are off to a good start for your scenario! To get some numbers I found
the a
Physics Stackexchange discussion useful. An estimate of the
distance between asteroids larger than 1 km is shown to be about 2
million miles (about 3 million km). This alone is sort of eye-opening—we
tend to think of the asteroid belt, ala sci-fi movies like Star Wars, a
dangerous densely-packed hazard whereas your rogue planet would be
extremely unlikely to encounter even a single one as it passed through
the belt. With a passage taking only a couple of months, the likelyhood
of even seeing a single large asteroid is small, let alone being close
enough to significantly alter its orbit. But, suppose that a large
asteroid is shaken loose. Now you have to think of the probability that
it will collide with the
earth; this will be even tinier than the probability of its having been
kicked out of the belt in the first place. I have not done any
quantitative probability calculations because I think the main point can
be made with very rough qualitative estimates as I have done here. The
bottom line is that space is unimaginably vast and even in places where
the density of objects is relatively large, it is still
extremely small in absolute terms. If this rogue planet were to appear,
I doubt that, on average, even a single asteroid collision with earth
would occur in a lifetime.
QUESTION:
ANSWER: mg down and air
drag D up and their sum must be the mass m times the
acceleration a of the tire: ma=-mg+D ; this is just
Newton's second law. When the tire is falling with constant velocity (v t ),
a =0 so mg=D . Now, what is D ? For this kind of
situation the drag is proportional to the square of the velocity —double
the speed and you quadruple the drag; a pretty good value of the
proportionality constant is A /4 where A is the area
presented to the onrushing air, so D ≈¼Av t 2
(this works only for SI units). What A is depends on how it is
falling; I will assume it falls with one face down, not edgewise, so
A=πR 2 =π x0.382 =0.45 m2 .
Now we can estimate the terminal velocity, v t ≈2√(mg /A )=2√(25x9.8/0.45)=47
m/s=105 mph. Now you want to know how high to drop it from. Well,
technically it is infinitely high because the speed approaches v t
and never really reaches it. So, to get a good idea we can ask how long
it takes to get to 0.99v t and how high h it
was dropped from. Solving the problem for y (t ) and v (t )
involves some pretty
tricky integration. The solutions are v (t )=v t tanh(gt /v t )
and y (t )=h -(v t 2 /g )ln[cosh(gt /v t )].
So, setting v (t )=0.99v t and
solving for t , I find t= (47/9.8)tanh-1 (0.99)=12.7
s. Finally, setting y (12.7)=0, we can get the height, h =((472 /9.8)ln[cosh(9.8x12.7/47)]=442
m=1450 ft.
Obviously, this math will be too much for your guys. It occurs to me
that they should be able to read a graph, though, so I have calculated
graphs of the functions below.
QUESTION:
ANSWER: F d of a bullet of speed v in air is well
represented by F d = ½Cρv 2
where ρ is the density of the air and C is a
constant determined only by the geometry of the bullet. As the density
of the air gets smaller, the drag on a bullet gets smaller, in accord
with your experience. There is really no need to invoke the ideal gas
law at all, but if you want to connect density to the ideal gas law,
PV=NRT , note that the density will be proportional to N /V ,
so ρ ∝ P /T . For
constant temperature, density is proportional to pressure—the
lower the pressure, the lower the drag.
QUESTION: 1 and m2 in space separated by a distance d, how long would it take for them to collide under gravity alone? I'd imagine it's a changing force/acceleration, so some calculus must be involved.
ANSWER: FAQ page. The one you
might find most useful is this
link . You will see that
the relevant time to collision is t=T /2=√(πμa 3 /K)
where μ =m 1 m 2 /(m 1 +m 2 )
is the reduced mass, a=d /2, and K=G m 1 m 2 .
Therefore, t = √[πd 3 /(8G (m 1 +m 2 )].
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: E =mγc 2 where γ= √(1-(v /c )2 ).
In your case, 2000 lb=907 kg, c =3x108 m/s,
γ= 1.005, 1.011, and 1.048. The energies in Joules are
E =8.204x1019 , 8.253x1019 , and 8.555x1019
J. There are about 4x109 J per ton of TNT, so the energies
are 20.51, 20.63, and 21.39 Megatons of TNT. I might add that this is
not actually very "realistic". Where are you going to get that much
energy (you have to supply it somehow) in the middle of empty space? Or,
look at it this way: I figure that for a mile-long gun the time to
accelerate the BFR to 0.1c is about t =10-4
s. During that time the required power is about 8x1019 /10-4 =8x1023
W=8x1014 GW; the largest power plant on the earth is about 6
GW! Also, don't forget about the recoil of the ship which would likely
destroy it.
QUESTION: 2014 discovery that the uncertainty principle and wave-particle duality are in fact the same thing?
ANSWER: obvious to me that wave-particle duality
is an inescapable consequence of the uncertainty principle. (This same
view is stated many times in the comments on this article.) When the
position wave function envelope narrows, the linear momentum wave
function broadens because the two are conjugate variables and thus
Fourier transforms of each other. I also
consider "a constant state of flux
between particle nature and wave-function" to be a misstatement; I
consider duality to mean that the particle/wave is simultaneously
particle and wave, not some dance between the two.
By making an appropriate measurement you can observe either one or the
other aspect.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: v (your 10 mph), and the top is moving forward
with speed 2v ; the belt is moving forward also with speed v .
I find this problem much easier to do if I transform into a coordinate
system which is moving with speed v to the right; in that
coordinate system the belt (upper surface) and tire are both at rest and
the top and bottom edges of the tire have speeds v as shown.
Before getting into the hard part, there is a special case which we can
get out of the way first: if there is no friction the tire will continue
on its merry way unchanged, both its speed and its angular velocity
unchanged because there are no net forces or torques on it.
f trying to accelerate it to the right so it will
start sliding along the belt (at rest in the frame we are using). The
frictional force will be f=μmg where μ is the
coefficient of kinetic friction, m is the mass of the tire, and
g is the acceleration due to gravity. Call v' the
speed which the center acquires in some time t . Then Newton's
second law for translational motion is m Δv=μmgt=mv' ,
so v'=μgt.
τ=fr=μmgr which acts opposite the
direction the tire is rotating. There will come a time when the bottom
edge of the tire will be at rest relative to the belt because of this
torque. At that instant, we can transform back into the original
coordinate system and the tire will be moving forward with speed
v+v' . Newton's second law for rotational motion is ΔL =τt=- (μmgr )(v' /μg )=-mv'r .
where L is the angular momentum of the system. The angular
momentum is the angular momentum about the center of mass plus the
angular momentum of the center of mass. L 1 =Iω 1 +0=Iv /r
where I is the moment of inertia about the center of mass;
L 2 =Iω 2 +mrv' =(v' /r )(I+mr 2 ).
Therefore, (v' /r )(I+mr 2 )-Iv /r=-mv'r
or, v'=Iv /(I+ 2mr 2 ). This is
surprisingly simple, particularly surprising that it does not depend on
μ. However, the time to stop sliding does depend on μ ,
t=v' /μg ; the slipperier the surface, the longer it
takes to stop slipping, as expected.
v as shown in the final figure. As an example, suppose
we model the tire as a uniform cylinder of mass m and moment of
inertial I =½mr 2 ; then v'=v /5,
or in your case, 2 mph, so the tire is rolling without slipping with a
speed of 12 mph.
ADDED
NOTE: ω'=v' /r
only a small fraction of the original value of ω=v /r.
This drop in angular velocity can be clearly seen in a
Mythbusters
episode showing a car driving onto a ramp from a moving truck, a
very similar situation to the conveyor belt problem in this question.
QUESTION:
ANSWER: g =9.8 m/s2 , it would be
wrong.
QUESTION:
ANSWER: v (as
seen by an outside observer at rest) the earth would recoil with a speed
of 2v . But, due to the gravitational attraction between the
two, they would go out to some distance and fall back to where they
started, again at rest.
QUESTION:
ANSWER: faq page.
QUESTION:
ANSWER: —air
would not enter at all.
QUESTION:
ANSWER: earlier
answer .
QUESTION:
ANSWER: …a
very strange thing indeed… x direction while the red-drawn
distances to the mirrors are at rest in the unprimed system. The rest
lengths of the two distances are L and the angle one makes with
the x -axis is θ . The round-trip time along the
x -axis is t 1 =2L /c and
along the other length is the same, t 2 =2L /c .
In the moving system all lengths along the the x' direction are
reduced by a factor √(1-β 2 ) where β=v /c ;
therefore L'=L √(1-β 2 ) so t 1 '=2L' /c =2L √(1-β 2 )/c=t 1 √(1-β 2 ).
For L" only its x' component is contracted, L" cosθ' =L √(1-β 2 )cosθ
while the y" component remains the same, L" sinθ' =L sinθ ;
from these you can easily show that L"=L √(1-β 2 cos2 θ )
and tanθ'= (tanθ )/√(1-β 2 ).
Finally, t 2 ' =2L" /c =[2L √(1-β 2 cos2 θ )]/c =2t 2 √(1-β 2 cos2 θ )≠t 1 ' .
ADDED
COMMENT: I see that I misread your question. I see that
you have put the lengths at rest in the moving frame whereas I have them
in the stationary frame. But in special relativity it makes no
difference which is moving and which is at rest. So, you have found that
the times in the frame not at rest relative to the lengths are not the
same, just as I have; I reiterate that this is expected, not surprising.
FOLLOWUP
QUESTION: Thanks a lot, your competent and prompt
answer is much appreciated!
I would like to reiterate that for two events that occur in the same place
One can think with some justification that this scenario means that the x'O'y' system is somehow more "special" than the xOy system, pretty much in contradiction, if not with Einstein relativity postulate proper, at least with its spirit. In other words, in my opinion, the relativity theory is not devoid of inner contradictions, or at least it leads to conclusions which seem to invalidate the most fundamental notions of cause and effect or of things being what they are irrespective of where we observe them. And to me it is debatable whether a physical theory, however legitimately counter-intuitive it may be understood to be, can lead to this type of conclusions and be nevertheless considered a valid description of the physical world.
ANSWER: not say
that the bomb will not detonate because the device which detonates it is
in the stationary frame and he will agree that the pulses in that frame
are indeed simultaneous. Is the stationary system in some way "special"?
Of course it is —that is where the apparatus resides and all
frames will agree that the bomb will be triggered in that frame.
QUESTION:
ANSWER: 2 /172 =0.0008. Compared to the force slowing down the vertical motion, the force
speeding up the horizontal motion is tiny.
QUESTION:
ANSWER: earlier answer ). If both particles are at
rest, each feels the electric field of the other as you state. If one
moves, the other sees only the electric field of the moving particle,
not its magnetic field; but the electric field of the moving field will
not be the same as when it was at rest so the force felt by the other
will be different; if the speed is small compared to the speed of light,
the change in electric field will be negligibly small.
QUESTION:
If a firefighter adds 45 pounds of gear to his overall weight, does it
increase the impact force if he has no choice but to jump out a window and
if so to what degree?
What is the effect on impact force of jumping out a second floor window
vs. 3rd floor and above?
This actually isn't homework. I'm 54 years old and advising a charitable organization that provides safety systems to firefighters. One of the challenges they face is fire departments who don't have high rise buildings and feel they don't need the bailout systems. So I'm trying to figure out what the impact is for a firefighter forced to jump out a second story or third story window. This will help inform how they talk to prospective donors.
If you could help me with that (understanding/identifying the increase in impact force from 1st to 2nd to 3rd floors and then up) it would be greatly appreciated. I've done a lot of googling around calculating impact and g forces, but its not making sense to me (I'm not being lazy, just not understanding).
ANSWER: v , a mass m , and stops in a time t ,
the average force F during the stopping time is given by
F=m (g+v /t ) where g= 32 ft/s2
h . Jumping from the second floor would result in a speed of v 2 = √(2gh );
jumping from the third floor would result in a
speed of v 3 =2 √(gh ) which is
about 1.4 times larger than v 2 . In general, jumping
from the n th floor would result in a speed v n n- 1)gh ].
Maybe some numerical examples would be useful to you. You prefer
Imperial units, which makes things a little complicated. The quantity
mg is the weight. I will take mg =160 lb and h =12
ft for my numerical calculations, so when I need the mass I will use 160
lb/32 ft/s2 =5 lb·s2 /ft. (The unit of mass
in Imperial units is called the slug, 1 slug=1 lb·s2 /ft.) The speeds from
second and third floors will then be v 2 =27.7
ft/s=18.9 mph and v 3 =39.2 ft/s=26.7 mph; these
speeds are independent of the mass. Finally we must approximate the
times to stop. If she lands feet first, she could extend, as
parachutists do, her stopping time by bending her knees into a squat; I
will estimate the distance s she will travel while stopping as
s =3 ft. The time may be shown to be t =2s /v
so t 2 =2x3/27.7=0.22 s and t 3 =2x3/39.2=0.15
s. Finally, the average forces during impact are F 2 =160+5x27.7/0.22=790
lb and F 3 =160+5x39.2/0.15=1467 lb. These are
approximately 5 and 9 times her weight (g-forces of about 5g
and 9g ).
QUESTION:
ANSWER:
Mandela effect . For readers who, like me, are similarly ignorant it
is a special case of a psychiatric phenomenon called
confabulation ,
defined as "the production of fabricated, distorted or misinterpreted memories about oneself or the world, without the conscious intention to deceive";
in other words, a "false memory". The Mandela effect is a confabulation
where a large number of people have the same same false memory; it is
apparently so called because there are many people who "remember" that
Nelson Mandela died in 1980, not 2013. So, where is the physics in all
this? A "paranormal consultant" named
Fiona Broome
suggested that such cases could be explained as "bleed through" from
parallel universes. She is not a physicist and her speculation, maybe
amusing, has no real basis in physics and there is no way to test it one
way or another; therefore it should not be labeled as a "theory" since
it is neither testable nor falsifiable.
Snopes.com has a long article on the Mandela effect which you can
read if you want to learn more.
QUESTION: 2 .
If it is N/m then which length will we have to take here.
ANSWER: F . It is
found, by doing many experiments, that F depends on L ,
but that the ratio F /L is always the same.
Furthermore, it makes no difference what the shape of the film to the
left of the slider is. We therefore define the surface tension for this
fluid to be γ=F /(2L ) as determined by
this experiment, the factor of ½ because there are two surfaces
we are stretching.
QUESTION:
ANSWER:
QUESTION:
ANSWER: faq
page .
QUESTION:
ANSWER: F W N f F=f and N=W . But, what you need is F
and to get it you need to sum the torques around the edge where N
and f act and sum to zero: Fh-Ww /2=0 or F=Ww /(2h ).
Using your numbers, F =22x6.40/(2x7.62)=9.24 tons. If you want
the force in Newtons, assuming that the 22 ton mass is metric tons
(22,000 kg), F =9.8x9240=90,500 N.
QUESTION:
ANSWER: earlier
answer , the gravitational force probably increases slightly or stays
about constant to a distance from the center of about 4000 km.
QUESTION:
ANSWER: …provide
mathematical equations…" I can tell you, though, that a proton
bound to a neutron, called a deuteron , exists in nature. It is
the nucleus of heavy hydrogen, 2 H. There is, however, no
bound state for two neutrons (referred to as a dineutron) —it
does not exist; the same is true for the diproton.
QUESTION:
ANSWER: …for
some weird reason gravity still works …" The earth's
gravity has nothing to do with its magnetic field.
QUESTION:
ANSWER:
QUESTION:
ANSWER: "a highly ionized gas containing an approximately equal number of positive ions and electrons"
(Dictionary.com ).
If you start with a molecular gas like water, the definition is
sufficiently vague that there are many answers to your question. If
temperatures are sufficiently high to start ionizing molecules, you will
also have dissociation taking place so possibilites for the positive
ions present would be H2 O+ , H+ , O+ ,
OH+ , O2 + , and H2 + ;
the gas also would have neutral molecules present of all these, the
fractions depending on the pressure and temperature. As the plasma
cooled down, there is no reason why all the ions and neutral molecules
would return to being water molecules.
QUESTION:
VIDEO
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: ½ in the case of a
neutrino (or electron, or proton, or neutron). A spin-½ particle
might also have an orbital angular momentum quantum number L
and its total angular momentum quantum number J=L ±½
could be different from its spin. Maybe that is how you are coming up
with 3/2. For example, if L =1 for the neutrino, J =½
or 3/2. (By the way, if you are just
learning about angular momentum, the actual angular momentum of an
object with angular momentum quantum number N is (h /2 π )√(N (N +1))
where h is Planck's constant.)
QUESTION:
ANSWER:
E=mc 2 ? Just as the particle has uncertain
momentum, it also has uncertain energy. Your last sentence makes no
sense since it takes no energy to be moving (Newton's first law).
QUESTION:
ANSWER: d traveled by an object
accelerating at a constant rate a for a time t is
d = ½at 2 . For your question, a =7x9.8=68.6
m/s2 and t =10 s, so d =½x68.6x102 =3,430
m=3.43 km.
QUESTION:
ANSWER: E Δt ≈ h /2π
where
h /2π -16 eV·s is the rationalized Planck's
constant. In order to measure the energy of a system, you must spend
some time; the more time you have to examine the system, the more
accurately you can determine its mass. But, if the system is unstable,
that is it has some lifetime before it decays to a different state, you
will get a different answer every time you try to measure its mass
(which is, equivalently, its energy). A proton is a stable particle and
therefore, since it lives forever its mass has zero uncertainty and is
why all protons have identically the same mass. However, you are wrong
that all neutrons have the same mass because the neutron beta-decays to
a proton+electron+neutrino with a half life of about 880 s. So, the
uncertainty in the neutron's rest mass energy is about ΔE ≈h /(2π Δt )=7.5x10-19
eV=7.5x10-28 GeV. The rest mass energy of a neutron is about
1 GeV, so the uncertainty is about 7.5x10-26 %! This is small
but not zero. These results have nothing to do with either the masses of
quarks or time quantization.
QUESTION:
ANSWER: 8 m
away and a beam rotating with a frequency of 1 cycle per second would
have a spot going across the moon with a speed of about 2.5x109
m/s, nearly 10 times the speed of light. Does this violate the law that
nothing can go faster than the speed of light? No, because this spot is
not anything and you cannot use the spot of light on the moon to
transmit information between two points on the moon.
QUESTION:
ANSWER:
Car Talk there was a puzzler
which asked "how much salt is there in salt water?" The answer was
"none". In solution, salt, NaCl, dissociates into Na+ and Cl- .
So, it is not such a good answer to simply say that salt has a higher
index. Understanding index of refraction on the atomic or molecular
lever is very difficult.
QUESTION:
ANSWER: earlier answer . Now,
infinite density does not mean infinite mass; rather the mass is some
finite number and the volume is zero. So, clearly, the rest of your
question does not apply to black holes since their mass is never
infinite. Super massive black holes have masses much larger than the
mass of a star.
QUESTION:
ANSWER: σ =dq/ da ;
(dq is the charge on an area da ). Another way to put
it is if you look at any area a anywhere on
the surface, you will find an amount of charge q =σa.
QUESTION:
ANSWER: vice versa .
QUESTION:
ANSWER:
QUESTION:
ANSWER:
tables where the R-value per inch of thickness is tabulated for
various materials. But, these blocks are fabricated from a random mix of
all classes of recyclable plastics. Furthermore, it is not clear how
much air is trapped in them. Plastics seem to have R-values in the range
of 3-5 ft2 ·°F·h/BTU/in, so you could
certainly take this range to estimate the value for a block of known
thickness. If you are interested in the details of the physics of the
R-value, you might be interested in an
earlier answer .
QUESTION:
ANSWER: E=mc 2 =9x109 J. The melting
point of iron is 1540°C and its specific heat is 449 J/kg·°C,
so the enegy to heat it from 20 to 1540°C is 1x1520x449=6.8x105
J. So, the ball would have increased its mass by 6.8x105 /9x109 =7.6x10-5
kg. You would be hard-pressed to actually measure so small a change in
mass.
QUESTION: 2 O. So is Deuterium oxide water with one Protium and one deuterium atom, or water with two deuterium atoms. If both hydrogens in water are actually deuterium, how did that come to be?
ANSWER: 2 )would be D2 O.
The term "heavy water" refers to D2 O and DHO is sometimes
referred to as semiheavy water. You would think it would be a hopeless
task to extract such a tiny fraction of D2 0 from natural
water. I was very interested to learn that there is a much more clever
way to get nearly pure heavy water. There are relatively many DHO
molecules so it is much more practicable to use standard isotope
separation techniques to get enriched DHO. So, for example, suppose that
you enrich the water such that the hydrogen atoms present are 50%
deuterium. Then it turns out that the water is 50% HDO, 25% H2 0,
and 25% D2 O. Where did all that D2 O come from? It
turns out that in water the hydrogen atoms (both isotopes) do not stay
attached to the same oxygen atom, rather they move more or less freely
around. By repeating whatever you did to get this far, you can get a
higher and higher percentage of deuterium. Enrichments of 99.75% D2 O
are routinely achieved.
QUESTION:
ANSWER: g s. One
g would correspond to the weight of the parachutist. Both graphs
end up after about 7 s at g force =1 which means that
the force up on him would be equal to his weight. For the "soft"
deployment, about 3g is the maximum, so you would have to be
able to hang from your hands from a stationary bar with three times your
weight attached to you; you could probably do this briefly and it would
certainly not tear the arm from its socket. For the "hard" deployment,
almost 6g is the maximum, so you would have to be able to hang
from your hands from a stationary bar with six times your weight
attached to you. You could probably not do this but it still would
probably not tear the arm from its socket. In any case, I would never
depend on being able to hang on with my hands!
QUESTION:
ANSWER: ρ =62.4 lb/ft3 =0.0361
lb/in3 . Therefore the
gauge pressure 2 ft down is P=ρh= 62.4x24=0.867 lb/in2
(psi). That pressure will just keep the tube filled with air all
the way to the bottom. Any pressure greater than this will expel air
into the water. This is probably the number you want; the force over the
area of the tube is F = 0.867·π· (0.25/2)2 =0.0426
lb=0.68 oz.
QUESTION:
ANSWER: W which
points down and and the force N the scale exerts up on him. (N
is what the scale will read.) If the person is accelerating upward,
Newton's second law states that N-W=ma where m is the
person's mass. Therefore, N=W+ma and so the scale reads more
than the person's weight.
QUESTION:
ANSWER: F G , which points
vertically down and causes the ball to accelerate downward (a
horizontally thrown 90 mph fastball falls about 4 feet on the way to the
plate); there is the drag F D which points opposite
the direction of flight and tends to slow the ball down (a
90 mph fastball
loses about 10 mph on the way to the plate); and there is the magnus
force F M which, for a ball with backspin ω
about a horizontal axis points perpendicular to the velocity and upward.
If the ball is moving horizontally the only way it could rise is for the
Magnus force to be larger than the weight. Measurements have been done
in wind tunnels and it is found that if the rotation is 1800 rpm, about
the most a pitcher could possibly put on it, the ball would have to be
going over 130 mph for the Magnus force to be equal to the weight. When
you say "thrown on a straight trajectory", you cannot mean it left his
hand horizontally because it would hit the ground before it got to the
plate; a fast pitch like that is impossible to accurately judge the
initial angle of the trajectory.
QUESTION:
ANSWER: V=NkT/P is the ideal gas law (IGL). I write it like this to
emphasize what happens if T decreases and/or P
increases while N stays the same—V decreases.
(You may write it as V=nRT/P where n is the number of
moles and R is the universal gas constant. For reasons of
clarity in my answer, N is the number of molecules and k
is Boltzmann's constant.) The IGL assumes that the volume occupied by
the molecules is insignificant and that the molecules do not interact
with each other. But, when the volume decreases the molecules get closer
together so that their interactions start to matter and the total
fraction of the volume which the molecules occupy is no longer a
negligible fraction of the volume of the container. (You are incorrect
when you say that "…the size of the molecules become much greater…";
the size stays the same but now occupies a larger fraction of the whole
volume.) The first thing to do to correct the IGL is to replace V
by V-Nv where v is the volume of a single molecule:
P (V-Nv )=NkT. Although this improves the
accuracy of the IGL, it still overestimates the pressure observed
experimentally. The reason is that the effects of the intermolecular
forces have not been included. It turns out by doing measurements that
the pressure is proportional to the square of the number density, P ∝(N /V )2 ;
this makes some sense because the molecules interact by pairs and there
are N (N -1)/2 ≈N 2 /2
pairs because N is very large. So, finally,
(P-c (N/V )2 (V-Nv )=NkT
where c is a proportionality constant which depends on what the
particular gas is. This is sometimes called the van der Waal equation.
QUESTION:
ANSWER: th birthday. It is
true that the traveler did not have any "access to" 70 years of his
brother's life, but that does not negate the fact that he had now,
spending 10 years of his life, been able to arrive at what would have
been his 70th birthday had he stayed at home. From this point
on, he is 60 years into the future because of having taken his trip. He
now has "access to" every second of that future. You may want
to read my discussion of the twin
paradox .
FOLLOWUP
QUESTION:
ANSWER: every second of that future, but he has to wait for it."
QUESTION:
ANSWER: g in
empty space and being in a gravitational field in which the acceleration
due to gravity is g .
QUESTION:
ANSWER: —the power is the product of the current and the
voltage. Both current and voltage are sinusoidal functions of time,
i (t )=I sin(ωt ) and v (t )=V sin(ωt+φ ),
so p (t )=IV sin(ωt )sin(ωt+φ ).
The graph above shows three choices for the phase φ between
i and v . For φ=π /2 the time average of
the power is zero, no energy flow; for φ=π and φ= 0
the time average of the power is negative and positive, respectively.
The motor in a mechanical meter turns in opposite directions for
different signs of the average power; in a digital meter the average
power is determined by an electronic circuit.
FOLLOWUP
QUESTION:
ANSWER:
QUESTION:
ANSWER: -10 =0.9999999993 seconds. You
may be sure that relativity has nothing whatever to do with observable
building decay here on earth!
QUESTION:
ANSWER:
QUESTION:
ANSWER: -10 m. So, if he halves N times, 2/2N =2x10-10
or 2N =1010 . The easiest way to do this is with
logarithms: log(2N)=N·log2=log(1010 )=10 or
N=10/log2=33.22. N should be an iteger, so the time it takes is about 33
seconds.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: M get lifted to h , I estimate that
the kinetic energy changes by ΔK≈ 2hK initial /R=MhRω 2
where R =6.4x106 m is the radius of the earth and ω =7.3x10-5
s-1 is the angular velocity of the earth. For example,
lifting 1 kg a height of 1 m requires 0.03 J of work to increase the
kinetic energy. But wait a minute! Once we acknowledge that the earth is
rotating, we have to recognize that the mass, being in a circular orbit,
has a centripetal acceleration a c =Rω 2
and therefore the net force on M is Mg-MRω 2 .
Therefore, the net work done is W ≈(Mg-MRω 2 )h +MhRω 2 =Mgh .
QUESTION:
ANSWER:
QUESTION:
ANSWER: 7 /3x105 =300
lb, and 14 miles at 22.5° would take you to 14sin22.5°=5.4
miles, higher than Mount Everest! Also, since we are working with a
fluid, pressure would be a more appropriate quantity to look at than
force. I will start from scratch with new
numbers: the density of crude oil is about ρ =900 kg/m3
and I will take a distance of about y =1000 m between the
top and the bottom, about the height of a small mountain; you need only
the drop, as we shall see below. Like you, I will assume that the oil in
the pipe is static which simplifies things. A good approximate way to
solve this kind of problem is to use Bernoulli's equation, P +ρgy+ ½ρv 2 = constant;
P is the pressure, g =9.8 m/s2 is the
acceleration due to gravity, and v (=0 for us) is the speed. At
the top, P top =P A and y top =1000
m and at the bottom P bottom =P+P A
and y bottom =0; here, P A is the
atmospheric pressure. Therefore, P A +900x9.8x1000=P+P A +0.
So the gauge pressure is P ≈107 N/m2 ≈100
atm=1450 PSI. The brief research I have done indicates that this pressure
is at the extreme upper limit of specifications for pipelines. To make
the oil move, you need to add a lot more pressure. My use of Bernoulli's
equation is a very crude approximation because it assumes a fluid with
no viscosity and no frictional forces with the walls, obviously not a
very good approximation.
QUESTION:
ANSWER: —the radiated sound, the drag of the string
moving through the air, and the friction associated with the string
bending and unbending. A guitar in a vacuum no longer has the first two
ways to lose energy, so only the bending friction would be converting
kinetic energy of the string to heat in the string. It would therefore
take longer for the string to stop than in the air, but the total energy
converted would be the same; if the string did not have a way to get rid
of its energy (it does, it could radiate or conduct to the pegs, but
let's forget that), it would end up hotter than it would in air.
QUESTION:
ANSWER: Yes, there is an equation. I will warn you, though, that this is quite
technical and pretty high-level. Refer to the figure to the left. The
charge q , with velocity v x t ) at time t ; we
wish to know the field at position r t . However, since the information about the field propogates at
the speed of light, c , the field at time t is
determined by where the charge was at some earlier time t ρ ;
|r x t ρ )|≡|ρ ρ =c (t-t ρ ).
After much calculation (see the detail in Chapter 10 of David Griffith's
book
Introduction to Electrodynamics ), the magnetic field is
B r t )=(v E r t ))/c 2
where the electric field is
E r t )=[q /(4πε 0 )][ρ /(ρ ·u 3 ][(c 2 -v 2 )u u u ≡c1 ρ -v 1 ρ is a unit vector in the
direction of the vector ρ .
A DDED
NOTE: The above expressions for electric and magnetic field are exactly
correct. You can write approximately correct equations for particles
with speed much less than the speed of light, v<<c :
E r t )=[q /(4πε 0 )][(r x |r -x 3 ]
and
B r t )= [ qμ 0 /(4π )][ v x(r x |r -x 3 ].
These are simply
Coulomb's law and the
Biot-Savart law .
QUESTION:
ANSWER: etc .)
have magnetic moments, but these are incredibly weak and do not normally
react to a magnetic field; if you just have an electric charge Q ,
it experiences zero force if at rest in a magnetic field. Yes, it is an
empirical fact (experimentally observed) that a charged particle Q
with velocity V
B experiences a force F =QV B
QUESTION:
ANSWER: M hits the ground with some
speed V and stops in time t , the average force over
the time is F=W+MV /t where W is the weight,
200 lb; to convert the
weight to mass, divide by the acceleration due to gravity, g =32
ft/s2 : M=W /g =(200 lb)/(32 ft/s2 )=6.25
ft ·lb/s2 . The speed at impact can be determined
from V =√(2gh )=√[2x(32
ft/s2 ) x(½ ft)]=5.7 ft/s. If we approximate that
the distance S over which your legs bend on landing as S =1
ft, the time to stop is t= 2S /V=(2x1 ft)/(5.7
ft/s)=0.35 s. So, finally, F =200 lb+(6.25
ft ·lb/s2 )(5.7 ft/s)/(0.35 s)=302 lb. This is
the force the scooter exerts on you which, by Newton's third law, is
equal the the force you exert on the scooter (but in opposite
direction). If you stop in ½ ft, the force would be 404 lb. Keep
in mind that this is a very approximate estimate of the average force.
QUESTION:
ANSWER: earlier
answer .
QUESTION:
ANSWER: velocity addition .
In classical physics, v CB =v CA +v AB ;
the notation is "v IJ is the velocity of I relative
to to J". I have written this so that it corresponds to your question —A
is you, B and C are the trains; it is more convenient for you if we
rewrite the equation as
v CB =v CA -v BA
which we can do because v AB =-v BA .
If the trains have speeds v in the same direction, v CB =v-v =0;
in opposite directions, v CB =v- (-v )= 2v— each
sees the other moving with speed 2v . But, this form of velocity
addition is wrong for very high speeds (see an
earlier answer ).
The relativistically correct velocity addition equation is
v CB =(v CA -v BA )/[(1-(v CA v BA /c 2 )]
which reduces to the classical equation for the speeds much less than
c . So, for the trains moving in opposite directions, v CB =2v /(1+v 2 /c 2 );
for example, if v =0.5c , v CB =c /(1+0.25)=0.8c .
Now, you seem
to think that if you double the speed, you double the time difference.
This is not correct— time dilation goes like the gamma
factor, 1/√(1-v 2 /c 2 ).
So, for your situation with v =0.5c , t A =1.33t B ,
t A =1.33t C , t C =1.67t B ,
and t B =1.67t C . These are
confusing, I admit. They are meant to denote, for example, that t B =1.67t C
means that B sees his clock tick out 1.67 s when C's clock ticks
out 1 s; B observes C's clock to be slow.
QUESTION:
ANSWER:
QUESTION:
ANSWER: B W N N N=W and f=B , equilibrium.
The biggest that the frictional force can be without the man's feet slipping is
f= μN where μ is the coefficient of
static friction between shoe soles and road surface. A typical value of
μ for rubber on asphald, for example,
μ≈ 1, so the biggest f could be is
approximately his weight W ;
this means that the largest force he could exert on the bus without
slipping would be
about equal to his weight. Taking W ≈200 lb, if the
frictional force on the bus is taken to be zero, the bus would
accelerate forward with an acceleration of a=Bg /16000=200x32/16000=0.4
ft/s2 where g =32 ft/s2 is the
acceleration due to gravity; this means that after 10 s the bus would be
moving forward with a speed 4 ft/s. If there were a 100 lb frictional
force acting on the bus, the acceleration would only be a=0.2 ft/s2 .
If there were a frictional force greater than 200 lb acting on the bus,
the man could not move it.
QUESTION:
ANSWER: twin paradox . Acceleration
just makes everything harder to understand. Basically, just assume
necessary accelerations (departing, turning around, landing) occur in a
vanishingly short time. But, you are not really interested in the twin
paradox, you are interested in how things appear in an accelerated
frame. I am sorry, but nothing in your question after " …clock
on Earth to SPEED UP…" makes any sense. First of all, the fact
that any observer will measure the speed of light to be c is a
law of physics and the general principle of relativity states the laws
of physics are the same in all (not just inertial) frames.
Second, to measure a speed in your frame of reference you must use your
clock, not somebody else's. And third, how another clock looks is really
irrelevant because how it appears to run and how it
is actually running are not the same.
I would like
to address how the clocks look from the perspective of the
Doppler effect . The relativistically correct Doppler effect is
usually expressed in terms of the frequency of the light; for our
purposes, it is more convenient to express it in terms of the periods,
T observer =T source √[(1+β )/(1-β )]
where β=v /c and β is positive for
the source and observer moving apart; it makes no difference which is
the observer—each twin will see the other's clock running at the
same relative rate. Let's illustrate with a specific example, β =-0.8,
the traveling twin coming in at 80% the speed of light; T observer =T source √[(1-0.8)/(1+0.8)]=T source /3
so the observer will see the source clock running fast by a factor of
three. But special relativity tells us that moving clocks run slow, not
fast; 1 second on the moving clock will be 1/√(1-0.82 )=1.67
seconds on the observer's clock. How the moving clock looks is thus
demonstrably not a measure of how fast it is actually running. Now, if
the incoming twin puts on the brakes such that he slows to β =-0.6,
T observer =T source √[(1-0.6)/(1+0.6)]=T source /2
so the observer will see the source clock running fast by a factor of
two, apparently slowing down. Now, 1 second on the moving clock
will be 1/√(1-0.62 )=1.25 seconds on the observer's
clock, speeding up compared to when β =-0.8.
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER: m moving in a
circle, has only a horizontal centripetal acceleration toward the center
of that circle, a c =v 2 /(L sin θ )
where L is the length of the string; applying Newton's second
law, T sinθ =m v 2 /(L θ ).
There is no acceleration vertically and so you can apply Newton's first
law,
T cosθ-mg= 0.
QUESTION:
Ok, so my question is...
How many 100 W light bulbs can I have ON at the same time with the energy coming from that falling weight? -- while the weight is falling, obviously.
if the weight falls for 1 hour
if the weight falls for 2 hours.
What's the formula?
Somebody asked the same question on some forum on the web. His weight was 200 tons falling 100 meters for 1 hour, and someone said that the solution is:
ANSWER: ·hr you have consumed; a
kW ·hr is a unit of energy, 1
kW ·hr=1000x3600 J=3,600,000 J.
The example
stated is correct but the units are not. It is fine up to the point
where the potential energy of the mass is 196.2 MJ (mega=M=106 ).
Now, if you let this mass drop over 3600 s, it is losing its energy at
the rate of (196.2x106 J)/(3600 s)=5.45x104
J/s=54.5 kW (not kW/hr). For your case, your mass has a potential energy
of 104 x9.81x5=4.9x105 J. If you deliver this
energy over an hour, the power is 4.9x105 /3600=136 W; You
could power one light bulb over this hour and have some energy left over
at the end (about ¼ of what you started with). Clearly, the power
delivered over two hours would only be half as much, not enough to power
even one 100 W bulb.
QUESTION:
ANSWER: mg ;
if I find the distance d between the electrons to not be very
small compared to the radius of the earth, I will have to start over
again and use the force as MmG /(R+d )2 . The
electrostatic force between the two electrons is ke 2 /d 2 =9x109 x(1.6x10-19 )2 /d 2 =2.3x10-28 /d 2 =mg =9x10-31 x9.8=8.8x10-30 ;
solving, d =5.1 m. Good, I don't have to start over!
QUESTION:
ANSWER: There is not enough information. How far do you have to pull the sling
shot to reach 500 lb? With that info I can get a good estimate without
including air drag which means without the wind. Air drag adds quite a
bit of difficulty to the problem.
FOLLOWUP QUESTION:
ANSWER:
This is a truly peculiar question! I will first neglect air drag and the
wind; then I will include an approximate calculation including air drag.
I will assume the pig launches at 45 º and returns to the
same level from which it was launched. You should know that scientists
prefer to work in SI units, so I will convert all your numbers and
convert back to English units at the end. 100 lb=45.4 kg, 500 lb=2224 N,
175 ft=53.3 m. So the spring constant k =2224/53.3=41.7 N/m. So
the energy stored in the spring is ½kx 2 =½x41.7x53.32 =59,200
J. This must equal the kinetic energy of the pig at launch, ½mv 2 =½45.4v 2 =22.7v 2 =59,200.
Solving for the launch speed of the pig, v =51.1 m/s. The range
of a projectile launched at 45º is R=v 2 /g =266
m=873 ft.
Wolfram . Without going into details, the graph shows the maximum
distance gone is about 120 m=394 ft and the launch angle is about 36º.
The red shows the path with air drag, the blue without. I emphasize that
this is a very rough calculation but it is useful to demonstrate that
air drag is important. To include the wind would be pointless since the
uncertainty in air drag is much bigger than any effects such a wind
would have; however, if you were to do it, you would need to specify the
direction of the wind.
QUESTION:
ANSWER: W=mg , where g=9.8 m/s2 .
But it doesn't really matter for this question because, as you will see
below, the stopping distance does not depend on what the weight is.) The
maximum force you can get when braking is the maximum frictional force
which, on level ground, is μW where μ is
the coefficient of static friction between the wheels and the road and
W is the weight. The work done by friction is equal to the
change in kinetic energy, -μWs =-½mv 2
where s is the stopping distance, v is the starting
speed, and m is the mass. But, the mass is the weight divided
by the acceleration due to gravity, m=W /g ; therefore
s=v 2 /(2μg ). As you can see, the minimum
stopping distance does not depend on weight, only on both having the
same wheels on the same surface, e.g ., rubber on asphalt.
QUESTION:
ANSWER: 2 ) and the dB
is proportional to the logarithm. So, if you want to specify the "dB of
the source" you need to specify some geometry. Suppose we say that we
will measure all the energy passing through a sphere of radius 1 m.
Then, you would measure for both sources that the total energy per
second passing through that sphere would be the same. Using the
definition of the
dB, 65 dB=106.5 10-12 =3.2x10-6 W. Now we
come to the tricky part; you should be able to see that how far we will
be able to hear this depends on how this power is distributed over the
sphere. If the source radiates equally in all directions, this power
will be evenly distributed over the sphere; as an example, let's assume
your "raw" is distributed that way. Then the intensity of the sound is
the power divided by the area of the sphere, 3.2x10-6 /(4 π 12 )≈2.5x10-7
W/m2 . Now, suppose that your amplified sound comes from a
speaker which only radiates in the forward direction; if you think about
it, the intensity at the 1 m distance will be about 5x10-7
W/m2 , twice as loud. As you moved farther away, you would
find that the intensity fell off like 1/r 2 where
r is your distance from the source; at 500 m away, the intensity of
the "raw" will be at 2.5x10-7 /5002 =10-12
W/m2 and the amplified will be at 2x10-12 W/m2 .
The "threshhold of hearing" is about 10-12 W/m2 ,
so, in principle, you could barely hear both, the amplified being
louder. Beyond this distance, and out to about 700 m, only the amplified
sound would be heard. These specific distances depend on the assumption
that there are no other damping mechanisims in the air. The bottom line
is that it all depends on the pattern of the radiation of both. You
might be interested in an
earlier answer .
QUESTION:
ANSWER:
QUESTION:
ANSWER:
FOLLOWUP
QUESTION :
ANSWER: y
motion is y =0=y 0 +v 0y t - ½gt 2 =200-0-½9.8t 2 =200-4.9t 2 ,
so t =6.39 s. The equation for x motion is x=x 0+vx t =0+60t =383
m.
QUESTION:
ANSWER:
QUESTION: novelty gift that purports to find the "balance point" of a golf ball by spinning it up to 10,000 rpm. After 10-20 seconds the ball reaches an "equilibrium" spin and a horizontal line is marked that indicates the "balance axis". The assumption that on tees and greens you orient the ball to put the line vertically along the intended path so the
center of gravity (CoG) is rolling/spinning over the target line and thus minimizing potential effects of the CoG being on the side and potentially causing a "wobble". Putting the ball at different starting orientations in the device doesn't matter. It does tend to find the same "equilibrium" spin after a time so it is consistent.
I decided to mark a dozen balls using the device and then put them in a container of salt water to compare it to finding the CoG using a buoyancy test. Put in enough salt and eventually the golf balls float and
NOTE FROM
THE PHYSICIST:
NEW
ANSWER: First, let's consider the physics of the spinning method. Clearly
the idea is that if the CoG is not at the geometrical center of the
ball, if you spin the ball about a vertical axis, the CoG will be
"thrown out" horizontally. But, will it end up in a horizontal plane
passing through the center of the ball? What I show here is that the
answer is no. If the CoG is a distance r from the
geometrical center of the ball, has a mass m ,
and is spinning with an angular velocity of ω , the CoG experiences
(in the rotating frame) three forces: the weight mg , the force T
holding it in place, and the centrifugal force C=mrω 2 .
In spite of all
i.e. r remains constant as the ball spins.
There will be an angle θ where the sphere is in equilibrium. Summing the
torques about the center of the sphere, mgr cosθ-mr 2 ω 2 sinθ =0
so tanθ=g /(rω 2 ).
Now, the questioner found that for most balls, θ ≈45º
so tanθ ≈1.
Taking g ≈10 m/s2 and ω 2 =(10,000
rpm)2 ≈106 s-2 ,
I find that r ≈10-5
m. This is very small, 1/100 mm, but demonstrates that the spinner will not find
the correct plane for very small r . Technically, it never finds the
correct plane, but for r >1 mm, θ <0.6º. Since it is my understanding that
modern golf balls are very homogenous, this device is not useful for most balls.
It would work better if the axis of rotation were horizontal.
Next I should address the question of whether such a small r will be detectable by the
floating method. We can use the same figure above except with C R =0.021
m. Taking r =10-5 m, the torque about the center of the ball is τ=mgr cosθ= 4.6x10-6 cosθ N·m. The moment of inertia of the sphere is I =2mR 2 /5=4.1x10-5
kg·m2 . So, the angular acceleration is α =τ /I= 0.11cosθ
s-2 . This means that, if you start at θ= 0º, after 1 s
the angular velocity would be about 0.11x180º/π ≈6º/s,
easily detectable I should think.
All this is truly academic, though, since I am sure nobody
really thinks that a ball with its CoG less than 1/10 mm from
the center of the sphere will behave in any measureable way
differently than a perfect ball. So the Check-Go Pro does no
harm to your game, it just does no good unless you happen to
have a really off-center CoG. If you do both measurements,
you can locate surprisingly precisely where the CoG is with θ
giving you r and the vertical giving you the direction
of the line between the CoG and the center of the sphere.
OLD ANSWER: g . (One "g " of force is equal to the weight of
the object.) Now, if your gizmo gets the ball spinning with an angular
velocity of ω =10,000 rpm ≈1000 s-1 ,
the centripetal acceleration is a=R ω 2 R is the distance from the axis of rotation. A point only
1 mm from the center would have a =1000 m/s=100g and
for a point on the surface where R ≈2 cm, a =2000g ;
the resulting forces on the ball will surely cause the ball to deform
into an oblate spheroid, flattened along the axis of rotation. (The
figure shows a much exaggerated flattening compared to what the actual
would be.) However, unless the CoG is at the center of the ball,
changing the shape of the ball will change the location of the CoG and
after the ball stops spinning the CoG will move back to its original
location. The CoG will never be very far from the center, so I would
guess that the errors in locating the plane in which it lies could be
quite large. I can see no reason why your floating method would not
work. Maybe you ought to market it!
FOLLOWUP
QUESTION: One follow up question (again purely academic) if I might. Since a golf ball spin rate could vary between 2500-9000 rpm depending on the club, would the "spinner" axis be more accurate/appropriate for a ball spinning at a high rate and the buoyancy axis more accurate for a ball rolling on a putting green?
Also, the buoyancy test wasn't my idea. I wish I were that clever. It's been around for a long time. I'm told that Ben Hogan checked his golf balls that way back in the 1950s when golf balls were quite a bit les
ANSWER:
FOLLOWUP
QUESTION: Your response to the second question makes this even more academic (if possible) because if the rolling ball is the only one affected then clearly the spinner is superfluous.
One last followup question. I promise. After your first response, I wondered if the change in ball shape is measurable at all. I mounted a laser on a level and shot it over the top of the ball at rest so that it barely touched the top of the ball. After spinning the ball up to full speed I couldn't perceive any change at all in the amount of light touching the top of the ball. Is the expected change in shape that small? If so, does that still explain a 45 degree axis differential in the measuring methods?
ANSWER:
I have to admit, though, as I get deeper into this problem, that I have
some reservations about my answers. Most modern balls use an inner core
made of
polybutadiene rubber which is what superballs are made of. It turns out that
this rubber with a bulk modulus K =1.5-2 GPa is nearly as
incompressible as water with K =2.2
GPa. Assuming your buoyancy results are reproducable, I am puzzled. My final conclusion would have to be that if the ball is very close to homogenous, it is impossible to make a really accurate measurement of the plane in which the CoM resides but not so hard to find the line on which it resides.
There are lots
of videos on youtube about this spinning device compared to the
saltwater method. Most are promos for a golfball called RealLine .
There is one
with a result similar to your results.
One shows a kid using it to mark a ball and the line is clearly off
the equatorial position (i.e. would not cut the ball in equal
halves). If you read the
comments on the spinner on Amazon, there are some enlightening
comments on what might go wrong. Unfortunately, I could not find any
comprehensive comparisons of the two methods like you have done.
QUESTION:
ANSWER:
QUESTION:
ANSWER: …outputs more Force…" does not really
mean anything. During the collision, Newton's third law requires that
the force the train exerts on the man must be equal and opposite to the
force the man exerts on the train. However, you could imagine the man
pushing with his very strong arm during the collision time such that he
adds a certain amount of energy to the system; this would result in a
different situation from if he just stood there. Nevertheless, Newton's
third law would still apply. The extremes of his just standing there are
The speeds of
each may be found easily and are classic introductory physics problems.
Because the mass of the train is much larger than the mass of the man,
the results are pretty simple:
For the
inelastic collision, the man and train continue with approximately
the speed that the train came in.
For the
elastic collision, the train continues at approximately the speed it
came in with and the man proceeds in the same direction with
approximately twice that speed.
If the collsions last some short
time t , the average forces on the man during that time will be
approximately mv /t for the inelastic collision and 2mv /t
for the elastic collision; of course, the forces on the train will
be the same.
Now suppose
the man (mass m ) adds just the right amount of energy to
stop the train (mass M ) coming in with speed v ; I
assume that there is no other energy lost or gained. Then the man's
speed after the collision would be equal to (M /m )v ,
a huge speed! In this case the average force on the man would be Mv /t ;
he will have to be amazingly strong to endure this! The energy he would
have to add to stop the train would be approximately (M /m ) times the
energy the train came in with.
QUESTION:
ANSWER: Hawking
radiation .
QUESTION:
ANSWER: mg down, and the normal force
which the bucket exerts on the water, N . So, Newton's second
law is mg+N=mv 2 /R where v is the
speed and R is the length of the rope; I have chosen down as
the positive direction. Solving, N=m [v 2 /R )-g ].
Now note that if v <√(gR ) then N is
negative which means that it would be a vector which points up since I
chose down as positive. But the bucket is unable to exert a force up on
the water and so the water would not stay in the bucket if going too
slowly.
QUESTION:
ANSWER: m =850 lb=390 kg, s =32
ft=9.8 m. I will use momentum conservation for the collision, 360v =390u
where v is the incoming speed of the motorcycle and u
is the outgoing speed of the 4-wheeler; so u =0.92v . As
the 4-wheeler slides, the friction of the brakes does work which takes
the energy of the 4-wheeler away. The energy is K = ½x390xu 2 =½x390x(0.92)2 v 2 =170v 2
J; the work done by the friction is W=-μmgs =-0.9x390x9.8x9.8=-34,000
J where μ= 0.9 is the approximate coefficient of static
friction of tires on dry asphalt and g =9.8 m/s2 is
the acceleration due to gravity. So, K+W =0=170v 2 -34,000
or v =14 m/s=31 mph.
it!
FOLLOWUP
QUESTION:
31 mph? That's it, really?
ANSWER:
QUESTION: 2 which is calculated using the masses of earth and the "falling" body and Newton's Law. Similarly I read that if I go and stand on the moon the "acceleration due to gravity" will be different because the masses are different? Again seems to be clear.
ANSWER: M and m , are separated
by a distance r . M exerts a force F mM
on m and m exerts a force F Mm
on M ; because of Newton's third law, these forces are equal and
opposite, F mM =-F Mm .
Because of Newton's law of universal gravitation the forces have
magnitude F Mn =F mM ≡F=mMG /r 2 .
Then, using Newton's second law, each mass will have an acceleration
independent of its own mass of a =[mMG /r 2 ]/m=MG/r2
and A =[mMG /r 2 ]/M=mG/r2 .
Note that I have been viewing this from outside the system; this frame
of reference is is called an inertial frame of reference. It is
important that we view the system from an inertial frame, because
otherwise Newton's laws are not correct.
So, what you
have been taught is correct only if you view things from outside the
two-body system. But what you have also probably been taught is that you
measure this constant acceleration relative to the surface of the earth
and that is technically incorrect because the earth is accelerating up
to meet the falling mass and is therefore not an inertial frame.
However, the earth's acceleration is extremely tiny, too small to
measure. If M is much much larger than m , which is
certainly the case for the earth and the moon, what you have been taught
is, for all intents and purposes, correct.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: K = ½mv 2 where m
is the mass and v is the speed. For example, if you have a mass
of 60 kg (about 130 lb) and are in an airplane going 500 mph=224 m/s,
your kinetic energy relative to the ground is K =½·60·2242 =1.5x106
Joules, a million and a half Joules! How did you get that energy?
During the takeoff the airplane pushed on you with some force to speed
you up and give you that energy. Suppose that the time to speed you up
to that speed was 10 minutes=600 seconds; then the power that the plane
had to deliver over that time would have been 1.5x106 /600
J/s=2,500 Watts. The power delivered by the airplane to you would be
enough to power 25 100 Watt light bulbs. During the 10 minutes, the
airplane would have been pushing on you with a force of about 22.4
Newtons=5 lb; that is not much of an effect on you. But suppose that you
had accelerated to 500 mph in 1 second instead of 600; then the force on
you would have been 13,400 N=3000 lb which would have crushed you. So
you see, it is the force when you are acquiring your kinetic energy
which "affects" you, not the kinetic energy itself.
QUESTION:
Where happens all the matter that enters a black hole?
Is all the matter in a black hole crushed into a "singularity"
What are the current theories you suggest I study?
If a black hole can consume matter without limits, could a super massive
black hole also consume the mass our universe?
So, could the "Big Bang" be from this "singularity", a singularity that
holds all the mass of our universe?
(This makes sense to my way of thinking and boggles my mind.)
What are White Holes? Have they been proven to exist yet?
Or like "worm holes" are they only theoretical today?
And does current thinking say White Hole evolve from Black Holes?
ANSWER: —almost nobody
does, even for my best, most creative answers. I do, however, recommend
that people wait for an answer before donating unless you like my site
so much that you just want to support it and I greatly appreciate that!
(Also, I have no way to "refund" contributions.) In your case, you did not read enough
on the site before submitting your
question(s). Two quotes from the site: "If your question is clearly astronomy or astrophysics, particularly detailed questions about black holes, stellar evolution, dark matter or dark energy, the big bang,
etc ., areas in which I am not expert, I may not answer" and "Please
submit single, concise, well-focused questions". Because of your contribution, I
will cut you a little slack this time.
The simplest idea of a
black hole is an object with mass, charge, and angular momentum
which has infinite density and where time has stopped. I believe the
latest thinking is that it is not really of infinite density.
Predictions of singularities are from classical general relativity;
if you view the situation quantum mechanically, then uncertertainty
ideas do not want zero size or stopped time because spacetime itself
should have a granularity. A discussion which you might find
accessible is at
Physics Forums .
This depends on how
things happen to be moving around. You could imagine a much older
universe composed of nothing but black holes and the radiation they
emit when they undergo
Hawking
radiation but moving in such a way that they will never
encounter each other.
There is a very
interesting book by Lee Smolin,
Time Reborn ,
where he reprises an idea originally by Wheeler and deWitt that
black holes are the seeds of new universes. There is a nice
interview with Smolin on
space.com and a
lecture by
him that you can watch.
White holes are
hypothetical and have never been observed. I know nothing about
them.
I hope I have earned your
donation!
QUESTION:
ANSWER:
QUESTION:
ANSWER: T has a component perpendicular to r
(T sinθ ) and therefore exerts a torque
τ=rT sinθ and therefore angular momentum is
not conserved; this does not make it a "fallacy", it just does not hold
for this particular problem. What is conserved, though, is energy.
Because the displacement (along the velocity vector) is always
perpendicular to the tension, the tension does no work so energy is
conserved, ½mv 1 2 =½mv 2 2
or v 1 =v 2 , the velocity stays
constant. So, if the m =50 g=0.05 kg marble starts at about 1 m
away from your finger with a frequency of ω =1 rev/s=2π
radians/s, its speed is v ≈ωL= 2π
m/s; then when L =5 cm=0.05 m, T=mv 2 /L =0.05x4π 2 /0.05=39.5
N=4.0 kg-force.
THE
PHYSICIST : link here to the
Off-the-Wall Hall of Fame .
QUESTION:
ANSWER: work function , the latter is
called the ionization potential . The idea is that when an
electron is removed from the metal, one positively charged atom is left
behind resulting in an attractive force trying to hold the electron in.
A point charge in front of a plane ideal conductor is a classic
electrostatics problem, usually solved using the
method
of images . If you then integrate from the size of an atom (~10-10
m) to infinity you can get a ball-park estimate of the work function,
W ≈3.6 eV.
QUESTION:
ANSWER:
QUESTION:
ANSWER: p was, indeed, defined as mv before
the advent of the theory of special relativity. If v is much
less than c , the speed of light, this is an excellent
approximation, but not exactly true. The correct expression for p
is m 0 v / √[1-(v 2 /c 2 )]
where m 0 is the mass of the object when at rest. The
relation among energy E , momentum p , and rest mass
m 0 is E 2 =p 2 c 2 +m0 2 c 4 .
So, you see, even if a particle has zero mass (like the photon) it still
has momentum if it has energy, p=E /c . I am also often
asked how a photon can have energy because E=mc 2 and
m =0 for a photon. You can find links on the
faq page to answers which
discuss this question.
QUESTION:
ANSWER: ω
about this point is v /R where R is the radius
of the wheel and v is the speed which the center of the wheel
is moving forward with speed v . Now, the point on the rim also
has the same angular velocity ω , but its distance from
the axis of rotation is R' . Because the angle which R'
makes with the ground is 45 º, it is easy to show that
R'=R √2. Therefore,
ω=v /R=v' /R'=v' /(R √2),
and so v'=v √2; the direction of the velocity
v'
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: earlier answer .
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: earlier answer . Also, the sound barrier
is irrelevant because there is no sound in space. Although not really
related to your main question, if you jump vertically upward in a
rotating coordinate system (like the earth spinning on its axis) you
will indeed not land exactly where you launched but the difference is
very small.
QUESTION:
ANSWER: trampoline
model " which illustrates (in a simplistic, not literal way) how
warping the space (the surface of a trampoline) by a massive object
(bowling ball) will cause a light object (marble) to be attracted to the
massive object.
QUESTION:
ANSWER: 2 ,
density of water is 1000 kg/m3 . I find that the total volume
to fall is about 2x1011 m3 , the mass is 2x1014
kg, and the potential energy of this mass is about 3x1016 J. If
you deliver that much energy over the course of a day, the average power
is about 350 GW. So, the power plants will only reduce the energy of the
tidal flow by less than 0.01%, a truly negligible amount.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
hyperphysics web site. The answer is about 42 minutes. However, as
discussed in an earlier answer , the earth is
by no means a uniform sphere so the answer would be different but not
easy to calculate.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
FOLLOWUP
QUESTION : I didn't realize I had posed this in the form of a homework-ish question. I'm a 47yo truck driver, tired of hearing all the lame-@$$ rhetoric of the motoring public. Long, quiet, empty night highway helped me come at this from a different angle. Without too much detail, one truck wheel wrangles slightly more than a single car's worth of mass; is why trucks do NOT stop better despite having more wheels.
ANSWER: μ , the maximum force on a level surface
is μW . where W is the weight of the vehicle. The
distance s traveled will be determined by the work done by this
force, μWs , which will equal the initial kinetic
energy, ½Mv 2 =½(W /g )v 2 ,
where M is the mass and g =32 ft/s2 . So,
s =(1/(2μg )v 2 . So here is the
interesting thing: the distance traveled does not depend on the weight
of the vehicle, only by the initial speed and the condition of the road.
The force applied does depend on the weight and the force per wheel
would be 0.7x4000/4=700 lb for the car and 0.7x80000/18=3111 lb; but the
larger force is simply because of the larger weight and the only thing
which is important is the distance to stop. Of course, this also depends
on whether you have antilock brakes which let you get the maximum amount
of friction when you are just on the verge of skidding; if you lock the
brakes and you skid, you go farther before stopping. For your example,
v =65 mph=95 ft/s and μ is approximately 0.7 for
tires on a dry road, so s ≈200 ft. I have neglected all
other forms of friction like rolling friction and air drag, but these
should be relatively unimportant. Bottom line—stopping distance
does not depend on number of tires nor on weight.
QUESTION:
ANSWER: A and the thickness is t , the volume is
V=At . Of course, you have to be careful to have A and
t measured in the same units. V is the volume
regardless of how you change its shape. Since you apparently want your
answer to be in cubic yards, you should probably convert everything to
yards at the outset. For the example you state, A =(100 ft2 )(1
yd/3 ft)2 =100/9 yd2 and t =(1 in)(1 yd/36
in)=1/36 yd. So V =(100/9)(1/36)=0.309 yd3 . You can
get more detailed if you know the length L and width W
of the rug so A=LW and V=LWt . If you roll it up to a
cylinder of length L , its volume will be V= πR 2 L =LWt
where R is the radius of the cylinder; therefore, R =Wt /π ).
So, if your rug is 10x10 ft2 , it will roll up with a radius
of R =√[(10/3)(1/36)/3.14]=0.172 yd=0.52 ft. All this
assumes that the rug does not compress when you roll it up.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: earlier question is
very similar to yours and you should read that answer to get more
detail.
QUESTION:
ANSWER: ≈1400
m/s. The drag force on an object with speed v in water may be
approximated as F drag =½CρAv 2 ;
here I will choose C≈ 1 (order of magnitude for most
shapes) is the drag coefficient which depends only on the shape, ρ ≈1000
kg/m3 is the density of water, and A ≈1 cm2 ≈10-4
m2 is the cross sectional area of the object. I want this
object to go as far as possible, so I have chosen A to be
relatively small; an object this small will have a relatively small
mass, certainly no larger than m ≈100 gm=0.1 kg. Knowing
all this, one can calculate the velocity v and position x
as functions of time t , v =v 0 /(1+kt )
and x =(v 0 /k )ln(1+kt )
where k =½CρAv 0 /m≈ 7000
s -1 and v 0 =1400 m/s is the
starting velocity. I find that at t =½ s, v≈ 0.4
m/s and x≈ 1.6 m; the object will have lost nearly all its
speed after having traveled only about a meter and a half. I can think
of no earthly way that any debris could propogate 2 km!
QUESTION:
ANSWER: r 2
where r is the distance from the center of the earth. If R= 6.4x106
m is the radius of the earth, W is your weight at the surface,
and W+ is your weight 1 mile=1.6x103 m above the
surface, then W + /W =R 2 /r 2 =0.9995;
your weight would be about 0.05% smaller. If you assume that the mass of
the earth is uniformly distributed throughout its entire volume, the
gravitational force falls off like r as you go toward the
center. Then W - /W =r /R= 0.99975;
your weight would be about 0.025% smaller. Given other factors like
local variations in the density of the surrounding earth, these would be
unmeasurably small; one mile (about 1600 m) is, after all, extremely
tiny realtive to the size of the earth. I should also note that
approximating the earth's density as uniform is a very poor
approximation; see an
earlier answer . If you had asked your weight 100 miles below the
surface the answer would be that your weight is nearly the same as at
the surface.
QUESTION:
ANSWER: independent measure of the distance of those
objects. When the two separate measurements were compared, it was found
that the velocity of the objects was a linear function of their
distances. The graph shows an example where distance was determined
using the brightness of type Ia supernovae. I do not understand your
question beginning with " …With Doppler the farther…"
QUESTION:
ANSWER: escape velocity ), it will never come back.
And, if you throw it with a velocity greater than the escape velocity,
it will never stop going. So, the ultimate fate of the universe is that
it will either expand forever or everything will turn around and fall
back together (the big crunch ). All that is out of date,
though, because it was discovered almost two decades ago that gravity is
not purely attractive or else there is some other force which is
operative as well (called
dark energy );
this results in the fact that the distant objects in the universe are
not just moving away from us, they are actually speeding up. This
repulsive force is not yet fully understood and the ultimate fate of the
universe is not known.
QUESTION:
ANSWER: M =2000 kg of
mercury. The density of mercury is ρ =13,600 kg/m3 ,
so the volume you must displace is V=M /ρ =0.15 m3 =150
l; this, I presume, is what is bothering you since only 28 l are used.
Suppose that the reservoir for the mercury is a cylinder of radius 1 m
and depth d ; to contain 0.15 m3 , the depth of the
container would have to be d =0.05 m=5 cm (estimating
π ≈3). So, I will make a container 6 cm high for an
extra 20%, 180 l and I will buy that 180 l to fill it up. Now, let's
make the pedestal on which the lens sits be a solid cylinder of radius
99 cm so that when you put it into the reservoir there will be 1 cm gap
all around. So as you lower it into the reservoir, mercury will spill
out the top and you will be sure to capture it. When you have captured
150 l, the whole thing will be floating on the mercury. You return the
extra 150 l and have floated the lens with only 180-150=30 l. Of course,
in the real world you would only buy 30 l, put the pedestal into the
empty reservoir, and add mercury until it floats. (I realize that the
shape of the pedestal and reservoir are probably not full cylinders,
since you said "trough", but my simple example wouldn't be so simple
with more complicated volumes. The idea is the same, though.)
QUESTION:
ANSWER:
QUESTION:
ANSWER: If the earth were
a cube… )
QUESTION:
ANSWER: m a distance R from the center of
the earth feels a force F=mΦ where Φ=GM/R 2
is the gravitational field and M is the mass of the earth.
But Newton's second law tells us that F=ma where a is
the acceleration of m ; therefore a =Φ
regardless what m is. All this assumes that the earth, which
also feels the same force as m , has a mass much larger than the
object, M>>m so that the acceleration of the earth is
negligible. If m were comparable to or larger than M ,
the earth would accelerate toward the object so the two would meet
earlier; but the the heavy object's acceleration would be just the same
as the feather's even in this scenario. You are right that the stone
feels more force but it also has a greater inertia which results in
identical accelerations.
QUESTION: X =Kx and the tension in Y is given by TY = 3Ky, where
K is a constant and x and y are the extensions of X and Y respectively. At what distance from A must a body of weight 5W be attached to the rod if the rod is to be horizontal?
ANSWER: T then the tension is Y is 3T . W acts at a
distance 0.1 m from X and 5W acts at a distance d from
X. You now need to apply the equilibrium conditions (Newton's first
law). The sum of the forces is zero, 4T -6W =0, so T =1.5W .
The sum of the torques about the left end is zero, (3x0.2)T -0.1W -5dW =0=0.8W -5dW
or d =0.16 m. (The solution which you attached, not shown here,
got the relative tensions in the springs correct, but you never applied
Newton's first law and never thought about torques at all.)
QUESTION:
ANSWER:
QUESTION:
ANSWER:
μ , the weight is W , and the angle of the incline is
θ . Then
the force to pull it with constant speed up the incline is F incline =W (sinθ+μ cosθ )
and to pull it along the floor is F floor =μW .
If there were no friction (like if the weight were on very well
lubricated wheels) and your 1000 lb weight were pulled up a 2.86º
(5%) slope, F floor =0 and F incline =49.9
lb. With μ= 0.5, F floor =500 lb and F incline =549
lb. The reason F incline is bigger is that, in
addition to pulling the mass in a horizontal direction, you are also
lifting it.
QUESTION:
ANSWER: E=mc 2
and is not taught in elementary chemistry courses. And, you should avoid
the notion that if you must add energy to make the reaction go it is
endothermic because you might get more energy out than you put in.
Fusion is such a case: in order to fuse two deuterons (2 H) to
one alpha (4 He) you must overcome their mutual Coulomb
repulsion; but if you compare the masses before and after the fusion,
you will find that their is much less after (an alpha is much more
tightly bound than a deuteron). Similarly, if you compare the masses
before and after a fission of a heavy nucleus, the total mass afterward
is measurably smaller. Fusion (fission) is exothermic for nuclei lighter
(heavier) than iron. That is why stars do not make elements heavier than
iron; heavier elements are created in supernova explosions or other
exotic astronomical events. You might have a look at an
earlier answer .
QUESTION:
ANSWER: atomic masses. The 14 N atom
left after the decay has only 6 electrons, not the 7 in a 14 N
atom, so you do not add the mass energy of the β -
(or, if you wish, you add it and subtract the mass energy of the
missing electron).
QUESTION:
ANSWER:
F which the neck would have to exert on an object of weight W
(lb) if the customer simply dropped the object from a height h
(ft) and then stopped it completly in a distance s (in): F =W [1+1.06 √(h /s 2 )].
For example, if a 10 lb ham dropped from 2 ft and stopped in 2 in, F =17.5
lb. That would be a pretty extreme case, though, since I would guess
most people would probably lower it at a lower speed than it would
achieve in free falling 2 ft. I think engineers like to insert a factor
of 2 safety factor. Overall, I would guess that the tag should to be
able to handle at least twice the weight of the product.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: parallax . For more distant
objects, it gets a bit harder and we are getting into the realm of
astronomy in which I am not an expert. By measuring the
redshift of the
spectra of stars we can deduce the distance if we know the
Hubble constant
because the farther away an object is from us, the faster it is going.
Another way to determine distance is to have a certain type of
astronomical object which has the same total lumninosity regardless of
where it is. Such an object is called a
standard candle
and an example is a type Ia supernova; knowing how bright something is
and how bright it looks can be used to determine how distant it is.
QUESTION:
ANSWER: -6 kg. If the
collision lasts 10 μs=10-5 s and the car is going
25 m/s, the mosquito felt an average force of 5 N; the car will feel a
backward force of 5 N and you know a mosquito could not lift a ½
kg mass!
QUESTION:
ANSWER: Interstellar ,
the time dilation is called gravitational time dilation which is a
feature of general relativity, not special relativity; here time runs more slowly because of a strong
gravitational field. In the movie a planet orbits close to a spinning
black hole and time is slowed. It is a bit of a stretch, though. See an
interview with
Kip Thorne who was the science advisor of the movie. The other kind
of time dilation is easier to understand. First, you need to understand
why moving clocks run slow. The best way to understand this is to use
the
"light clock" as an example which I have done in many earlier
answers. In order to find the light clock useful, you must accept that
the speed of light is the same in all reference frames; see my
faq
page for answers which address this. Finally, the twin paradox is
easy to understand; again, see an
earlier answer . Actually, the twin paradox relies more on length
contraction than time dilation to understand because the traveling twin
sees the distance to the distant star as smaller than the earth-bound
twin does although both agree on the traveling twin's speed.
QUESTION:
ANSWER: Wikepedia article .
QUESTION:
ANSWER: h above the ground and L to the right of the
near side of the object. The
smallest force F you can apply is at the top, H above the ground (31 ft
for you, but if you want to solve for F at a different height, just use
a smaller H in the final equation I will give you). The weight
of the object is W , the normal force is N and the
frictional force is f . The object, just about to tip, has N
acting at the right hand lower edge of the object, so N and
f are acting there. Newton's first law tells you that N=W
and f=F . Finally, sum the torques about the right hand lower
edge: FH-W (d-L )=0 or F =W (d-L )/H .
For your case, if the weight is uniformly distributed, i.e .
h=H /2 and L=d /2, then F=Wd /(2H )=22x12/(2x31)=4.26
tons. You must push it until the center of gravity is over the right
hand lower edge and from there it will fall on its own. Also note that
if the floor is too slippery it will slide before it tips.
QUESTION:
ANSWER:
Mythbusters
episode which perfectly answers your question.
QUESTION:
ANSWER: earlier answer where I
examine the problem of whether you could play catch in such an
environment. Another
earlier answer addresses what happens if you jump "straight up".
Perhaps the most important thing to take from those answers is that this
is not really like gravity at all because the instant you lose contact
with the surface you experience no forces at all so you move in a
straight line; nevertheless, with a large enough radius, the behavior of
the path as viewed by an observer on the surface can be quite similar to
how it would be on earth, particularly for the straight-up jump. Also we
will need to know that the angular velocity ω for the centripetal
acceleration of a cylinder of radius r 0 to be g =9.8
m/s2 is ω =√(g /r 0
Now, why does a hot-air baloon
rise? It is because of the buoyant force and the buoyant force arises
because the pressure on the bottom of the balloon is larger than the
pressure on the top. So the first question we need to answer is whether
there is a pressure gradient in the space station (i.e. pressure
decreases with distance from the surface) and how that compares with the
pressure gradient in a gravitational field. In the space station there
is certainly a gradient because (as you note) it is simply a giant
centrifuge so air will tend to move to larger radii. For an
incompressible fluid there is a centrifuge equation for pressure a
distance r from the center, P (r )=P 0 -½ρω 2 (r 0 2 -r 2 )=P 0 -½(ρg /r 0 )(r 0 2 -r 2 );
here r 0 is the distance from the center to the outer
surface, P 0 =P (r 0 ), and
ρ is the fluid density. Technically, this is not correct
because the density varies with r , but we are really only
interested in situations where the density is fairly constant; for
example, if r 0 =1000 m, r =900 m, and P 0 =105
N/m2 (approximately atmospheric), P (r )=0.99x105
N/m2 . In a uniform gravitational field there is an
empirical equation to calculate pressure as a function of altitude
h , P (h )=P 0 [1-(Lh /T 0 )]gM /(RL )P (h =100
m)=0.98x105 N/m2 ; we can therefore conclude that
the buoyant force in the space station will be very similar to that on
earth, at least for the first few hundred meters.
I want to look at what the balloon
does from two perspectives—an observer outside the space station
and the guy inside who releases the balloon.
From outside, the instant that
the balloon is released it experiences no force except the buoyant
force B v =r 0 ω =√(g r 0 B B v v B
When looking from inside the
rotating cylinder there are two fictitious forces (see
earlier answer for
more detail), the centrifugal force F r =mg
and the Coriolis force F c =2mv √(g /r ),
in addition to B
F c will get larger and change
direction as v turns to the right and gets larger and r
decreases. So, if there were a ceiling (space station like a torus)
it would definitely not strike directly above the launch point.r 0 were large enough, the effect would be
minimal because F c would
never get large enough to cause a significant deflection because of
the 1/√r term.
Note that I have not considered another possible
force, air drag. I believe this would have a small effect and would not
alter the conclusions above. You were right to speculate that the result
would be different from on earth but angular momentum has nothing to do
with it; it is simply due to the Coriolis force.
FOLLOWUP
QUESTION: What would happen if an observer got into a car and accelerated beyond the rotational velocity of the
cylinder in the same direction, once the vehicle lifted from the surface? Would it then glide for a certain period until, from the perspective of another observer inside the centrifuge, descending back onto the surface at a slightly slower speed? Also, the second part to
the question is, would the driver experience weightlessness if he was slowed enough while airborne? Would any airborne object, for that matter, potentially experience weightlessness if slowed or misdirected?
ANSWER: u
in the same direction as the cylinder is rotating as you stipulated.
There is one real force acting on
the car, the normal force N F r
and the Coriolis F c , both pointing
radially outward (the Coriolis force is F c =2mu xω
Now the Coriolis force points
radially inward, so there will be a speed where N =0. Now, it gets a little tricky to
understand what happens as u increases. The first thing you
have to understand is that the car, in the frame of the rotating
cylinder, is going around the circle; so the sum of all the forces must
add to the centripetal force: mu 2 /r 0 =N-mg +2mu √(g /r 0 ).
So, when N =0, u 2 +gr 0 -2uv =0=u 2 -uv.
So, when u=v , N =0. This makes total sense because
if you look at the car from outside, it is at rest and, since this is
taking place in empty space, there are no forces whatever on the car.
In this situation, any small force with a radial component could launch
the car and, as seen from outside, it would simply move in a straight
line with constant speed (no forces on it once it loses contact with the
wall) until it encountered the wall; the car and all its passengers are
weightless. But there will never be a "lift off" situation;
if you view the situation from from the inertial (outside) frame, for
any u≠v the car will be traveling in a circle and therefore
require N ≠0.
I must confess that the fact that
all the forces on the car must add to the centripetal force mu 2 /r 0
eluded me at first. If you set N-F r +F c =0
you find u =½v for N =0 and this just
did not seem right to me. I therefore posted a question on
Physics Forums and got a great answer from Simon Bridge which
set me straight.
FOLLOWUP
QUESTION: This is my final question about being airborne in a large atmospheric centrifuge in space: If you decided to locate yourself in the very center of the sphere, having no rotation, and you were then pushed toward the rotational equator at a walking pace, what would happen to you before you reached the surface, or rather how do you figure it out? Toward the surface, wind current would be rotating in the centrifugal direction, and the resistance would surely begin to alter your trajectory from "vertically, straight down" to "diagonally," from an observer on the equator, on a curve toward the surface. Assuming that the sphere is large enough that the accelerating effect of wind has enough time, would you ultimately be able to meet the surface on a landing curve that merges into the circumferential curve? And, if you accelerate too much, would you skip like a stone or a soccer ball as you collide with surface, slowing each time until you reach the rotational velocity? Ideally, you would be wearing a protective suit.
(Note from The Physicist: in earlier parts of this discussion I have
edited out "sphere" in the questions many times and replaced with
"cylinder" because a sphere is not a good model for pseudo gravity
because only at the equator of a sphere would the apparent gravity mimic
earth's. My answer below replaces "sphere" by "cylinder" and "center" by
"axis")
ANSWER: v =√(gr o )≈100
m/s=≈220 mph; this would be be the "wind speed" at the rim as
seen by an observer not rotating with the cylinder, including you
starting at the center; this speed would decrease linearly to zero on
the axis. It is easiest to view your gedanken from outside, but
I will discuss both observers. I will also discuss two scenarios, with
and without air.
If there were no air in the cylinder:
If viewed from outside, you would move
in a straight line away from the axis with a constant velocity
2.2 m/s until you hit the rim (in about 1000/2.2=455 s=7.6 min)
moving tangentially relative to you with speed 100 m/s—bad
news for you! If viewed from inside, the centrifugal force would would be
radially out and the Coriolis force would push you azimuthally
in the direction opposite the direction the cylinder rotates.
You would still hit the rim with a relative tangential speed of
100 m/s and a radial speed of 2.2 m/s (which implies your
landing speed in the rotating cylinder frame is about 100 m/s)
after 7.6 minutes. Since the period of the cylinder's rotation
is about 63 s (T =2πr 0 /v ),
your path would spiral around 7.2 times before landing.The paths
might look something like the picture below, the blue being the
outside view and the black the inside view.
But, there
is air, so you would have two real forces acting on you, the buoyant
force acting toward the axis and the air drag acting opposite your
velocity. Earlier in this discussion it was concluded that the
buoyant force would be comparable to that on earth, and you
certainly would not expect the buoyant force to have much effect on
you here; it would be a very small force pointing toward the axis
causing you to slightly slow down vertically. But the wind is a
different story.
If
viewed from outside, as you started falling air drag would be
very small because the wind speed would be small, your speed is
small, and because the air density would be low. It is too difficult to try to calculate
this with any precision, but clearly what is going to happen is
that the outside observer would see you spiraling out from the
axis in the same direction the cylinder was rotating. Given the
large speeds near the end (more than a category 4 hurricane), it is indeed likely that you would
land with a speed close to the speed of the rim; this would be a
much better scenario on landing than if there were no air. You
might think that the
force due to 220 mph winds would hurt you, but as you
acquire more and more tangential speed, the wind speed you see
becomes smaller and smaller. The total fall time would be much
longer than 455 s because of the large tangential velocity you
acquire; the average acceleration over the whole trip would be
much less than 100/455=0.22 m/s2 =g /45
because the time is much longer.
If viewed from the inside,
the centrifugal force would
give you an acceleration g radially out (toward the
rim). Early in the
descent the Coriolis force would accelerate you in the direction
opposite the direction of rotation; as the velocity acquired an
azimuthal component, the Coriolis force would have a radial component
radially inward, opposite the centrifugal force. The air drag
would always act opposite the velocity vector, so you would
never reverse the direction (you never rotate with the
cylinder). It is interesting that the inside observer sees you
going against the rotation but the outside observer sees you
going with the rotation; this would mean that your angular
velocity is always smaller in magnitude than the angular
velocity of the space station.
QUESTION:
ANSWER:
earlier answer .
QUESTION:
ANSWER:
QUESTION:
VIDEO
ANSWER: down ; this is due to Newton's third law.
You can conclude that all the forces which the four exert on each other
cancel out if you take the ensemble of 4 as the object to focus on. As
long an nobody tips over, nothing will happen.
QUESTION:
ANSWER: d 2 where d is the
distance from the source. Eventually, the intensity will drop below the
sensitivity of the ear. At some intensity the medium will no longer be
able to support the sound wave, but when that happened would be
complicated to understand and depend on the medium properties and the
frequency of the sound.
QUESTION:
ANSWER: τ motor ;
if the radius of the gear or pulley attached to the motor shaft is r,
this results in a tension in the chain or belt of F =τ motor /r.
Now, if the gear or pulley on the drive shaft of the conveyor is R,
the resulting torque on that shaft will be τ conveyor =FR =τ motor (R /r ).
So, yes, increasing the size of the drive gear increases the torque.
This is the physics of an ideal situation. I am not an engineer!
QUESTION:
ANSWER: R 1 at time
t 1 and a position R 2
at time t 2 , then the displacement vector
D is defined as D ≡R 2 -R 1
and average velocity vector is defined as v avg ≡D t 2 -t 1 ).
As you can see, the average velocity and the displacement are in the
same direction. To get instantaneous velocity you find the limit of the
average velocity as the time difference approaches zero. If you know
calculus, the instantaneous velocity is v R t ;
again, velocity v R
FOLLOWUP QUESTION:
ANSWER:
You are correct but your example is
meaningless. It makes no sense to compare the displacement over two
different times with the instantaneous velocity at the second time—you
could get any answer. Read my first answer carefully. The relationship
between velocity and the corresponding displacement over any time
interval is clearly defined, even in the limit that the time interval
shrinks to zero.
QUESTION:
ANSWER: microphones .
QUESTION:
ANSWER: general relativity linked to
on the faq page. Incidentally, only photons inside the
Schwartzchild radius R= 2GM /c 2
will be captured.
QUESTION:
ANSWER: f =5
revolutions/s. If the radius of the circle is R =3 ft, the
velocity is v =2 πRf= 2x3.14x3x5=94 ft/s=64
mph.
QUESTION:
ANSWER: —like in the vicinity of
a black hole or neutron star.
QUESTION:
ANSWER:
QUESTION:
I'm not asking you to solve this. I did it myself.
But the physical meaning eludes me.
[Here the questioner adds a bunch of mathematics discussion which is not
really relevant to the physics, but the crux of what eludes him is,
since this is a quadratic equation, what are the physical meanings of
the two solutions and does their average have any physical
significance.]
ANSWER:
x on the same side of the equation as the
kinetic energy term?
The equation clearly is intended
to represent modeling the problem of a m =46 kg object (child)
moving with a speed v= 2.4 m/s when landing on an ideal spring
with spring constant k =1.94x104 N/m. This is a
standard introductory physics energy conservation problem. If you choose
y =0 at the top of the spring and the +y direction as
vertically up, the energy conservation equation is ½mv 2 =½ky 2 +mgy
where y is the position(s) where the mass is at rest. This
says that the kinetic energy of m when y= 0 (initial
energy) equals the potential energy of the spring plus the gravitational
potential energy of the mass when the object is at rest. Note that this
equation assumes that the mass is always attached to the spring and so
the two solutions are how far the mass travels in the negative direction
(y <0 solution) and how far it rebounds in the positive
direction (y >0 solution). These two solutions are the extremes
of the resulting oscillation of the mass and the oscillations are
symmetrically about the location where the mass would be at rest if not
moving, y rest =-mg /k . Since this is
the center of the oscillation, ½(y 1 +y 2 )=y rest .
The solutions for the problem at hand are y 2 =+0.096
m, y 1 =-0.142 m, y rest =-0.023 m;
note that ½(y 1 +y 2 )=y rest .
If you solve the equation in the problem as stated, you find that the
positive solution is greater in magnitude than the negative solution and
that the average is not the equilibrium condition. I believe that what
the author intended was that the stack of mattresses was 2.8 m high, the
floor was chosen as the zero of gravitational potential energy, and the
final gravitational potential energy was moved to the left side for
compactness. This would mean that the final potential energy assumed by
the author was mg (-x ) and therefore x must
mean the distance that the mass went down. But that is wrong—the
final potential energy should have been mg (2.80-x ).
This would then reduce identically to my equation except that x =0
at the position of the unstretched spring and the +x direction
is vertically down. The problem as stated is incorrect.
QUESTION:
ANSWER:
h =v 2 /(2g ),
would be about 68 ft. If air drag is included, I estimate that the
terminal velocity (see earlier answer ) would be
about 262 mph. Since this is so much larger than 45 mph, I judge that
air drag would not be important and 68 ft would be your answer.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
2 .
But, to discover how convoluted the measurement of intensity can be, see
an earlier answser .
QUESTION:
ANSWER:
pack
together. As you can see by carefully examining the figures above, each
orange is pushed down by three oranges from above, is pushed up by three
forces from oranges below, and is pushed horizontally by six oranges in
the same layer. These twelve forces all are directed toward the center
of the orange and all add up to zero net force, but there is a
net pressure over the surface of each orange approximately trying to
squeeze it into a smaller orange. But an orange is mostly water which is
nearly incompressible so the orange does not get crushed. Think of a
nicely packed snow ball: if you try to crush it into a smaller ball by
squeezing with cupped hands you will fail; to crush it you will have to
flatten it by pushing it with diametrically opposed hands. You might
think the bottom-most layer would get crushed because there is just one
force pushing up; but, each orange has six others in the same layer
pushing toward its center and these keep it from getting flattened.
Here is a little more about sphere packing. There are two possible
packings which achieve the maximum density of π /(2√3)≈74%:
hexagonal close-packed (HCP) face-centered cubic (FCC); these are
compared in the figure above (HCP on the left).
FOLLOWUP
QUESTION: What would happen to the oranges once the force exerted upon them reached a critical strength which they couldn't bear? Does the entire group of oranges burst simultaneously in a flood of juice?
ANSWER:
QUESTION:
ANSWER:
The simplest
place to start is your first question: does the mass of the rotor have an
effect on the torque on it? Typically, the turbine has three blades. I
will just analyze a single one and the same arguments could be made for
the other two. Call the length of the blade L and assume that
the force on it due to the wind is approximately uniform along the
length of the blade (the force on a tiny piece of the blade near the
center is the same as the force on an identical tiny piece near the
end). Then the total force F due to the wind will depend on the
length of the blade, but the force per unit length, Φ=F /L
will be more useful because it will depend only on how hard the wind is
blowing. It is now pretty easy to show that the torque due to the wind
is τ wind =½ΦL 2 .
So, the answer to your question is no, mass does not affect the
torque; the torque depends only on how hard the wind is blowing and how
long the blade is.
Your second question is how can
you decrease speed by changing the mass M . If I model the blade
as a uniform thin stick of length L , its moment of inertia is
I=ML 2 /3. If it has an angular velocity ω 1 ,
its angular momentum is L 1 =Iω 1 =Mω 1 L 2 /3.
If you increase the mass to M+m , the moment of inertia will
increase to I' =(M+m )L 2 /3 and its
angular velocity will change to ω 2 . But, the
angular momentum will not be changed, Iω 1 =I'ω 2 ;
you can then solve this for the new angular velocity, ω 2 =(I /I' )ω 1 =[M /(M+m )]ω 1
which is smaller. However, the rotational kinetic energy E of
the blade is now lower, E 1 =½Iω 1 2
and E 2 =½I'ω 2 2 =½I {(M+m )/M}{[M /(M+m )]ω 1 }2
or E 2 =[M /(M+m )]E 1 .
On the other hand, if you wanted to add mass but keep the energy the
same, E 2 /E 1 =1=I'ω 2 2 /Iω 1 2
or ω 2 =ω 1 √[M /(M+m )];
in this case, the angular momentum will have changed.
Your third question is moot since we have established that torque does
not depend on mass.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
Nucleon number
means number of nucleons. Electrons are not nucleons.
QUESTION:
FOLLOWUP QUESTION :
ANSWER:
h =5/6=0.83 ft=10 inches above her head. All this
assumes that the ball is thrown straight up. Note that I have not really
answered your question because you asked for force and the energy input
depends both on force F and the distance s over which it is
applied. The energy input W could be written as W=Fs ,
So, if you assume that she throws it the same way and pushes as hard as
she can, the force need not be known.
QUESTION:
ANSWER:
slice
and one which curves to the left is called a hook ; these have
opposite spins. Neglecting the possibility of wind, the reason that a
ball curves is because it has spin. But now it gets complicated because:
the hard-hit ball is in the air much longer than the softly-hit ball;
t he
lateral force causing the curve depends on both the rate of spin and
the speed of the ball, so the hard-hit ball will experience more
lateral force than the softly-hit one if they have the same spin;
even if the slow ball
has a bigger lateral force, the fast ball is likely to be deflected
a greater distance because of its longer flight time;
a lateral wind will
exert the same force on both, but the fast ball will be deflected
farther because of the longer time.
So, you see, there is no
simple answer. To avoid curving, learn to hit the ball without imparting
significant spin!
QUESTION:
ANSWER:
QUESTION:
FOLLOWUP
QUESTION:
ANSWER:
v= √(2gh )=√(2x9.8x560)=105
m/s=235 mph! Back to the drawing board! There is some friction due to
the wheels and bearings and I estimate that this is probably not more
than about 15 lb; this would only slow him down to about 99 m/s=220 mph.
Back to the drawing board! Finally, since the speed is going to get
pretty big, we need to take air drag into consideration because the drag
force is proportional to the square of the speed. A rough estimate would
be that the force is about F drag ≈¼Av 2
where A is the area his body presents to the onrushing wind. When F drag
is equal the net force down the incline (component of weight minus
friction, which I estimate to be about 117 lb=520 N), he will stop
accelerating. Taking his area to be about 2 m2 , you can then solve ¼Av 2 =520 to get
v =32 m/s=70 mph. This is all very rough but should give you an
order-of-magnitude estimate. (I still find it hard to believe that he
went down a half mile, 45º slope without braking at all!)
QUESTION:
ANSWER:
QUESTION:
ANSWER:
via an attractive force
(gravity in this case), any one of them can only speed up when moving
toward their center of mass. The details of the motion would be
determined by the initial conditions. If all the objects were moving
away from some common point at some time, the only possible motions
would be
for all to move forever away from each other, but forever slowing
down;
for all to slow down and eventually turn around and speed up back
together; or
for some to come
together and some to keep going.
The simple reason is that
the potential energy of such a system increases as the objects get
farther apart so the kinetic energy must decrease to conserve energy.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
gravitons and
general relativity .
QUESTION:
ANSWER:
reentry angle were too small that they would "skip off" the
atmosphere into space.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
virtual photons which means that they pop into and out of
existence very quickly, too quickly for you to observe them
QUESTION:
ANSWER:
QUESTION: 2 ."
Thank you very much. I am in Gr 5!! I want to by a physicist.
ANSWER:
t =0)
when you let go of the stone, u =0 because u in your
equation is the speed at t =0. Also this equation assumes that
S =0 at t =0 and that S increases in a downward
direction. So, at the end of 5 seconds (the beginning of the 6th
second) the position is S 5 =2 =400
ft; at the end of 6 seconds the position is S 6 = ½x32x62 =576
ft. So the total distance traveled is 176 ft. I trust you will not
present this work to your uncle as your own.
QUESTION:
ANSWER:
f doing negative work and gravity mg sinθ
doing positive work; if f <mg sinθ
the box will be speeding up.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
v t , does not depend
on the altitude from which you drop your car. This can be a very tricky
problem because v t does depend on the density of the
air which changes greatly from sea level to 30,000 ft. So to get a first
estimate, I will just assume sea level density everywhere. There is an
estimate for the drag force in sea level air which is good for a rough
estimate, F D = ¼Av 2
where A is the cross sectional area and v is the
speed. From this you can show that v t =2mg /A ).
In SI units, m =4300 lb=1950 kg, g =9.8 m/s2 ,
and A ≈2x4=8 m2 (estimating the car as 2 m wide
and 4 m long). Then v t
I guess we should now ask
whether we expect it to reach terminal velocity before it hits the
ground. Actually, it will technically never really reach terminal
velocity, only approach it—see an
earlier answer . I will
calculate how far it falls before it reaches 99% of v t .
In the earlier answer , I
show that the height from which you must drop it for it to reach
terminal velocity with no air drag is h no drag =v t 2 /(2g ),
and the height from which you drop it for it to reach 99% of terminal
velocity with air drag is h drag =1.96vt 2 /g
(derived from the expression v /v t =0.99=√[1-exp(-2gh /v t 2 )].
So, for your case, h no drag ≈510 m and h drag ≈2000
m. At 2000 m (around 6000 ft) the air is about 85% the density of
sea-level air, so I believe that my approximation assuming constant
density is pretty good and the car would probably reach 99% of the
terminal velocity by the time it hit the ground. To actually put in the
change in density with altitude would make this a much more difficult
problem.
QUESTION:
ANSWER:
anno domini , (in the year of the Lord i n Latin) and BC is
before Christ . The dividing line is, supposedly, either the
birth or conception of Jesus. Times before are labeled BC and those
after are labeled AD. There is no year zero, so the first year after
this time is labeled 1 AD and the first year before is labeled 1 BC.
Hence, the time from, e.g ., January 1, 10 BC to January 1, 10
AD is 20 years; but, the time from January 1, 1 AD to January 1, 20 AD
is 19 years.
QUESTION: -14 in space?
My guess is that we are capturing 99% of the effective force differential using the chamber, so not much more to expect from the vacuum of space.
ANSWER:
QUESTION: 5159703
and I wanna know how you can record it because a regular microphone doesn't pick it up.
ANSWER:
QUESTION:
ANSWER:
F D
on an object may be approximated for every day objects, masses, and
speeds as F D = ½C D ρv 2 A
where ρ is the density of the air, v is the speed,
A
is the cross-sectional area presented to the onrushing air, and C D
is called the drag coefficient. C D depends only on
the shape of the object. C D ≈0.5 for a sphere
and C D ≈0.8 for a cube (falling with one face
to the wind). So, if their areas are about the same, the drag on the
sphere will be smaller and it will go faster. Of course if the sphere
area were ten times bigger than the cube area, that would be more
important than the somewhat smaller drag coefficient and the cube would
win.
QUESTION:
ANSWER:
Mr Thompkins books by George Gamov.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION: link .
Is it correct?
ANSWER:
The traveler, as you correctly surmise, observes the trip to
take 0.604 years. However, you have not taken length contraction into
account. The traveler observes the distance she has to travel to
be 4.24/γ =0.598 ly, so she perceives
her speed to be 0.99c .
QUESTION:
ANSWER:
E=mc 2 , is not lost and the black hole
becomes more massive by the amount of the mass energy. When light is
swallowed by the black hole, the mass increases even though the light
does not have any mass because light does have energy and the energy
shows up as increased mass of the black hole, m=E /c 2 .
QUESTION:
ANSWER:
QUESTION:
ANSWER:
mv 2 =½x3.83x107 x2.34x106 =4.48x1013
J. Note that the composition, size, and shape are irrelevant. Since
everything else was in SI units, I assumed that ton is metric tonne, 1
tonne=1000 kg. This must be some kind of projectile in a computer game.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER:
L
with weight W N F F x )
and perpendicular (y ) to the ramp. Next write the three
equations of equilibrium, x and y forces and the
torques; this will give you the force you need to apply
to move it up the ramp with constant speed.
ΣF x =0=F x -W sinθ F y =0=F y +N-W cosθ Στ= 0=½WL cos(θ+φ )-NL cos(φ ).
I summed torques about the end where you are
pulling. Putting in W =400 lb, θ= 35º, and
φ= 20º, I find F x =229 lb, F y =206
lb, and N =122 lb. Note that you do not need to know the length
L . The net force you have to exert is F =√[(F x )2 +(F x )2 ]=308
lb. If someone were at the wheel pushing up the ramp with a force
B F x
and F y . This would change the equations to
ΣF x =0=F x -W sinθ +B F y =0=F y +N-W cosθ Στ= 0=½WL cos(θ+φ )-NL cos(φ )+BL sin(φ ).
For example, if B =100 lb, the solutions
would be F x =129 lb, F y =169 lb,
and N =159 lb; so your force would be F =213 lb.
QUESTION:
Imagine if you wrapped a rope tightly around the earth. How much longer would you have to make the rope if you wanted
it to be exactly one foot above the surface all the way around?
ANSWER: C and another of
circumference C+
δ ; that
is not really physics. But, it is easy enough to do. If C is the
circumference of the earth, then C= 2πR where R
is the radius of the earth and
C+ δ =2πR' where
R' is the radius of the circle your rope would make if δ =1
ft. Then δ= 2π (R'-R )=2π =6.28 ft.
Note that δ depends only on how high the rope is above ground,
not how big the earth is: if the earth were 1 ft in radius and you increased
the length of the rope by 6.28 ft, the rope would be 1 ft above the surface!
QUESTION:
ANSWER:
fissile . If a thermal neutron is absorbed by a fissile
nucleus, it will fission and result in more neutrons leading to more
fissions and the reaction can be self-sustaining. Thorium, which is 100%
232 Th, absorbs a neutron to become 233 Th which β -decays to 233 Pa (half life 22 minutes)which
β -decays to 233 U (half life 27 days). The
233 U is fissile and is nearly stable (α -decay half
life 160,000 years). Thorium is said to be fertile , absorption
of a neutron results in production of a fissile fuel. So a thorium
reactor is a breeder reactor, a reactor whose purpose is create fuel. At
the startup one needs a starter fissile fuel as well as the
thorium to provide neutrons to create the 233 U. As the
233 U builds up, it becomes the fuel.
QUESTION:
ANSWER:
F of air drag on an
object of mass m , speed v , and cross sectional area A is F =Av 2 ;
this works only in SI units. The speed v when F=mg is called
the terminal velocity. I estimated a reasonable terminal velocity would
be the speed the tank would have if you dropped it from about 10 m,
v ≈14 m/s. The mass of the
MBT-70 (KPz-70)
is about 4.5x104 kg. Putting it all together, I find that
A ≈104 m2 , a square about 100 m on a side or a
circle of radius about 30 m. You could do a more accurate calculation
but this gives you a reasonable estimate.
QUESTION:
ANSWER:
earlier answer .
QUESTION:
ANSWER:
30 kg, quite a bit bigger than the mass of
fuel in a fusion reactor. Therefore the rate of fusion in the sun can be
low but the energy output would still be huge. Increasing the
temperature would increase the rate. The density of the sun's core is
about 150 g/cm3 , 150 times more dense than water. In a fusion
reactor, the practical densities are many orders of magnitude below
this. Again, the rate of fusion would be dependent on the density of the
fuel.
QUESTION:
ANSWER:
recent answer for more detail.
However, the vacuum does not contain mass in the sense you normally
think of it
QUESTION:
ANSWER:
ħ L (L +1)] where L is an
integer. (For our purposes here, "not normalizable" means that the wave
function becomes infinite somewhere.) It also turns out that L
is called the orbital angular momentum quantum number and its being an
integer has nothing to do with spin except that spin is also an angular
momentum. To visualize spin, read two earlier answer (#1
and #2 ). In relativistic quantum mechanics, the
Dirac equation replaces the Schrödinger
equation. When you write the Dirac equation for an electron, it turns
out that spin is indeed predicted to have an angular momentum quantum
number of ½ as is observed experimentally.
QUESTION: link
to an article from the L.H.C people at Cern that shows this.
I'm amazed and confused because I thought that matter and anti matter particles would annihilate each other. Why don't they? And, if they do, how is the 'zillions of other quarks' balance maintained?
ANSWER:
Lamb shift , for
example). The CERN explanation should have included zillions of
positrons and electrons and photons inside the nucleus also. Just as a
hydrogen atom is pretty well described as a proton and an electron as a first
approximation, a proton is pretty well described as three quarks as a
first approximation.
QUESTION:
ANSWER:
N of anything they will, at some time t ,
have a time rate of change R =dN /dt . If R <0,
N is getting smaller (as in radioactivity) and if R >0,
N is getting larger (like bacteria growth). For a great many
cases in nature, it turns out that the rate is proportional to the
number, R N . Radioactivity turns out to
behave that way: if you measure R for 2x1020
radioactive nuclei it will be twice as big than when you measure R
for 1020 nuclei. You can then solve for what R is for a given
situation: dN /dt=-λN where
λ is the proportionality constant, called the decay
constant; the minus sign is put there so that λ will be
a positive number since the rate is negative. If you know differential
equations, you will find that N=N 0 e-λt
where N 0 is the number when t =0. The decay
constant is related to the half life τ ½ =ln(2)/λ.
However, this all depends on there being a large number to start.
In the extreme case, if N 0 =1, they would all decay
at once! You just could not predict when. If N 0 =2,
they might both go at once or else one might go before the other but not
necessarily at τ ½ .
QUESTION:
ANSWER:
F such
that the springs are stretched by a distance s . Then each
spring will be stretched by s . If k is
the common spring constant, F =½ks . Therefore
the two together behave like a single spring with spring constant ½k.
QUESTION:
ANSWER:
etc . because all these things affect your
velocity; e.g. , relative to the earth's axis, you are moving in
opposite directions at noon and midnight. There is a nice qualitative
discussion of this question at
Curious about Astronomy ; adding up all the various velocities, she
finds a maximum speed relative to the center of our supercluster of
about 900 km/s. Given the roughness of this calculation, trying to
fine-tune it by adding in the expansion of space-time is pointless.
QUESTION:
ANSWER:
mv 2 2 -mMG /r 2 2 =W +½mv 1 2 -mMG /r 1 2 .
Of course you will have to figure out what the velocities are in terms
of the radii, but that should be a piece of cake for AP students!
QUESTION:
ANSWER:
QUESTION:
ANSWER:
-6
m/s2 . For comparison, if you are in a car with an
acceleration from 0 to 60 mph in 10 s, your acceleration is about 3 m/s2 ;
this means that you "feel" pushed back into your seat with a force about
one third of your weight. The force you "feel" due to the orbital motion
of the earth is about 1/600,000 of your weight, far less than you could
"feel". The acceleration due to the earth's rotation on its axis is
larger than this but still way smaller than you could "feel".
QUESTION:
Questions:
Since a lot of the mass of a proton comes from energy, doesn't this 'binding energy' add mass to the He4 nucleus?
The protons that formed this nucleus are made up of Quarks held together by energy in the form of the exchange of Gluon particles. To convert to a neutron, an Up Quark has become a Down Quark with more mass. What mass has the He4 Nucleus lost?
Do the Quarks in the He4 nucleus have less mass than those in the H1 since the Quarks appear to be the only mass in the process that can convert to energy? Surely if the energy came from the Gluons it would still just be energy and hence not have the mass/energy conversion ratio?
If the energy did come from the Gluons, are Quarks in He4 less tightly bound than those in H1?
Does this continue to occur as you move up the fusion sequence to higher mass elements?
Has the creation of matter from energy (e.g. the electron produced in this fusion) ever been actually achieved? Or are we working purely on theory based on remote observation of the Sun?
ANSWER:
trying
to understand a relationship between nuclear physics and particle
physics", I will ignore all questions relating to quarks and gluons
because if the nuclear physics itself is not understood, then there is
no point in trying to relate it to particle physics. To understand the
fundamentals of nuclear physics, you do not need to even know that
quarks and gluons exist, only that a strong nuclear interaction exists.
In the opening paragraph the mass of the 4 He nucleus is
compared to the mass of four protons. This is a meaningless comparison
because the constituents are two neutrons and two protons. (See below*
for how protons are converted into 4 He nuclei.) Indeed, the
mass of the 4 He nucleus is less than the mass of its
constituents. If one knows E=mc 2 , the reason for
this mass difference is intuitive. Does it take work to pull a proton or
neutron out of the 4 He nucleus? Of course it does, otherwise
the 4 He nucleus would not stay together. So energy must be
added to disassemble the nucleus and this energy resides in the greater
mass of the constituents. I find this much easier to understand than to
say that the excess mass of the constituents is "converted" to binding
energy.
As you move
up to create heavier and heavier nuclei via fusion, the fusions continue
to release energy but in decreasing quantities until the final product
is iron. Thereafter, if you want to fuse nuclei, you must add energy.
So, elements heavier than iron are not produced in stars. Heavier
elements are produced in supernova explosions.
You ask if "…creation of matter from energy…" has ever
been observed. It happens all the time. But that is really the wrong
question because matter and energy are not different things, matter is
simply a form of energy. So, for example, if you fused two 32 S
nuclei to a single 64 Ge nucleus, the Ge would have more mass
than the two S. Any time you have an endothermic chemical reaction you
end up with more mass than you started with (although almost immesurably
small because chemistry is so inefficient).
*As can be seen in the figure above, the proton-proton cycle requires an
input of 6 protons and ends with one 4 He and 2 protons. So the net input
is 4 protons. However, there are numerous outputs—2 positrons, 2
neutrinos, and 2 photons, all of which carry energy away. That is why
comparing the net input with the final product is not meaningful.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
-11 m. The atoms are all randomly
vibrating so that there always as many atoms moving in one direction as
in the opposite direction. The frequencies of the vibrations are ~1013
Hz, far above a frequency which you could feel, hear, or see.
QUESTION:
ANSWER:
cosmological constant , which can add a repulsive force to the
attractive gravitational force. I do not quite see how this could be a
route to quantizing gravity, but who knows? Good luck!
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER:
E is defined to be the force F
felt by a charge Q divided by Q . The electric
potential V is defined as E times distance d
over which it acts. V=Ed=Fd /Q =[J/C]
QUESTION:
ANSWER:
earlier answer . Gravitons are
not predicted by general relativity. Physics inside the event horizon is
not well understood.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
Z>N nucleus is 3 He. The
reason that all stable isotopes heavier than helium have the same or
more neutrons than protons is complicated because the nuclear force is
complicated. One important reason is that the repulsive Coulomb force as
well as the attractive nuclear force is present trying to blow the
protons apart and the more protons you add, the more important this
becomes; adding neutrons tends to push the protons farther apart to
reduce the repulsive force. Neutrons are sort of "insulation" for the
protons' electric repulsion from each other.
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER:
r 2 force just like the
gravitational force, so if the atom were a classical system Kepler's
laws could be used for an atom. In fact, the
Bohr-Sommerfeld model , the first extension of the Bohr's circular
orbit model, essentially does this by including elliptical orbits and
appropriate quantization. Of course, the atom is not a classical system
and although such models can be instructive, they are not strictly
accurate. What you refer to as the "electron cloud model" would be the
proper solutions to the Schrodinger equation, not having well-defined
orbits.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER:
fictitious force . If
the car is accelerating with acceleration a (which points
opposite the velocity v when braking), Newton's first law will
be valid in the car frame if a force F=-ma acting at the center
of mass (COM) is introduced. The "real" forces on the car are a
normal forces up on each wheel by the road, the frictional forces
backwards on each wheel by the road, and the weight mg which
acts at the COM. In the drawing I have only labeled the weight, the
normal force on the rear wheel, and the fictitious force because those
are all you need to answer your question. If you wish, you could find
both normal forces and the sum of the two frictional forces in terms of
a ; if all wheels were locked you could find the individual
frictional forces if you knew the coefficients of kinetic friction. Now,
I will sum torques τ about the point where the front wheel
touches the ground, Στ=mah+NL-mgs= 0 where h
is the height of the center of mass above the ground, L is the
horizontal distace from the front axle to the COM, and s is the
horizontal distance between the wheels. You can now solve for N ,
N =(mgs-mah )/L . Now think about N ; if
N <0, the road would have to pull down on the back wheels to
hold the car from rotating forward about the front wheels. N
will be zero when the car is just about to "fly up"; therefore, if a >g (s /h ),
the car will "fly up".
QUESTION:
ANSWER:
earlier answer . If you
calculate the angle for earth instead of the sun, φ =4GM /(c 2 r )=0.002
arcseconds, just about impossible to observe.
QUESTION:
ANSWER:
E=mc 2 =m 0 c 2 /v /c )2 ]
where m 0 is the mass at rest. For example, if v /c
increases from 0.99 to 0.999, that is only a 1% change in speed; but the
energy increases from 7.1m 0 c 2
to 22.4m 0 c 2 , more than triple.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
article by
Kadoya, Matsunama, and Nagashima. Pressure is given in MPa and 0.10
MPa=0.987 atm
QUESTION:
generated the waves? 2). Since the earth is trapped in orbit around the sun by the sun's gravity, how can a wave moving out from an object in turn pull a second object back towards the object that generated the wave? In other words, how can gravity waves moving out from the sun at the same time pull the earth towards the sun?
ANSWER:
recent answer . In your second answer, you are confusing gravity with
gravitational waves. Our picture of gravity is that mass (like the sun)
deforms the space around as in the animation above. Something like the
earth orbiting the sun is not really feeling a force, it is following
the contour of the space. (Do not take this
simplistic model too seriously, it is really the four-dimensional
space-time which is "deformed".) But, if something accelerates, like the
earth going around in its orbit, it will send out ripples and those are
the gravitational waves. On the scale of this animation, the waves the
orbiting earth is sending out are far too tiny to be seen. But, if the
objects orbiting are much more massive, like the two orbiting black
holes which were observed in the recently reported
observation of the waves, the waves are much bigger.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION: φ =4GM /(c 2 r )=1.75
arcseconds is total deflection when light source is opposite side of the Sun?
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
FOLLOWUP
QUESTION:
ANSWER:
τ =Iα where α is the
angular acceleration, τ is
the torque, I is the moment of inertia. In your
case, the angular acceleration would be the final angular speed divided
by the time to get there; so if you want to get up to speed in 10
seconds, the angular acceleration would be 5.8/10=0.58 rpm/s=0.061
radians/s2 . You see, now, why there is no answer to your
question: the torque will depend on how quickly you want to get up to
speed. The moment of inertial of the rope is I =½M (a 2 +b 2 )=½(2300)(12 +0.32 )=1250
kg·m2 . Finally you can estimate the torque
τ = 1250x0.061=76 N·m=56 ft·lb.
A longer spin-up time would need a smaller torque.
QUESTION: 8 m/s) or 186,000 Miles per second. Can anything you know of travel at the speed of light especially radio waves?
ANSWER:
QUESTION:
ANSWER:
or a wave, it is a particle
and a wave. This is called wave-particle duality. If you design an
experiment to observe a particle, you will observe a particle; and, if
you design an experiment to observe a wave, you will observe a wave.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
speed at which a gravitational field propogates should be the same
but it has never been measured either; this would determine, for
example, if the sun suddenly disappeared how long it would be until the
earth stopped orbiting. The theory of general relativity, which predicts
gravitational waves, say that the speed of gravity should be the same as
the speed of light. The recent
observation of gravitational waves determined the distance to the
source to be about 1.33 billion light years, but the uncertainty was
very large, about 40%, so it really provides no measurement of the
speed. Whatever the speed, there is no reason to think that it would not
be constant everywhere in empty space.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
-10 m and you just cannot make physical slits
that close because that is like the distance between atoms. What is
actually done is to shoot electrons at a single crystal and you get a
double-slit like experiment. For diffraction to be observable the slit
spacing and slit width must be small compared to the wavelength of the
wave.
QUESTION:
ANSWER:
youtube video . You will see that what matters more than the added
weight is where you put it. (Thanks to my son Andy for pointing me to
this video; his son and my grandson Finn placed second in the Cub Scout
pinewood derby last year on the strength of the tips here!)
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
faq page for
links discussing this. The best way to see this, I think, is to consider
the light clock discussed in an
earlier answer . To
understand the light clock, you must accept that the speed of light is
the same for all observers; see the
faq page . Also, to help you with understanding time dilation, read
about the twin paradox .
QUESTION:
ANSWER:
lacross ball
and a baseball . I
will make the same assumption that I did in those answers that the
amount by which the ball will fall will be very small compared to its
horizontal distance and the speed acquired in the vertical direction
will be very small compared to its horizontal speed. So I will ignore
the small effect which the vertical motion will have on the horizontal
motion. As discussed in the
earlier answer ,
the horizontal distance x and speed v are given by
v=v 0 /(1+kt ) and x =(v 0 /k )ln(1+kt )
where v 0 is the initial velocity and k is a constant
determined by the mass, geometry, and initial velocity of the ball. For
v 0 =340 m/s (speed of sound) and a baseball
(mass=0.15 kg, diameter=0.075 m) these become v =340/(1+2.8t )
and x =121ln(1+2.8t ) and are plotted below.
In one second the ball goes about 160 m and slows down to
less than 100 m/s. During the same time, the ball will fall
approximately 5 m and so, if launched horizontally from a height of 5 m
will hit the ground in one second as shown below. Be sure to note the
difference in horizontal and vertical scales; an insert shows the trajectory drawn to
scale. The small distance fallen is the justification for my
approximations above.
QUESTION:
ANSWER:
particles
popping into and out of existence…" is what a vacuum is. However,
space does have lots of atoms and molecules in it. But for light to be
affected at all, it must interact with an atom. The atoms are so far
apart that the light seldom interacts with them. If that were not the
case, we would not be able to see distant stars and galaxies.
Furthermore, what we usually call the change of speed of light in a more
dense medium is due to its average interactions with the electrons in
the material and not that the light is homogenously slowed down; the
photons wend their way through as you would going through a thick
forest.
QUESTION:
ANSWER:
…cos2
of half of the angle…", but you seem to not know what, e.g. ,
spin
½ means. When you say spin s , this means that
the spin angular momentum quantum number is s ; this means that
the magnitude of that spin angular momentum is
S=ħ √[s (s +1)]
where ħ
is the rationalized Planck constant. But, there is another quantum
number, m s =-s , -s+1, -s +2, …,
s -1, s which is the quantum number for the z -component
of S , S z =m s ħ . Hence, the angle θ which the vector makes with the
positive z -axis is θ= cos-1 (S z /S ).
For s =½, θ= cos-1 (½/√(3/4))=54.70
and θ= cos-1 (-½/√(3/4))=125.30 ;
similarly, for s= 1, θ= 450 , 900 ,
1350 .
QUESTION:
ANSWER:
QUESTION: music video in which a group of performers appear to be in an aeroplane cabin in free-fall for 2 minutes and 40 seconds. The choreography is spectacular, and it appears to have been done in ONE take!
How far would the plane have had to descend to maintain zero gravity conditions for 160 seconds?
ANSWER:
2 , but I think you can see that this would not be a
very safe situation. Anyhow, back to your question, I could not find
reference to any such parabolic flight lasting longer than 30 s, so the
video must have been shot in more than one take. Because this is such an
unfamiliar environment, I cannot believe that, even with a lot of
practice, it could be done in a single take without errors. And, if the
plane were simply falling for 160 s, it would have to have started at an
altitude of about 80 mi (far higher than a plane can actually fly) and
would end with a speed of about 3600 mph (far faster than the plane
could fly without disintegrating).
QUESTION(submitted by my daughter!): this exciting for you?? I dont really understand it but I'm trying to...it seems cool!
ANSWER:
direct evidence for gravitational waves. Indirect evidence was found when observing a pair of stars orbiting each other and spiraling in toward a collision. The loss of energy turned out to be exactly
equal to the amount of energy they would lose if radiating gravitational waves. A nobel prize in physics was awarded in 1993 for this observation.
The animation below shows the predicted gravitational waves from two
neutron stars orbiting each other.
QUESTION:
ANSWER:
never does work because of its very nature. Although the presence of
magnetic fields is imperative, work is always done by electric fields
they produce. Most
electric motors have "brushes", contacts which slide on the rotating
armature, and small sparks are inevitable when electrical contact is
made or broken. The Tesla electric car, however, employs a type of motor
called an
induction motor which is brushless and is therefore probably
sparkless.
QUESTION:
ANSWER:
web
site which seems to have made it easy for me by including a
calculator. Frankly, I have no idea what the roughness coefficient is,
but it suggests a value of 150 for plastic. The result is below. As you
can see, to get a flow rate of about 2200 gpm would require a pipe with
diameter of about 10 in.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
B and the drag force D W f +W c +B+D . If
the object is falling with constant speed, it is in equilibrium and so
B+D=W o and the scale reads the total weight of
container, object, and fluid. If the object is accelerating down,
B+D<W o and the scale reads less than the total weight of
container, object, and fluid. There is an
earlier question
similar to yours except the object is rising instead of
sinking.
QUESTION:
ANSWER:
deepest hole ever drilled is only about 12 km deep. I could
not find any reference to attempts to measure g at various depths
down
this hole. Since the radius is about 6.4x103 km, you would
only expect about a 0.2% variation over that distance. There are models
of the density of the earth, though, which have been determined by
observing waves transmitted through the earth during earthquakes or
nuclear bomb tests; these are believed to be a pretty good
representation of the radial density and can be used to calculate g .
The two figures above show the deduced density distribution and the
calculated g .
Usually in introductory physics
classes we talk about the earth as having constant density, but as you
can see, that is far from true—the core is much more dense than
the mantles and crust. If it were true, g would decrease
linearly to zero inside the earth. Instead, it increases slightly first
to around 10 m/s2 and remains nearly constant until you are
at a depth of around 2000 km. There is little likelihood that g will
ever actually be measured deep inside the earth because the temperature
increases greatly as you go deeper, already to near 200ºC at 12 km.
However, if you have detailed information on density distribution, there
is really no need to measure g .
QUESTION:
ANSWER:
on the
chair . The reaction force is the force which the chair exerts
on you . Only forces on the chair
determine how the chair will move.
QUESTION:
ANSWER:
π
m for the rim and π m for the second point; the second point
has half the speed (π m/s) of the first (2π
m/s). Don't go nuts!
QUESTION:
ANSWER:
earlier answer .
Obviously, if you stopped this orbiting, anything would fall into the
black hole.
QUESTION:
ANSWER:
f depends on only two things:
what the surfaces which are sliding on each other are (conveyer material
and your plastic film) and the force N which presses the two
surfaces together; normally, on a level surface, the force N is
simply the weight of the object (suitcase in your case). There are two
kinds of COFs, kinetic and static. The kinetic coefficient, μ k ,
allows you to determine the frictional force on objects which are
sliding. In that case, f =μ k N.
The static coefficient, μ s , allows you to
determine how hard you have to push on the suitcase in order for it to
start sliding; in this case f max =μ s N
where f max is the greatest frictional force you
can get. Since you are being asked to prove that it is not too "sticky",
it is the static, not the kinetic, coefficient which you need; measuring
μ s is quite easy. The only problem is that μ s
depends on the surfaces so you must have a piece of the material
from which the conveyer belts are made to make a measurement. Once you
have that, use it as an incline on which to place a wrapped suitcase.
Slowly increase the slope of the incline until the suitcase just begins
to slide. Your COF (μ s ) is equal to the tangent
of the angle of the incline which (see diagram above) is simply μs =H /L .
QUESTION:
ANSWER:
can only give you a rough answer. I will calculate the power needed to
lift two large people (total mass 200 kg) up the hill. I will work in SI
units because that is the system in which the watt is the unit of power.
Your speed is v =4 ft/s=1.22 m/s so the rate at which the load
is rising is v sin250 =0.516 m/s. The rate at which
the energy of this mass is increasing is mgv =1011 J/s≈1
kW. I guess I would throw in a safety factor to account for friction and
other energy losses, so maybe a 2 kW motor would do you. The voltage
(240) and gear box are not really relevant in determining the power.
Disclaimer: I am not an engineer, so you should get a second opinion!
QUESTION:
ANSWER:
N +F cosθ -W=0, f-F sinθ =0,
and ½LW -LF cosθ =½W -F cosθ =0;
here F is the force you exert, W is the weight, N
is the force the floor exerts vertically, and f is the
frictional force exerted on the floor. Solving these I find that N =½W ,
F =½W /cosθ , and f =½W tanθ .
I suspect that the case you are interested in is when you first lift it
off the ground, θ= 0. F =750 lb, half the total
weight. Note that this analysis is valid as long as the floor is not too
smooth, that is the box does not start sliding at some angle; the angle
is less than θ= tan-1 (L /H )
because at that angle the center of gravity is directly over the pivot
side.
Of course there are lots of other ways you could
lift it which would be more efficient if your aim was to tip it over;
for example, you could start pushing horizontally once you got it off
the ground so that the floor would hold up all the weight rather than
half the weight. There is an
old answer very
similar to yours that you might be interested in.
QUESTION:
ANSWER:
…"
means. Imagine that we just drop each from some height. If air friction
is ignored, they hit the ground simultaneously; this would approximately
be the case if they were dropped from a few meters. However, air drag
becomes more important as the speed increases. The drag force f
may be approximated as f ≈¼Av 2
where A is the cross sectional area (πR 2 )
and v is the speed (you must use SI units). So, for a ball,
f-mg=ma where m is the mass, a is the acceleration, and
g is the acceleration due to gravity. Since f gets bigger as
v
gets bigger, eventually f=mg and the ball will stop accelerating and
fall with a constant speed. The velocity at which this happens is called
the terminal velocity v t =√[4mg /(πR 2 )].
The mass and radius of a tennis ball are m= 0.059 kg and R =0.069
m; the mass and radius of a soccer ball are m =0.43 kg and R =0.22
m. So, I find v t tennis ≈50 m/s for
the tennis ball and v t soccer ≈11 m/s
for the soccer ball. The tennis ball will easily win the race because it
continues accelerating long after the soccer ball stopped accelerating.
QUESTION:
I love hearing about new discoveries from particle accelerators, but one aspect of them confuses me. If we accelerate particles to 99% of the speed of light, shouldn't that make the matter appear to go slower to us, a la the twins paradox? Why don't these particles approach infinite mass and compress time from our perspective?
ANSWER:
c . You would observe it to live for 1/2 )=7.1
s. The accelerated particles do indeed approach infinite mass but they
have a really long way to go to get there even in the most powerful
accelerators; see an
earlier answer .
QUESTION:
My question relates to my own voluntary time involved with the rope/rigging community. It is not of any commercial interest to me, but the answer to my question has eluded me for a long time despite research and attempts to calculate it ! If you could at least explain to me the correct method of attempting to work our the answer I would greatly appreciate it.
Question background - horizontal restraint line.
A 7.6 m long horizontal rope is anchored at both ends. It is 2m horizontally above ground level (AGL). It is pre-tensioned to 50kg.
A 20kg mass is suspended vertical above the horizontal rope and attached to the rope by a leash of 0.7 m length.
The object/leash attachment point is mid-way along the 7.6 m horizontal rope i.e.
3.8 m from either anchor.
The 20 kg object is then dropped.
At the time of peak impact force, the sag in the horizontal tension line is 1.0 m. Despite my best efforts I do not seem to be able to calculate theoretical (i) peak impact forces on the anchors (ii) peak impact force on the object.
I do however have peak impact force load cells that have recorded average (i) peak impact forces on the anchor 220
kg and (ii) peak impact force on the object 115 kg.
Is it possible for you to detail the correct formula/method for at least working the theoretical approach to calculating these forces ?
QUERY:
REPLY :
I
sin θ=s/ √(L 2 +s 2 )=0.41/√(3.82 +0.412 )=0.11,
so F =20=0.22T ; thus T =91 kg which is in
reasonable agreement with the measured value of 100 kg. In the case of
the maximum s ,
sin θ= 0.25 and F =0.51T .
Taking the measured value of F =115 kg, T =225 kg, again
in good agreement with the measured value of T =220 kg.
At this point what a physicist normally does is to try to understand the
situation in terms of a simple spring model. If the rope is like a
simple spring, i.e. its tension is proportional to its stretch,
you can usually make approximations which would result in the simple
model that F≈ks where k is a constant. For the
equilibrium situation, then, 20≈ 0.41k
or k≈ 49 kg/m. This would then predict that the force
when s =1 m would be F =49 kg which is too small by more
than a factor of 2.
The previous try indicates that the rope is probably not approximated as
an ideal spring. My last attempt is to try to treat the rope more
explicitly as an ideal spring, not using small approximations used
above. So, looking at one half the rope, I will write T =50+kδ
where
δ= √(L 2 +s 2 )-L
is the amount by which this half of the rope is stretched relative
to L . Note that k is not the same as in the
approximation above where s was the stretch parameter rather
than δ. Calculating δ for the at rest
situation where T =100 kg, k =2270 kg/m. Then,
calculating T for the maximum s using this value of
k , T =50+2270x0.129=344 kg which is too large by nearly 60%
compared to the measured value of 220.
I conclude that the rope is poorly approximated as an ideal spring and
that to do any more detailed analysis of this problem would require
measuring s as a function of the load by varying the load for
the equilibrium situations.
FOLLOWUP
QUESTION :
If possible, without the aid of the load cell values which I get by practically performing the drop test. How would you derive the theoretical values for the tension/force at the both the anchor and on the body in the drop test scenario. I understand how to calculate forces etc for the system at rest, but it is the dynamic falling system, deceleration etc that I am unable to clarify.
To summarise, if you have the load/object, placed at any height above or below the horizontal line (depending on the leash length), without loading the horizontal line, and then you drop the object, what would be the correct formula method to determine the approximate theoretical peak impact forces on the anchors and body, without the luxury of any load cell info. when it comes to rest after being decelerated by the horizontal line.
ANSWER:
a =(115-20)g so
a =4.75g =46.6 m/s2 .
Then I tried to model the rope as a simple spring, the tension is proportional to the stretch. I failed to do so in both attempts. You do not know the force which this rope will exert given a certain amount of stretch. Physics may be powerful, but you cannot do any predictive calculation if you are ignorant of the force. I have two data points but that is not enough if the force is not linear. The last paragraph suggests the only hope for having a predictive analysis: you need to measure the tension in the rope as a function of what you call sag. To do this you need to use many weights, say at 2 kg steps, and measure load#1 and sag. This would give me the information necessary to do predictive calculations. Life in the real world is not always simple and analytical. It occurs to me you do not really need to hang varying weights because you have the load#2 device; just pull down on it until it reads 2 kg, 4 kg, 6 kg, etc.
It is, of course, simple to calculate the speed the weight has just
before the leash goes taught: v =√(2gh ) where is
the distance fallen.
QUESTION:
A conductor, even though it is carrying a current, has zero net charge. why then does a magnetic field exert a force on it?
ANSWER:
QUESTION:
Do all five balls on a Newton's Cradle have to weigh the exact same amount in order for it work correctly?
ANSWER:
QUESTION:
If F=Q1.Q1/R^2, does it mean that two unlike charges touching each other have infinite forces of attraction between each other? If so, how are we able to separate 2 oppositely charges objects which are stuck together with our bare hands?
ANSWER:
R is the distance between the centers of
the spheres. So, if you have two charged insulating spheres of radii 1
cm and 2 cm which are touching, R =3 cm.
QUESTION:
What is inertia? From what I understand, the inertia for a sphere is: I = (2/5)MR^2. Why is the radius included? Why is the volume of the sphere related to the resistance to change motion? As long as any 2 objects have the same mass, why should they have different inertias?
ANSWER:
a (like a car speeding up or slowing down), and
rotational acceleration α like the rotational speed of an
engine's flywheel speeding up or slowing down. Inertia for resisting
a is simply mass which you can see from Newton's second law,
a=F /m. If you double the mass on which a given force acts,
you halve the acceleration; mass is sometimes called inertial mass for
this reason. For rotational acceleration, the rotational inertia depends
not just on how much mass there is, it also depends on how it is
distributed; for example, it requires much more effort to get a wheel
with radius 1 m spinning than it does to spin a wheel with the same mass
but a much smaller radius. The rotational form of Newton's second law is
α=τ /I where τ is the torque
applied to the object whose moment of inertia is I . In
rotational physics torque plays the role of force and moment of inertia
plays the role of mass.
QUESTION:
If friction acts perpendicular to the direction of motion, if i place an object against a wall and let go, since the object is falling downwards, will friction from the wall cause the object to move away from the wall? Also if I am standing still, gravity results in a downwards force so friction should act leftwards or rightwards. Which one is the correct one (assuming that i am not moving)?
ANSWER:
QUESTION:
I am an engineer working on the reconstruction of a traffic accident where it is alleged that a car traveling over a railroad crossing became airborne at a speed lower than the posted speed of the road. The information that I have available includes the type and make of the car and the geometry of the road. Nothing in the engineering literature that I can find addresses this issue. Despite the five quarters of physics that I took long ago, I am having trouble finding the information that I need to model this.
Any suggestions?
ANSWER:
R ,
the weight of the car is mg , the angle specifying the current
position of the car is θ , the force which is
causing the car to move with some constant speed v is F ,
and the normal force of the road on the car is N . Note that
although the car has a constant speed, it has a centripetal acceleration
a=v 2 /R toward the center of the circle.
Applying Newton's laws, F-mg cosθ mg sinθ N =mv 2 /R.
The first equation tells you what force you need to keep it moving at a
constant speed, F=mg cosθ which is really not of
interest to you; note that the force is in the direction of v
on the way up and opposite on the way down since the cosine changes sign
at 900 . The second equation tells you what N is for
any position of the car, N=mg sinθ-mv2 /R ;
note that if N is negative it points toward the center but the
road cannot pull down, only push up, so the car could not stay on the
road at that speed and angle. What is really of interest is under what
conditions would N =0=g sinθ-v2 /R
or v 2 =Rg sinθ ; this would
tell you the speed (angle) at which a car with a particular angle
(speed) would leave the road. Note that it is independent of the mass.
Notice also that v 2 /Rg< 1 because the sine
function cannot be larger than 1.0. For example, for what speed will the
car leave the road at 450 if gR =300 (m/s)2
(R v =√(300x0.707)=14.6
m/s=32.7 mph. Or, at what angle will a car with speed 35 mph=15.6 m/s
leave the road?
θ= arcsin(15.62 /300)=54.20 .
One thing which occurs to me though is, since you know the car and
railroad crossing, why don't you just do the experiment and drive it
over the crossing?
QUESTION:
When i place an object in between my 2 palms, why does it not fall? My palms only supply horizontal force, where does the vertical force come from to hold the object between my palms without falling?
ANSWER:
QUESTION:
Say there is a cylinder on a ramp and the friction force from the ramp cancels out the parallel component of gravity. Therefore, the cylinder should be in linear equilibrium. However, from the reference point of the center axis of the cylinder, there is a net torque exerted by the friction force. Additionally, there is also a net torque exerted by the gravitational force from the reference point of the point of contact of the cylinder and the ramp. Therefore, it is not in rotational equilibrium and should start to rotate, correct? How is this possible, because if the cylinder starts to roll how can it also be in linear equilibrium?
ANSWER:
-f+mg sinθ=ma where θ is
the angle of the incline and a is the acceleration of the center of mass
down the incline.
FOLLOWUP QUESTION:
Thank you very much for your response. However, I think you may have misunderstood my question. I was asking what would happen in a case where the frictional force is set to cancel out the parallel component of weight. It seems as if the center of mass cannot move, but the cylinder needs to rotate. Therefore, it would appear as if the only outcome of this situation would be a cylinder rotating in place on a ramp, which does not seem possible. I think that the cylinder would have to roll down the ramp, but I can't see how this would be consistent with linear equilibrium.
ANSWER:
mg sin θ
where
θ is the angle of the incline and m is the mass
of the cylinder; imagine that the incline is smooth (frictionless). Now
there will be a net torque about the center of mass of mgR sinθ=Iα
where I is the moment of inertia and
α is the angular acceleration of the cylinder. The
cylinder will spin in place. Your hand will have an acceleration of
a =mgR 2 sinθ /I ; for a uniform
solid cylinder with I = ½mR 2 ,
a =2g sinθ .
μ k = tanθ ,
the cylinder will slide down the incline with constant speed because the
frictional force will be f=μ k N =(tanθ )(mg cosθ )=mg sinθ.
So, if you start it sliding and not rolling, it will begin spinning
about its center of mass because of the torque due to the friction and
have an angular acceleration α=fR /I ; it will continue
sliding down the plane with constant speed, though.
FOLLOWUP QUESTION:
Thanks again but I am still a bit confused. It makes sense that the center of mass will move at a constant velocity while the cylinder is rolling, but how did it acquire that velocity in the first place if I start the cylinder at rest and not sliding as you wrote in the additional thought. In other words, what happens if the coefficient of static friction is equal to mgsinθ and the cylinder starts with no velocity of any kind?
ANSWER:
μ k = tan θ
(coefficient of static friction will be larger),
the cylinder will roll without slipping. If you solve the dynamics for
the cylinder rolling without slipping you will find that (see the figure
above) f= (mg sinθ )/3 and a =(2g sinθ )/3
where f is the frictional force and a is the
acceleration of the center of mass. Since f <mg sinθ ,
if you simply let it go, it will roll without slipping. So, if you want
it to slide down with constant speed, you must give it a shove to start
it slipping.
QUESTION:
Does mass of an object increase its mass exponentially as it approaches infinitely close to the speed of light?
ANSWER:
f /dq =Aq for some function f (q ),
where A is a positive constant, f is said to be
exponentially increasing with q . For your question, f
is m and q is v . The expression for m (v )
is m=m 0 /v 2 /c 2 ]
where m 0 is the rest mass. Calculating the
derivative, dm /dv =mv /[c 2 -v 2 ].
Thus, although m increases without bound as v
increases, it does not increase exponentially.
QUESTION:
Can sustaining enough angular momentum help us stay in a black hole for longer(near the event horizon)? If yes, how much energy would be required to sustain it for ten minutes.
ANSWER:
earlier answer ). This
is a result of general relativity. No stable orbits inside this exist
and anything inside would fall into the black hole.
QUESTION:
Hi! I have been told that hypothetically speaking when a metal bar or something other object travels in space at the speed of light it shrinks.
So my question is if it is correct and why yes or no?
ANSWER:
appear
to be shorter, they actually are shorter. You might look at an
earlier answer about length contraction.
QUESTION:
We have material that generates electricity when exposed to light, or force.
Why haven't not found one that does this when exposed to atomic radiation. Imagine almost permanent batteries.
ANSWER:
batteries which get their primary energy from radioactive decay.
Atomic batteries are routinely used in heart pacemakers and low-power
requirements in space probes.
QUESTION:
My question is about the theory of relativity and time dilation. It appears to only work in one direction and I'm not sure why. For example, take a person on earth and compare to a person in a rocket going near the speed of light. The person on earth observes the rocket man going near light speed and aging slower. The man in the rocket observes earth moving past him near the speed of light except the earth man ages quicker.
ANSWER:
faq page.
QUESTION:
Can a ball bounce higher than the height it was dropped? I know that air resistance would slow it down but is it possible?
ANSWER:
QUESTION:
An ice-hockey player throws his stick on the ice. The stick translates and
rotates. Before it stops, it always rotates and translates - never only
rotates or translates. Why do both motions occur simultaneously like this?
ANSWER:
FOLLOWUP QUESTION: uppose it isn't thrown intentionally in such a way so as to just make it spin or just make it translate.
ANSWER:
x-y plane) may be broken into two components,
motion of the center of mass (translation) and motion about
the center of mass (rotation). There are, in your example, no forces in
the x - or y -direction once the stick has been thrown
if you neglect friction. If the force you throw it with is directed
through the center of mass, it will not rotate because there is no
torque about the center of mass to get it rotating. If the net force you
throw it with is zero (you would need to use two hands to do this) it
will not translate. If you just randomly grabbed it and threw it, there
would be a net force which would result in the center of mass
accelerating during the time you were throwing it; and there would be a
net torque which would result in the stick having a rotational
acceleration during the time you were throwing it. Once you let go,
there is no force and no torque on the stick and so both linear and
angular momentum would be conserved meaning it would continue moving the
way it was moving when you released it.
QUESTION:
Suppose a container full of water is kept in a certain area. Let no external force
be applied. Now the question is: Will there be an overall circulation of molecules of water?
Or will the rest position of water perfectly 100% at rest?
ANSWER:
QUESTION:
If I were to drop two round balls of different mass under water would they both fall to the bottom at the same velocity or would one reach the bottom first?
ANSWER:
mg down; the
buoyant force B up which would be equal in magnitude to the
weight of an equal volume of water; and the drag force f up
which would depend on the size the ball and its speed. The net force
F would be F=-mg+B+f ; as the ball went faster and faster,
f would get bigger and bigger until eventually F =0 and
the ball would go down with a constant speed called the terminal
velocity. The larger the mass, the larger the terminal velocity. Without
specifying the the sizes of the balls, your question cannot be answered.
If they had identical sizes, the heavier ball would reach the bottom
first because it would have a larger terminal velocity. Of course, if
B>mg the ball would float!
QUESTION:
If you performed the double slit experiment in outer space and were observing the electron particles would you get a diffraction pattern or two rows of particles. Does the earth's atmosphere have any effect on the experiment?
ANSWER:
QUESTION:
Many Science Fiction ships have for protection a shield that stops projectiles and saves the ship from a lethal impact. Such an example would be the energy shields possessed by the faction known as Covenant from Halo.
So given the laws of Newton and the third law about there always being an equal force countering the force that was exerted first, wouldn't this just mean that whenever the energy shield takes a hit by a projectile, the force would ''travel'' from the shield to whatever generator generates the shield.
Like the UNSC Frigates shoot a 600 ton slug at 30km/s. They're 30ft long and around 7ft wide. So they have a lot of force and momentum behind them.
So wouldn't an impact like this just cause the shield generator to be violently thrown off its attachments and ''fly off'' through the compartments of the ship destroying a lot of the ship?
ANSWER:
"So they have a lot of force and momentum behind them." Yes, these
projectiles have a lot of momentum, but saying that "…there is a
lot of force… behind them…"
has no meaning. The momentum a slug has is p=mv =(6x105
kg)(3x104 m/s)=1.8x1010 kg m/s. When this hits
something, it will certainly exert a force, but the magnitude of that
force will depend on how long it took to stop. I have no idea how big it
is, but suppose it has a radius of 10 km from the ship; I will think of
it as very flexible and suppose that it stretches inward just stopping
before it hits the ship. I also will assume that the ship is much more
massive than the slug; (elsewise how could a comparably-sized ship carry
a bunch of them and fire them without huge recoil?) If it stops in 10
km=104 m with uniform acceleration, I can apply simple
kinematics, -3x104 =at and 104 =3x104 + ½at 2 ,
and find that t =1.5 s, the time for the slug to stop. The
average force felt by the slug is (Newton's second law) the rate of
change of the momentum, F =1.8x1010 /1.5=1.2x1010
N. You are certainly right, this very large force will be felt by the
ship because of Newton's third law. But suppose that the ship is 100,000
times more massive than the slug, 6x1010 kg; in that case,
the final velocity of the ship after being hit will be found from F =mv /t =6x1010 v /1.5=1.2x1010
N, so v =0.3 m/s, not so bad! If the shield were very rigid, it
would be catastrophe for the ship. I have never played these games but I
expect the shield is shown as stopping the slug almost instantly.
A little should be said about the other end,
launching the slug in the first place. In this case, unless the cannon
has a 10 km barrel, the recoil force on the ship will be huge. You would
be interested in similar Q&As
along the same line as yours.
QUESTION:
I'm trying to find a fast/easy way to test whether a sealed, consistently-dimensioned rectangular box is sufficiently "stable" for transport on a (small) 2-wheeled bicycle trailer. The box is pretty tall. If it's over-weighted and top-heavy, it'll flip the trailer around turns (which are sufficiently tight/quick). I figure there might be a quick, static "tip test" with a combination pull gauge, inclinometer and scale, but my math skills are primitive. Is there a simple way to ascertain whether, for a given object, a target stability threshold is met?
ANSWER:
earlier
answer about a bicycle making a turn. It would be helpful for you to
read that first. The easiest way
to do this problem of your trailer turning a curve is to introduce a fictitious
centrifugal force which I will call C mv 2 /R where
m is the mass of the box plus trailer, v
is its speed, and R is the radius of the curve. The picture to the right
shows all the forces on the box plus trailer: W is the weight and the green
x is the
center of gravity (COG) of the box plus trailer; f 1 and f 2 are the
frictional forces exerted by the road on the inside and outside wheels
respectively; N 1 and N 2 are the normal
forces exerted by the road on the inside and outside wheels respectively;
the center of gravity of the box plus trailer is a distance H above the road and the wheel base is
2L (with the center of gravity halfway between the wheels). Newton's
equations yield:
f 1 +f 1 =C
for equilibrium of horizontal forces;
N 1 +N 2 =W
for equilibrium of vertical forces;
CH+L (N 1 -N 2 )=0
for equilibrium of torques about the red x .
If you work this out, you find the normal forces
which are indicative of the weight the wheels support: N 1 =½(W-C (H /L ))
and N 2 =½ (W+C (H /L )). A few things to
note are:
the outer wheel supports more weight,
if C =0 (you are not turning), the inner
and outer wheels each support half the weight,
at a high enough speed C will become so large
that N 1 =0 and if you go any faster you will tip over, and
if the road cannot provide enough friction you
will skid before you will tip over.
Now we come to your question. You first want the
maximum speed without tipping. Solving for v in the N 1 =0
equation gives
v max = √[RWL /(mH )]=√[RgL /H ]
where g =9.8 m/s2 =32 ft/s2
is the acceleration due to gravity. For example, suppose that R =7
ft, L =17 in=1.42 ft, H =30 in=2.5 ft. Then
v max = √[7x32x1.42/(2.5)]=11.3 ft/s=7.7
mph.
Be sure to note that the assumptions of a level
road (not banked) and wheels not slipping are used in my calculations.
Also be sure to note that W is the weight
of both box and trailer and 2L is the wheel base, not
the box width.
One more thing is that you might not know how to
find the COG of the trailer plus box. If the COG of the trailer is H trailer
above the ground (probably close to the axle) and the COG of the
box is H box above the ground , then H =(H box xW box +H trailer xW trailer W .
ADDED
THOUGHT:
N 2 =W
and f 2 = μN 2 =μ W ,
where
μ is the coefficient of static friction. If you work it out,
the minimum value the μ must have to keep the trailer from
skidding is μ min =L /H . For the
example worked out above, μ min = 0.57. For
comparison, μ for rubber on dry asphalt is about
0.9, so the trailer would not skid.
QUESTION:
ANSWER:
Fusion requires
bringing two charged nuclei close together and the energy required to
bring them close enough together to fuse is proportional to the
product of their charges. Nitrogen would require 49 and oxygen 64
times the energy to fuse compared with hydrogen.
T he
thermonuclear device is designed to compress and confine the fusion
fuel using the fission device energy to do this. If you just had a
fission bomb, the surrounding air molecules would be blown apart from
each other rather than compressed.
There is also
efficiency to consider. In hydrogen fusion, about 1% of the mass of
the fuel is converted to energy but for oxygen or nitrogen, only about
0.1% is.
Finally, you have to ask
whether the desired reaction will go at all and what its probability
(cross section) is. My guess is that there would be a very low
probability for 14 N+14 N —>28 Si
to occur.
QUESTION:
ANSWER: faq page.
QUESTION:
ANSWER: earlier worked out the
velocity of something which would correspond to occupants of your spaceship
experiencing a force equal to their own weight due to the acceleration which
I will adapt to your case later. (See the graph above.) First, though, I
will work out the (incorrect) Newtonian calculation which is presumably what
you want. The appropriate equation would be v=at where v =3x108
m/s, a =1.2g =11.8 m/s2 , and t is the time to
reach v ; the solution is 2.5x107 s=0.79 years. For the
correct calculation, you cannot reach the speed of light; from the graph
above (black curve), though, you can see that you would reach more than 99%
of c when
gt /c =3.
To make this your problem, we simply replace g by 1.2g and
solve for t . I find that t =7.7x107 s=2.4 years.
QUESTION:
ANSWER: ψ*ψ over all
space is then the probability of finding the particle somewhere which, of
course, must be 1.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
this article .
QUESTION:
ANSWER:
FOLLOWUP QUESTION: this site
it gives me force on the car 7. Tones with just the weight of car one and 2
meters of course it refers to 1 car hitting a tree, but the force of impact
must be close.
Does it sound reasonable to you?
ANSWER:
FOLLOWUP QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER: v after the collision has finished of
the cars may be estimated from momentum conservation: 1400x11=(1400+1100)v ,
so v =6.2 m/s. The average force on car B is the rate of change in
its momentum F =(6.2x1100)/t= 6800/t where t
is the time of the collision. Since we know the acceleration a is
6.2/t , we can use the kinematic equation for distance to get the
time: 0.25= ½at 2 =3.1t , so the time
of collision is t =0.081 s; therefore F= 84,000 N=19,000
lb=8.5 ton.
QUESTION:
ANSWER:
earlier answer .
QUESTION:
ANSWER: —regardless
of how it interacts with something else, it never speeds up nor slows down;
the speed of light in a vacuum is a constant of nature. If you look on my
faq page you will find links to discussions regarding
why the speed is constant and why it has the value it does. When you throw a
ball up in the air, it slows down; you throw it down from a tall building,
it speeds up. Photons don't do that but they do change their energies by
changing to a longer, redder (shorter, bluer) wavelength when going up
(down) in a gravitational field.
QUESTION:
ANSWER: 12 J. The energy of the Nagasaki bomb, a small
bomb by today's standards, was about 1014 J, 100 times bigger.
Two things to consider are:
T he
bomb number represents total energy delivered whereas I would guess
that likely less than half the energy content of the gasoline would
actually be delivered.
The time over
which the energy is delivered is likely much longer for the gasoline
explosion than the nuclear explosion. As an example, just to
illustrate the importance, suppose the bomb exploded in 1 ms=10-3
s and the tank in 1 s. Then the power delivered by each is 108
GW for the bomb and 103 GW for the tank. As a result the
destructive power of the bomb would be much bigger.
QUESTION:
ANSWER: fictitious force .
So, for you in the accelerating car we add a force, let's call it the
"impulsive force" in the forward direction, which is your mass times the
magnitude of the car's acceleration. You can then say that you accelerate
forward with the same acceleration as the car is accelerating backward
because you are being pushed by the impulsive force.
Centrifugal force is a fictitious force. If you are on a
merry-go-round you feel a force trying to push you off; there is no such
force. But if you pretend there is, you can do use Newton's first law:
there is a centrifugal force outward which is mv 2 /R
and you are at rest; the force which is keeping you from accelerating
outward is friction with the floor, so it must be equal to mv 2 /R.
If someone on the outside analyzes the problem she will say that you
have an acceleration v 2 /R inward so there
must be a force (centripetal) inward to provide that acceleration, F =mv 2 /R
which is the same frictional force we found on the merry-go-round. Clearly
the centrifugal and centripetal forces in these two cases are acting on
the same body —you—so your claim that texts say
otherwise is strange. In 40 years of teaching I have never heard such a
claim. It is true that both centripetal and centrifugal forces never act
in the same solution to a problem; if you choose a person on the
merry-go-round to analyze, you never have both a centripetal and
centrifugal force acting on that person. Perhaps you misunderstood and
that is what is meant by never acting on the same object.
You can learn more about
fictitious forces in an earlier answer and the links it leads to.
QUESTION:
ANSWER: earlier answer for an
explanation. (Note that the earlier answer has acceleration rather than
braking.) The air balloon will move forward because it feels no buoyant
force and therefore its inertia will cause it to want to crash into the
windshield.
QUESTION:
ANSWER: 18 J of energy to the collision
which is equivalent to about 10,000 Nagasaki atomic bombs. Not only would
the man and elephant be obliterated, also anything within miles would be.
QUESTION:
What exactly is momentum and how is it different from force?
I understand that p = mv and F = ma but they seem so similar in application that I haven't fully wrapped my head around it.
Additionally, what does it mean for something to have momentum in the first place, and why must it be conserved?
How does light have momentum if it, by nature, has 0 mass (I presume E =mc^2 comes into play somewhere here)?
Finally, what is the significance in the fact that light does have momentum (if it does)?
ANSWER:
For your first question, just say that p=mv .
My answer to #2 should clarify how force and linear momentum are
related. Force and momentum have to be different because they are not
even measured in the same units—momentum is mass*length/time and
force is mass*length/time2 .
What is acceleration? It is rate of change of velocity,
a= (v 2 -v 1 )/t
where t is the time to change speed from v 1
to v 2 . So one could write that F= (mv 2 -mv 1 )/t=(p 2 -p 1 )/t ;
so force may be thought of as the rate of change of momentum. Newton
actually stated his second law this way, not as F=ma . It is
the second law which is a fundamental law of physics, momentum is just
defined because of its simple and natural relationship to the second
law.
It doesn't "mean" anything for something to have
momentum, it is just a definition. However, consider the second law if
there is no force acting; then F =0=(p 2 -p 1 )/t.
In other words, p 2 =p 1
which simply means that momentum does not change (is conserved). The
condition for conservation of momentum of a system is that there be no
external forces on it. For example; suppose you look at an isolated
galaxy which has billions of stars in it all interacting with each
other and there are negligible forces from the outside; if you sum up
all the momenta of all the stars today and in 10 years from now, you
would get the same answer even though the shape and orientation of the
galaxy would change.
It turns out that if v is very large,
comparable to the speed of light c , Newtonian mechanics is
incorrect. (You could say that Newtonian mechanics is wrong but a
superb approximation for low speeds.) If you say that p=mv it
turns out that momentum is not conserved for an isolated system at
very large speeds. However, since momentum conservation is such a
powerful way to solve problems, we redefine momentum (in the theory of
special relativity), to be p=mv /√(1-(v 2 /c 2 )),
momentum is conserved again and we still have p ≈mv
for small v . It also turns out that, as a result of this new
definition of p , we can write that E =√(p 2 c 2 +m 2 c 4 )
where E is the energy of a particle of mass m . So if
the particle is at rest, p =0 and E=mc 2 ;
if the particle has no mass, p=E /c .
No more
significant than if a billiard ball has momentum—it just does.
I have deleted
your question about angular momentum—it is off topic.
QUESTION: article
about relativity. The article stated that the "The combined speed of any object's motion through space and its motion through time is always precisely equal to the speed of light." For an object moving at 90% the speed of light, it should only be moving at 10% the speed of light through time. I assume that means that time should pass only at 10% the actual speed of time for the object (correct me if I am wrong). However based on the Lorenz factor, time passes at 0.43 the actual speed of time for the object. Why is this so?
Furthermore when time dilation occurs it is only seen by someone else in a stationary frame of reference. In the moving object's frame of reference time does not slow down at all. Does this mean that the combined speed of the moving object's motion through space and time can be more than the speed of light in the moving object's frame of reference? Can the moving object go beyond the speed limit in its frame of reference?
ANSWER: To get a fuller sense of what Einstein found, imagine that Bart has a skateboard with a maximum speed of 65 miles per hour. If he heads due north at top speed—reading, whistling, yawning, and occasionally glancing at the road—and then merges onto a highway pointing in a northeasterly direction, his speed in the northward direction will be less than 65 miles per hour. The reason is clear. Initially, all his speed was devoted to northward motion, but when he shifted direction some of that speed was diverted into eastward motion, leaving a little less for heading north. "
The speed in the northward direction will now be v N =65cos450 =46
mph; his velocity in the eastward direction will be v E =65sin450 =46
mph. His total velocity will be √(v N 2 +v E 2 )=65
mph, not (v N +v E x and t for the
stationary observer and x' and t' for the position and
time of the moving observer as seen by the stationary observer. The catch,
though, (which I presume is why the author of your article avoided these
details) is that time and space are slightly different from space (N) and
space (E) in that to get the length of the vector you calculate the square
root of the difference of the squares instead of the sum. So the speed
through spacetime would be √(c2 -v2 ).
So, the stationary observer will see a "spacetime speed" of v spacetime =√(c2- v2 )
and the "space time distance traveled" by the moving observer as seen by the
stationary observer would be d spacetime =v spacetime t'
where t'=t / √(1-(v 2 /c 2 )).
If you do your algebra you will see that d spacetime =ct
indicating that the stationary observer sees the moving observer having
a speed of c through spacetime.
QUESTION:
ANSWER: R =6.4x106 m is the radius of the earth, d =1
in=0.024 m is your 1 inch, θ is the angle which
subtends the arc length, call it s , you seek.
From the triangle you can write cosθ =R /(R+d )=[1+(d /R )]-1 ≈1-(d /R )+…
where I have done a binomial expansion of [1+(d /R )]-1
because (d /R ) is extremely small. Now, because
θ is also very small, I can represent cosθ by
the expansion cosθ≈ 1-½θ 2 +…
so (d /R )≈½θ 2 .
Finally we can write that θ=s /R. If you now solve
for s you will find s ≈554 m.
QUESTION:
ANSWER:
QUESTION:
ANSWER: y -direction,
B =j BE q is in the x -direction, v =i vF k qvB
in the z -direction. In the moving frame the new fields would be
B' =j γB and E' k γvBγ= 1/ √(1-(v /c )2 ).
Note that
E' =v x B' F' E' v x B' k γqvBF' ≠F
γ ; this is because force is said to be not
Lorentz invariant
and it is not really a useful quantity in relativity.
QUESTION:
ANSWER: 8 m/s. So, nothing has to "…keep that
photon moving forward…" But, space is not really a vacuum and not
100% of photons make it over such vast distances. Some interact with the
occasional molecule they might encounter and are changed or scattered. Some
find themselves striking a star or planet and being removed from the stream
coming from the source. Some are deflected by the strong gravity near
galaxies. Some encounter a black hole and disappear altogether. Because
space is so empty, though, the majority come through unchanged.
QUESTION:
ANSWER: —your
photons.
Q&A OF THE WEEK, 2/5-2/11 2017
QUESTION:
ANSWER:
the rocket
always goes straight up;
the rocket
stops moving vertically when the rope is taught;
the rope
is cut the instant that the rocket stops;
fuel and
the weight of the rope are not issues; and
the launch
is from the equator. This makes things much simpler and I will briefly
talk about a similar launch from some latitude at the end.
The
thing to appreciate is that even though the rocket goes straight up, it
will have the same angular velocity ω as the earth so its
speed will be ω (L+R ) where L is the
length of the rope and R is the radius of the earth. The angular
velocity is ω =[(2π radians)/(24 hours)]x[(1 hour)/(3600
seconds)]=7.27x10-5 s-1 . If L is just
right, the rocket will assume an orbit like the
geosynchronous communication satellites ; this turns out to be if L =5.6R .
So, if the rope happens to be 5.6 times larger than the radius of the
earth, the rocket will remain (apparently) stationary above its launch
point; it is actually going in a circular orbit with a period of 24 hours.
(The animation above illustrates this, although not to the correct scale.)
For any other L the orbit will be an elipse with the center of
the earth being at the focus. Visualize .
If L >5.6R , the starting
point of the rocket will be the perigee (closest point to the earth) of
its orbit. Visualize .
If L <5.6R , the starting point of the rocket
will be the apogee (farthest point from the earth) of its orbit.
Visualize .
For L <5.6R , though, there will be some critical distance
L=Lc when the perigee of that orbit is
exactly equal to R; in that case the orbit will just skim the surface of
the earth. After some laborious algebra I found that L c ≈3.7R ,
about 2.2 earth radii inside the geosynchronous orbit.
Visualize .
For L<Lc ,
the rocket will crash back into
the earth but not where it was launched from because it is a projectile
which has a horizontal speed greater than that of the earth's surface.
Visualize .
Finally,
if the horizontal speed of the rocket ω (L+R )
is greater than or equal the escape v e =√[2MG /(R+L )],
where M is the mass of the earth, the rocket will escape the
earth and never come back. I calculated this to be when L =7.6R ,
two earth radii beyond the geosynchronous orbit.
Visualize .
For
any other latitude θ the speed of the satellite will be ω (L+R )sinθ.
The resulting orbits, while all elliptical, are much more difficult
to visualize and maybe we should save that for another day! However, the
simplest launch of all would be from the north or south pole (θ= 00
or 1800 ) because it acquires no horizontal velocity (ω (L+R )sinθ= 0)
where the rocket would fall straight back down regardless of how high
it went.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: ice . You can get a clearer
picture there. The thing to notice is that all the ices except for ice I and
ice XI occur at extremely high pressures —1000 atmospheres or
higher. So, you cannot really "look" at them and say what they look like
since they are formed in a containment vessel of some sort. Ice XI occurs at
atmospheric pressure but only at very low temperatures. If you look at the
Phases section of the
article, you will find links to separate articles for all 16 forms of ice
where you can get information about their properties. Although the
terminology for the crystal structure is pretty "science speak" as you say,
you will find usually pretty comprehensible pictures of these crystal
structures. I think you would find these articles about as understandable as
you will find.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: ≈2.8
W. The gym would probably consume a few kilowatts so you would probably need
a couple of thousand folks to generate that much power. I don't think you
are on to something here!
QUESTION:
ANSWER:
QUESTION:
ANSWER: a=F /m where a is the acceleration
of an object of mass m when acted on by a force F . Suppose
you exert a 1 N force on an electron (m ≈10-30 kg)
and a space ship
(m ≈106 kg). Then the acceleration of the
electron (initially) is 1030 m/s2 and that of the
space ship is 10-6
m/s2 .
QUESTION:
ANSWER: v there
is always another going in the opposite direction with speed v . The
average velocity of the air in the airplane as measured by an observer on
the ground would be 500 mph. It could be thought of as like a wind in which
there is a net flow of air in the direction the wind is blowing.
QUESTION:
ANSWER: twin paradox and for how fast
clocks appear to run.
QUESTION:
ANSWER: d /t is the average velocity,
not the starting velocity; the velocity is changing the whole time. In this
case, uniform acceleration, the average velocity is ½v 0
so your acceleration is wrong by a factor of 2; a =-3.12 m/s2 .
The correct expression for the stopping distance is d =½v 0 2 /a
where v 0 is the initial velocity and a is the
magnitude of the acceleration. The force which stops the car depends only on
the nature of the rubbing surfaces and the weight of the car. But the weight
of the car cancels out when you calculate the acceleration, so the
acceleration is independent of both the weight and the initial speed.
Therefore your guess that you can use one datum to determine the
acceleration and then use that for all other speeds is correct; the added
bonus is that it is also independent of how heavy the car is as long as it
is the same road surface with the same tires.
QUESTION:
ANSWER: earlier answer .
FOLLOWUP QUESTION:
ANSWER: earlier answer .
Finally, there are two places on earth where the fictitious forces are
zero and therefore the two objects will move identically—at the
north and south poles.
QUESTION:
ANSWER:
QUESTION:
ANSWER: r 2 . Because a black hole is a
singularity (zero size) it is possible to get very close to the center where
the field will be huge. Also, many black holes are very much more massive
than a typical star. A regular star might have a very big mass, but it is
spread out over a very large volume; inside the star itself the field
decreases linearly to zero at the center.
QUESTION:
ANSWER: …like it would here on earth…"
QUESTION:
ANSWER: p
is not conserved if momentum is defined as p=m 0 v
where m 0 is the inertial mass at rest. If you
redefine p to be p=m 0 v / √(1-(v 2 /c 2 )),
momentum of an isolated system is constant. You can interpret this as simply
a redefinition of p or that the mass has increased so that
m =m 0 /√(1-(v 2 /c 2 )).
That said, let's answer your question assuming that mass increases.
Length contraction occurs only along the direction of
motion. Since mass increases by a factor
1/ √(1-(v 2 /c 2 ))) and
volume decreases by a factor of
1/ √(1-(v 2 /c 2 )),
density increases by a factor
1/ (1-(v 2 /c 2 )).
QUESTION:
ANSWER:
QUESTION:
ANSWER: 1
and 2 ) for more
details. Keep in mind that all cats are not the same size, weight, or
fuzziness, so this can only be an approximate answer. It is interesting that
about 10% of cats falling from 5 stories are killed but fewer from higher
stories because, once reaching terminal velocity, they can relax, spread out
(to slow down) and get ready to hit.
QUESTION:
ANSWER: d to specify the length of the side on
any particular tile. However, due to some fundamental law of the universe, none of
these tiles is allowed to have to have d>d max =4 cm. Now
suppose that I write an equation for the area of any of these tiles, A=d 2 .
Now one of these tiles has a size of d =3 cm (not violating our
fundamental law) and so its area is A =9 cm2 . But wait,
how can that be? Since d 2 is more than twice as large as
d max , this must violate our fundamental law of the universe, right? Of
course not. The area of a square is simply a different thing from the length
of the side of a square and does not even have the same units (cm vs.
cm2 ).
QUESTION:
ANSWER: earlier answer .
QUESTION: V 1 in one reference system and
V 1
+ V' in another one. Now let's say this body starts slow down and its
velocity change is ΔV in both systems. Simple algebra shows that the
kinetic energy change in the 1st system will be
ΔE 1 = m ΔV /2(2V 1 +ΔV )
and in the 2nd system
ΔE 2 = (m ΔV /2)(2V 1 +ΔV +2V' )
i.e. its different in different reference systems. Now if we imagine that
this slow down and kinetic energy change was due to the friction it follows
that the amount of heat released is different for different reference
systems! Nonsence! It should be a mistake somewhere.
ANSWER: v has kinetic energy K = ½mv 2
but zero kinetic energy in a frame moving with it. Therefore, the change
in kinetic energy will be different in different frames. However, the
work-energy theorem (ΔK=W where W is work done by
external forces) will be true in all frames, just not the same numbers in
all frames.
In the first frame in your example,
where m has speed v 1 , the floor is, I assume,
at rest. To make the algebra simpler and the explanation easier to grasp,
I am going to let m come to a complete stop, Δv=-v 1 .
Then, when m moves some distance s and the frictional
force is f , the work done is W=-fs and so ΔK=- ½mv 1 2 =-fs .
You can deduce that the heat generated is ½mv 1 2 ;
all other frames will see this amount of heat. Now, in a frame with speed
v' moving in the same direction as v , you can easily
show that ΔK'=- ½mv 1 2 +mvv' .
You already know that - ½mv 1 2
is the heat and mvv' is simply the amount that ΔK
is different in this frame. If you calculate the work done in this frame,
it is different because the distance m slides is different; but,
if you do that you will find, as expected that W =- ½mv 1 2 +mvv' .
One more special case. Suppose
v' =½v 1 . You should be able to convince yourself that
m originally is catching up with the moving frame but then stops
and turns around and ends up where it started from; thereafter it moves
with constant speed v' . The total distance it has traveled in the
moving frame is zero and so W =0 and therefore ΔK'= 0.
So the heat is not anywhere evident here even though you would feel it. In
this case you would find W =- ½mv 1 2 +½mv 1 2 .
Now you can see where that heat is hiding, right?
So, if you want
to most conveniently determine the amount of heat generated, you apply the
work-energy theorem in the frame where the surface is at rest. This is
because to the moving observer the surface is moving as well as m . The
moving observer has to correct for that motion by calculating m 's motion relative to the surface.
m 's motion relative to the second observer is not
really relevant in calculating the heat generated.
Thanks to R. M.
Wood and A. K. Edwards for helpful comments.
QUESTION:
ANSWER:
W 1 is the weight of
the parent,
W 2 is the weight of
the core,
W 3 =W 1 -W 2
is the weight of the paper in the parent,
R 1 is the radius of
the parent,
R 2 is the radius of
the core,
L is the
length of the roll,
V 1 =πL (R 1 2 -R 2 2 )
is the volume of the paper on the parent,
W is the weight of the
paper on the partially used roll,
R
is the radius of
the partially used roll, and
V=πL (R 2 -R 2 2 )
is the volume of the paper on the
partially used roll.
You may be interested only in the final
answer, but I will outline the solution for anyone interested.
The
density of the paper is ρ=W 3 /V 1 =W /V .
Therefore, W=W 3 (V /V 1 )=W 3 (R 2 -R 2 2 )/(R 1 2 -R 2 2 ).
What you
want, is the weight of the whole partially used roll, W+W 2 =W 2 +W 3 (R 2 -R 2 2 )/(R 1 2 -R 2 2 ).
So, just
plug in your measured values and you will have the weight of the partially
used roll. Incidentally, you say "distance from the edge of the core". If
this (call it D) is more conveniently measured you can replace R
in the equation above by (D+R 2 ) which results in
2 =W 2 +W 3 (D 2 +2DR 2 )/(R 1 2 -R 2 2 ).
QUESTION:
IDEA:
A distant planet picks up broadcasts from Earth, so the inhabitants have some knowledge of us, and know that life is out there. A "man" on the alien planet gets part of his consciousness encoded into a light beam that is sent out into space.
When it hits earth -It is able to project itself and appear to be human. When the light fades the man vanishes, leaving a message that he is long dead. (It's a love story.) But the question is what broadcasts would he have received time wise. His knowledge of earth would be limited, ending at a certain time...
This is the end/explanation.
Thank you in advance for your help!
"
My people developed a holographic beam so powerful it could be projected through space. Through light-years. It is more advanced than anything on Earth, so advanced that some actual particles of my being were intertwined with the particles of light being projected. So it was like I was actually here, next to you, talking to you, laughing with you. But I was not. I am not.
When we sent the beam of light traveling into space my particles were woven into it, not all of them, of course. But enough particles of my consciousness were intertwined with the particles of light so that I could interact with anyone I discovered. ...So that I could learn more about the world we'd previously only known through television broadcasts. ..
I traveled in a spaceship made of light. Now that the light has died so have the particles of my consciousness that were interwoven with the beam.... The real 'me,' the corporeal 'me,' died HOW MANY? light years ago."
Long ago, men went to sea. And women waited for them, peering out into black waters, searching for a tiny speck of light on the horizon. Now I too wait, looking out into the vast blackness of space, searching for my love, my heart, HOW MANY? light-years dead.
ANSWER:
One other detail which probably isn't that important would
be you could not really carry 'particles' of
consciousness (whatever that might be!) on a light beam if they had any
mass at all. Nevertheless you could envision the information regarding his
'consciousness' could be carried by light.
ADDED
THOUGHTS: en
route and incorporated into his consciousness somehow; other
timing would be pretty much the same. Or, if you simply put him on that
spaceship, he would age only about half a year during the trip which the
earth would see as 99.999 years. But then you would not have your dramatic
sad ending!
QUESTION:
ANSWER: t =1 s and
the second at t =6 s, an elapsed time of 5 s.
FOLLOWUP QUESTION:
ANSWER: 8 m high to keep the numbers from becoming
impossibly small. Indeed, general relativity tells us that the stronger a
gravitational field is, the more slowly a clock will run. This is called
gravitational time dilation (and is directly related to gravitational red
shift). The rate at which a clock runs slower in a gravitational field (MG /r 2 )
is √[1-(2MG /r )] where M is the mass of the
source (the earth in your case), G =6.67x10-11 N·m2 /kg2
is the universal gravitation constant, and r is the distance from
the center of the source. If some time t elapses on a clock in zero
gravitational field, the time which will elapse for other clocks is t /√[1-(2MG /r )].
For our purposes I will assume that the zero-field clock will measure 1 s
for a pulse traversing your building.
After
doing some arithmetic I find that if t elapses on the
zero-field clock, t'=t [1-(4.4x10-3 /r )]
elapses on another clock. So, for the first pulse where both clocks
have begun at t' =0, the elapsed time of the two clocks will
be t' bottom =(1-6.9x10-10 ) s
and t' top =(1-1.5x10-11 ) s. The upper
clock will will measure a shorter time by about 6.8x10-10
s=68 ns.
Now, we
need to specify who determines when 5 s have elapsed. The most
sensible choice would be the guy at the bottom of the building who
will be triggering the laser. When his time is t' bottom =5
s, the zero-field time is t =5/(1-6.9x10-10 )≈(5+3.5x10-9 )
s and (this is tricky) the top clock will measure t' top =(5+3.5x10-9 -6.8x10-10 x5)=(5-1.0x10-10 )s.
We know
from the first pulse the additional time on both clocks, so we can now
calculate what each clock reads when the second pulse hits: t' bottom =(5+1-6.9x10-10 )=(6-6.9x10-10 )
s and t' top =(5-1.0x10-10 +1-1.5x10-11 )=(6-1.2x10-10 )
s.
The top clock
will see 57 ns more time elapse than the bottom clock. I believe that if
you use the 250 m tall building, things would simply scale and time
differences would be 250/3x108 =8.3x10-7 times
smaller.
QUESTION:
ANSWER: r cm of the
center of mass for a collection of point masses is defined as r cm =[ Σm i r i ]/Σm i
where r i is the position of the point
mass m i . If you have a continuous mass distribution with
mass density ρ (r
r cm : r cm =[∫r ρr r ρ (r r
QUESTION:
ANSWER:
F hand , your belt pulls down with a force F belt ,
and the earth pulls down on the scale (its weight) with a force W scale .
These three have to add to zero which means that F hand =F belt +W scale .
Now, the scale is calibrated so that it reads zero when F belt your weight, never
appears; everything would be just the same if you weighed 1000 lb. Regarding
your second question, it is really no different from the first question:
simply replace 'belt' by 'bar' everywhere. In either case you would observe the scale reading your weight when you pulled up with a force equal
to your own weight.
FOLLOWUP QUESTION:
ANSWER: was folly, total folly! Doing this experiment gives
you absolutely no information about your weight. Only if you already knew
your weight and were able to pull with a force that hard would the scale
read your weight.
QUESTION:
ANSWER:
A voltage large
enough to drive a large enough current through your heart will kill
you.
A
conductor moving through a magnetic field will have a voltage induced
across its ends. (Details depend on direction of the field and the
direction of the velocity of the conductor. Just assume that you have
a straight wire moving perpendicular to the field.)
So the wire
will have a voltage across its ends and will therefore look like a
battery. Similarly, the body will have a voltage of the same polarity (say
positive at the mouth and positive at the end of the wire in your mouth)
but with a much smaller voltage than the wire; this would create a circuit
looking like two batteries, one weak and one strong, which would drive a
current through the weaker battery (your body). So, the idea works if you
move fast enough. But, the earth's magnetic field is really weak and the
speeds would be impossible to achieve without burning everything up in the
atmosphere I would bet. Let's just do a rough calculation. The voltage
V is about V=BLv where B is the field, L
is the length, and v is the velocity. I will take L =1 m,
v =18,000 mph≈8000 m/s (low earth orbital speed), and B ≈5x10-5
T. So the voltage would be less than half a volt! Have I been colloquial
enough for you?
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: FAQ
page.
QUESTION:
ANSWER:
QUESTION:
ANSWER: f . Initially
an energy of hf is removed from the star. But, by the time that
the photon is very far away the star, it has lost some amount of energy Δhf ,
so the net loss to the star and its field is
hf -Δhf . At the instant the photon is created, the
mass of the star is reduced by hf/c 2 . But when the
photon, which is losing its energy to the field, is far away, the energy
of the field will have increased by Δhf /c 2 ;
I would not use the terminology "mass of the field" since the field has
energy density, not mass. But now, it seems to me (not a cosmology/general
relativity expert) that there is a bit of a paradox: we always associate a
given mass with a particular gravitational field, so the field should have
the energy content associated with a mass M'=M -hf/c 2 .
But, in fact, the field would have energy content associated with a mass
M'=M -hf/c 2 +Δhf /c 2 .
I am guessing that the field and the mass somehow "equilibrate" so that the
final mass of the star is consistent with the energy of the final field. (I
could easily be wrong! Perhaps it is only meaningful to look at the total
energy of the star and its field. Whatever the case, I would not talk about
mass of the field.)
QUESTION:
ANSWER:
QUESTION: Hawking radiation trapped in a black hole's gravitational pull and eventually sucked right back into the black hole?
ANSWER: Schwartzschild
radius . One particle of the pair is outside that radius and can
therefore escape.
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER:
Energy conservation:
Incorrect: the total energy of any system is conserved. In fact
you can change the energy of a system by doing work on it.
Correct statement: the total energy of any system is conserved if
no external forces do work on it.
Electric
charge conservation:
Incorrect: The
total electric charge of a system must remain constant. In fact,
you can always add or subtract charge from a system.
Correct statement: The total electric charge of an isolated system
must remain constant.
These laws are laws because they
are always found to be true and it is rather pointless to ask "what if".
If energy conservation were not true, then energy could suddenly appear
out of nothing.
Only if there
is some overriding physical law could another be broken, but only within
certain constraints. For example, the Heisenberg uncertainty principle
says that it is impossible to precisely know the energy of a system for a
short enough time, ΔE Δt ~10-34 J⋅s
where ΔE is the amount by which you change the energy of a
system and Δt is the time during which the energy is
changed by that amount. For example, 10-20 J could appear out
of nothing for as long as 10-14 s; after the time had elapsed,
though, it would have to disappear again.
QUESTION:
ANSWER: strobe effect . The frequencies
of light are enormously larger, hundreds of THz (1012 Hz).
QUESTION:
ANSWER:
QUESTION:
ANSWER: d in the figure here which is easy to calculate.
If you go 8000 m along the surface, that is the arc length subtended by the
angle θ . Therefore, θ=8000/R =8000/6.4x106 =1.25x10-3
radians=0.0716º. Finally, you can write cosθ =R /(R+d )
or d =R (1-cosθ )=5
m.
QUESTION:
ANSWER:
QUESTION: g=GM /d 2 then why does acceleration due to gravity decreases as we go deeper into earth?
ANSWER: outside a
spherically-symmetric mass distribution. If you go inside the mass
distribution, the equation is g=GM' /d 2 where
M'
is the mass inside d .
QUESTION:
ANSWER:
This one is probably the best. Another good one to read is
this one . So, to understand current you have to appreciate that it is
statistical; there are such a huge number of electrons moving that all it
makes sense to talk about is the average velocity of the electrons which is
referred to as the drift velocity . When a collision with an atom
occurs, kinetic energy is turned into heat energy. On average, electrons
have constant kinetic energy but lose potential energy as they move along in
the electric field. This energy lost to heat is supplied by the power source
which is why you have to change batteries now and then or burn coal in a
power plant somewhere to supply the energy. The drift velocity is, as
explained in the second reference above, extremely slow, not the speed of
light. When you turn on the potential difference, the electric field
establishes itself along the wire at about the speed of light which is why a
light seems to come on instantaneously.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: m originally separated by a distance
r is v escape =2 √(mG /r ) where G =6.67x10-11
N·m2 /kg2 is the gravitational constant.
QUESTION:
ANSWER: U at the
highest point must be equal to that added by the jump, and U=mgh
where m is the mass (400 lb), h is the height (2 in on earth), and
g is the acceleration due to gravity (9.8 m/s2
on earth, 1.6 m/s2 on the moon). The mass is the same both
places, so h earth g earth =h moon g moon . Solving,
h moon =12.25 in.
QUESTION:
ANSWER: ≈26
m high; the student survived without serious injury. The second person also
wanted to know if I could estimate the force experienced on impact. First I
will calculate the speed he would hit the ground if there were no air drag.
The appropriate equations of motion are y (t )=26-½gt 2 . and
v (t )=gt where y (t ) is the height above the ground
at time t , and g is
acceleration due to gravity which I will take to be g ≈10
m/s2 . The time when the ground (y =0) is reached is found
from the y equation, 0=26-5t 2 or t =√(26/5)=2.3
s. Therefore v =10x2.3=23 m/s (about 51 mph). The terminal velocity of a falling
human is approximately 55 m/s, more than double the speed here, so the
effects of air drag are small and can be neglected for our purposes of
estimating. (If there is air drag, terminal velocity is the speed
which will eventually be reached when the drag becomes equal to the weight.)
Estimating the force this guy experienced when he hit the ground is a bit
trickier, because what really matters is how quickly he stopped. Keep in
mind that this is only a rough estimate because I do not know the exact
nature of how the ground behaved when he hit it. The main principle is
Newton's second law which may be stated as F=m Δv /Δt
where m is the mass, Δt is the time to stop, Δv= 23
m/s is the change in speed over that time, and F is the average force
experienced over Δt. You can see that the shorter the time,
the greater the force; he will be hurt a lot more falling on concrete than on a
pile of
mattresses. I was told that his weight was 156 lb which is m =71 kg
and he fell onto about 2" of pine straw; that was
probably over relatively soft earth which would have compressed a couple of
more inches. So let's say he stopped over a distance of about 4"≈0.1 m. We
can estimate the stopping time from the stopping distance by assuming that
the decceleration is constant; without going into details, this results in
the approximate time Δt ≈0.01 s. Putting all that into
the equation above for F , F≈71x23/0.01=163,000 N≈37,000
lb. This is a very large force, but keep in mind that if he hits flat it
is spread out over his whole body, so we should really think about
pressure; estimating his total area to be about 2 m2 , I find
that this results in a pressure of about 82,000 N/m2 =12 lb/in2 .
That is still a pretty big force but you could certainly endure a force of
12 lb exerted over one square inch of your body pretty easily.
Another possibility is that the victim employed some variation of the
technique parachuters use when hitting the ground, going feet first and
using bending of the knees to lengthen the time of collision. Supposing
that he has about 0.8 m of leg and body bending to apply, his stopping
distance is about eight times as large which would result in in an eight
times smaller average force, about 5,000 lb.
Falling from a third story window (about 32 feet, say) would result in a
speed of about 14 m/s (31 mph) so the force would be reduced by a factor
of a little less than a half.
ADDED NOTE:
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: d . The magnitude of the magnetic
field B of a monopole is inversely proportional to the square of the
distance r from the charge, B=kq /r 2 where
k is some constant. In the drawing above the field at point p is B=B -q +B +q =kq [(1/(r-d )2 -(1/r+d )2 ]=4kqrd /[(r+d )2 (r-d )2 ].
Now, look at the field when r>>d : B ≈4kqd /r 3 .
The field does not look like a monopole because it falls off like 1/r 3 ,
not 1/r 2 .
QUESTION:
ANSWER: M =15 kg and F =45 N and therefore a=F /M =3
m/s2 . Next I choose B as the body. The only force on it is the
force which A exerts on it, F BA . Since we know that m B =10
kg and a B =3 m/s2 , F BA = m B a B =30
N. Finally choose A as the body. Two forces act on A, F= 45 N and the
force which B exerts on it, F AB =-F BA =-30
N; its mass is m 5 =5 kg and therefore a =(F+F AB )/m 5 =(45-30)/5=3
m/s2 . The reason that there is not a 45 N force on B is because
your finger is not touching B, only A is touching B. The reason A "knows" to
exert a force on A is that it has no choice since A's acceleration and mass
are already fixed. Once you know the "reaction" force you automatically know
the "action" force because of Newton's third law, F AB =-F BA =-30
N for this problem. You could also have chosen A as the body before you
chose B as the body. Two forces act on A, F= 45 N and the force which
B exerts on it, F AB . Its mass is m 5 =5 kg
and its acceleration is 3 m/s2 ; therefore (45+F AB )= m 5 a= 15
N or F AB =15-45=-30 N.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
First of all, they are fictional with not much information available
regarding their power. However, a quick search of the web reveals that many
Star Wars enthusiasts have tried to make estimates of their properties from
detailed examination of the movies. First observation is that they are not
lasers because you can see their trajectories and Han Solo is seen to jump
out of the way in time to not get hit; light is just a lot faster. (You
should not write them off just " …because
it's just light… "
light, though; lasers can melt steel.) Also, if the pulses are any kind of
material (super-heated plasma, charged particle beams, etc.), they could not
propogate very far in air which they do. But the thing that makes them, in
my view, impossible is the energy each pulse carries on the highest setting;
one site I found estimated the energy of a single pulse to be on the order
of several megajoules. Suppose that a single pulse lasts 1
ms=10-3 s and carries an energy of 1 MJ=106 J; the
power carried by that pulse would be 106 /10-3 =109
W=1 GW. This is the power output of a typical power plant. Where are you
going to get this kind of power in some compact form you can carry around
with you? No way!
QUESTION:
ANSWER:
Scientific American article. The "purpose" for anything is not within
the purview of physics.
QUESTION:
ANSWER: earlier answer about the
Bonneville Flats where a similar question to yours was looking to observe
curvature at the 10-mile level. It takes several hundred miles to be able to
do an accurate measurement of the earth's curvature; the
ancient Greeks did this around 200 BC.
QUESTION:
ANSWER: v ≈6.7
m/s. The equations of motion are x =6.7t and
y =-4.9t 2
where x and y are the horizontal and ver tical positions
relative to the edge of the cliff and t is the time after jumping.
Solving the y equation for t when y =-80 m (the ground),
t =√(80/4.9)=4.04 s, so x =6.7x4.04=27.1 m. That is the answer
neglecting air drag. Note that it is independent of the mass m .
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: πR 2 )
is getting bigger so the energy of that wave is getting more and more spread
out meaning that the light is getting dimmer.
QUESTION:
ANSWER: c . That does not
mean that the light is identical. If the source has a huge speed, the light
has a huge red shift, its wavelength is greatly larger than that observed if
at rest relative to the source. It would likely be invisible to the unaided
eye.
QUESTION:
ANSWER: R H )
is proportional to their volume. But, as soon as they start heating up, they
start cooling off by radiating from their surfaces; the rate of cooling (R C )
is proportional to the area. Roughly speaking, the volume is proportional
the the L 3 and the area is proportional to L 2
where L is a measure of the size of the pile of beans. The
ratio R H /R C =KL 3 /L
2 =KL
determines the net rate of heating; K is some constant. For example,
suppose your beans were contained in a spheres of radii L and 1.26L ;
I chose those because the larger radius sphere would have twice the volume
of the other. Then, the larger sphere will heat up 1.26 times faster than
the smaller (two spheres). Assuming that the microwave energy is constant
with time, the longer time it takes the more energy is used.
The problem with my
analysis, though, is that it depends on the power of the microwave. If it
has a relatively low power, the radiation will be mostly absorbed in the
beans and the assumption that R H is proportional to the
volume becomes untrue. Only if a small fraction of the radiation is absorbed
by the beans
will my analysis be approximately correct. You should do an experiment, try
it!
QUESTION:
ANSWER:
QUESTION:
ANSWER: F between masses M and m whose centers are
separated by a distance R is F=GMm /R 2 . But,
the acceleration a of m is a =F /m =GM /R 2 .
So, as you ask in your last sentence, the inertia (m in a =F /m )
cancels out the mass (m in F=GMm /R 2 ) in the
acceleration. This is actually a profound finding for the following reason.
In Newton's second law, mass is, as you say, inertia (inertial mass, m I );
in Newtons universal law of gravitation, mass is the ability to feel or
create gravity (gravitational mass M G ). The fact that
gravitational accelerations are the same implies that m I =m G ,
not something you would necessarily expect. Why these are equal was not
really understood until the general theory of relativity was proposed by
Einstein.
QUESTION:
ANSWER: answer for a very similar question to yours that
you should look at.
QUESTION:
ANSWER:
particle-antiparticle pair
(PAP):
Vacuum fluxuations where a particle-antiparticle pair is spontaneously
created just outside the event horizon, violating energy conservation.
However, the uncertainty principle (UP) allows this violation but only
for a very short time. If one escapes, it adds energy to the universe so
the captured particle must have negative total energy thereby decreasing
the mass of the black hole and maintaining energy conservation. If both
particles are captured, they are required by the UP to recombine thereby
keeping the mass of the black hole constant.
When
a virtual pair is created just outside the event horizon, the intense
gravitational field of the black hole can provide the energy for the PAP
to become real rather than virtual. In this case, the energy acquired by
the pair will be lost by the black hole reducing its mass. If only one
of these is captured, the energy of the escaped particle will be added
to the universe but the energy of the captured particle will be added to
the (already reduced) energy of the black hole resulting in a net loss
of energy of the black hole equal to the energy of the escaped particle.
Energy
of the whole system is conserved.
QUESTION::
ANSWER: ρ
is not the only thing which determines the speed of sound. In general, the
speed of sound v may be written as v = √(K /ρ )
where K is a parameter which specifies the "elasticity" of the
material. (Note that v is inversely proportional to √ρ , not ρ
as you state.) For example, just qualitatively, you might say that steel is
much denser than air but it is also much more elastic. The way K is
specified for gasses is quite different than for solids. For an ideal gas,
K=γP where P is the pressure of the gas and γ= 5/3 for
monatomic gasses and 7/5 for diatomic gasses. On the other hand, for solids
K=B +4G /3 where B is the bulk modulus and G is
the shear modulus. For solids K is enormously bigger (~1011
N/m2 ) than γP for gasses (~105 N/m2 ).
That answers your question. So, let's just do a rough calculation to see if
I get reasonable values for air and steel:
Taking air as mainly diatomic
γ≈ 7/5 ,
atmospheric pressure P ≈105 N/m2 , ρ ≈1
kg/m3 , v ≈ 374
m/s; a little high, but close.
For
steel, B ≈ 1.6x1011
N/m2 , G≈ 7.93x1010 N/m2 ,
ρ ≈7900
kg/m3 , v ≈5800 m/s; this is exactly the speed of sound
in stainless steel!
QUESTION:
ANSWER: g at the equator. This effect
is smaller than the effect due to the earth's rotation which results in
about a 0.34% reduction at the equator (due to centrifugal acceleration).
QUESTION:
ANSWER: F which each student must exert
depends on the weight w he is pushing and the coefficient of friction
#956;
between the sled and the ground, F = μw .
The work W done in
pushing the sled a distance d is W =Fd = μwd .
The power generated if W
is delivered in a time t is P=W /t =μwd /t .
Both students have the same
μ
and
t , so W A /W B =P A /P B =d A w A /d B w
B =(160x340)/(80x480)=1.42.
So student A did 42% more work, generated 42% more power, than student B.
From a physics point of view, B demonstrated more strength, A demonstrated
more power. I would judge that this is not a fair way to assign a grade. It
would be interesting to see if A (B) could move B's (A's) sled 80 (160)
yards.
QUESTION:
ANSWER: twin paradox .
The number of bottle caps produced by the moving factory will be N √(1-0.92 )=0.44N
where N is the number produced on earth.
QUESTION:
ANSWER: Newtonian fluid ,
Stokes showed that A (d )=A 0 e-αd where
A (d ) is the sound amplitude a distance d into the
material, A 0 =A (d= 0), and
α =8π 2 η f 2 ρv 3 );
here
η
is viscosity of the fluid, f is frequency of the sound, ρ is
the density of the fluid, and v is the speed of sound in the fluid.
The important thing to notice is that the amplitude decreases like exp(-βdf 2 )
for any Newtonian fluid. For example suppose you compare f 1 =20
Hz with f 2 =1000 Hz for a thickness of d =1 cm=0.01 m
and A 1 (0.01)=e-αd 0 =A 0 e-β· 0.01·400 =A 0 e-4β
which can be solved for β , β= 0.35 s2 /m; then, for
f 2 , A 1 (0.01)=A 0 e-0.35·0.01·1,000,000 =A 0 e-3500 = A 0 x 9x10-1521 ≈0.
Finally, for any material, Stokes' law of attenuation still may be written
as
A (d )=A 0 e- αd
except α =βf
ε where 0<ε <2. Water and
most metals have ε≈ 2. But even if ε is smaller, it is always
positive which means that the attenuation is always greater for a higher
frequency.
QUESTION:
ANSWER: 0 .
But that does not mean that everything is near that plane. Many of the dwarf
planets are much out of the plane of our orbit; Pluto, for example, is about
180 inclined relative to our orbital plane. Outside the most
distant planets is the
Oort cloud , a collection of rocky and icy bodies which is distributed
roughly equally in all directions; comets are thought to originate in the
Oort cloud and when they enter the inner solar system they come from all
directions relative to the orbital plane. The reason that the planets are
all nearly on the same plane is that when the sun first formed from a huge
cloud of dust and rocks, it had angular momentum and as the system collapsed
under its own gravity, it spun faster and faster and the resulting
centifugal force caused it to flatten like a giant pizza dough.
QUESTION:
ANSWER: e.g. a
mirror or a lens, they never meet.
QUESTION:
ANSWER:
QUESTION:
ANSWER: total
internal reflection . If the angle which the light hits the surface
between the water and the air is glancing enough, the light will be
reflected back into the water rather than be refracted into the air. Total
internal reflection can happen whenever light strikes the interface with a
material of smaller index of refraction; the minimum angle for which it can
occur depends on the relative indices of refraction.
QUESTION:
ANSWER: FAQ page .
QUESTION:
ANSWER: v y =2
m/s. The train is moving with speed v x =100 km/h=27.78 m/s
so the ball and train move a distance 27.78 m horizontally during the one
second the ball is moving. But the distance the ball moves is not
27.78+2=29.78 m but rather
2√(
13.89 2 +12 )=27.85
m because the ball will take a zig-zag path. But the ball also has a
different speed as seen from the station, v =√ (v x 2 +v y 2 )=27.85
m/s. We therefore conclude that the time it takes the ball to travel 27.85 m
is 1 s, so everything hangs together. Your error was to ignore the fact that
the ball has a horizontal component of its velocity in the frame of the
station. You are right that relativity plays no role here.
QUESTION:
ANSWER: q and point P the net
charge enclosed is q but that does not mean that the field due to the
dipole is zero everywhere on that surface. All you can say is that the net
electric flux passing through that surface is q / ε 0 .
Gauss's law is usually useful in determining a field only if the field can
be argued to have constant magnitude and normal to the surface everywhere.
Superposition, not Gauss's law, should be used for this problem.
QUESTION:
ANSWER:
earlier answer help you? One thing is clear is that you do not really
appreciate that the speed of light in vacuum is independent of the motion of
either the observer or the source. This is one of the main postulates of
special relativity. See the FAQ page for questions related to this.
QUESTION:
ANSWER:
If you
are interested, I will give here the electric and magnetic fields for one of
the charges moving with velocity v x direction.
E' i Ex +
γ (j Ey +k Ez )
and B' =-v E' c 2
where
γ= 1/√[1-(v /c)2 ] and i j k E q is at rest and E' B' q is moving.
QUESTION:
ANSWER: in vacuum . What matters in the theory is the universal constant
c , not the speed in the medium you happen to be looking at.
QUESTION:
ANSWER: … i.e. she will have aged less when she
returns to earth. You should read FAQ Q&As on the
twin paradox ,
how clocks run , and the
light clock .
QUESTION:
ANSWER:
QUESTION:
ANSWER: FAQ
page), so let us change " same kicking force" to "same kicking
impulse"; impulse is essentially force multiplied by the time it was
applied. That is also more convenient because the change in linear momentum
(mv ) is equal to the impulse (J ). So, with equal impulses, the
lighter ball has a larger initial speed: v light =(m heavy /m light )v heavy .
So, if both balls are launched at the same angle and if air drag can be
neglected , the lighter one will go farther because it started with a
larger speed. Maybe that is all you wanted. But, it is altogether possible
that air drag will not be negligible. The force of air drag may be roughly
approximated as F =¼Av 2 for a sphere where A
is the cross-sectional area of the balls. (This is true only for SI units
since
¼ a=F /m so even if the two balls
experienced the same force, the less massive ball would have a larger
magnitude of acceleration, slow down faster. Now it becomes complicated
because you have to know the masses, the impulse, and the angle of the
launch to calculate which would go farther.
QUESTION:
ANSWER: A , the depth of the bottom is
D , and the density of water is
ρ .
At some time after uncapping, the distance up to the current level of the
water in the tube I will call z . The force up on the column of water
would be F bottom =(P atm +ρg D) A ,
the weight of the column of water is W =ρgzA ,
and the force down on the top is
F top =P atm A. The net force on the
column is F = ρgA (D-z )=ma = ρAza
where a is the
acceleration and m= ρAz
is the mass. So, a = ρgA (D-z )/ ρAz=g [(D /z )-1].
Suppose we take g ≈10 m/s2 and D ≈104 m;
then a ≈10[(104 /z )-1] m/s2 . The graph of
the acceleration is shown to the right; note the logarithmic scale for
acceleration to show the huge variation from bottom to top. Note that it is
huge, about 105 m/s2 for 1 m and falls by four orders
of magnitude when the column of water is halfway to the surface, finally
falling to zero when the pipe is full. Suppose we start with z =1 m
and the column at rest (where a ≈105 m/s2 ).
After 10 milliseconds (10-2 s) the column will have acquired a
speed of approximately v=at =1000 m/s and the water level would have
risen approximately to z =1+½at 2 =6 m. Now, the speed
is almost immediately huge and so the assumption of no frictional forces
becomes impossible. The drag of due to the water moving in the tube for the
column of water going 1000 m/s will be huge
and keep the water from speeding up so rapidly. Also, the air above the
column of water will not just get pushed up will get compressed and
therefore push down on the column with a pressure greater than atmospheric.
It is therefore clear that my simple calculation above is only a best case
scenario and the drag and added pressure will take energy away. The drag
might get big enough that the water in the tube would boil and release steam
to futher impede its progress. I believe that you will end
up with column of water slowing down as it approaches the surface of the
ocean and it will just come to rest there. In any case, when the tube was
filled, even if still moving, the column of water would have zero net force
on it. If you tried to have it do work on your turbine, it would soon loose
all its kinetic energy and stop. Alas, no clean energy solution!
ADDED
THOUGHT: W =½ρgAD 2 .
Neglecting any energy loss, this would have to be equal to the kinetic
energy of the column of rising water when it got to the top, K =½ ρDAv 2 ,
so v =√(gD ). For the case above, v =316 m/s, pretty
modest. And keep in mind that all the water in the tube is now in
equilibrium, so as the water goes above the surface, it is slowing down as
it gains potential energy. And, if you imagine the tube projecting high
above the surface, you can calculate how high h the water will rise
before falling back. The center of mass of the column of stopped water is
y =½ ρhA , so mgy ==½ h 2 ρgA= ½ ρDAv 2 ,
so i>h =v √(D /g )=D . Therefore, with no friction, the
water would rise up and oscillate back and forth for ever. But, as argued
above, friction will be extremely important so this oscillation will quickly
damp out and will never rise up to a height equal to the depth of the ocean.
See the video in the answer to the followup question below as an example for
a miniature of this.
FOLLOWUP
QUESTION:
ANSWER: You have so many misconceptions that I am not even going to try to
convince you with the physics arguments. A couple of comments: The reason
that the pressure is so high deep in the ocean is that the water deep down
must hold up the weight of all the water above it; to to try to ignore
weight of the water in the tube misses the whole point. Suppose you had a
tube 10 km long full of water (not underwater). The pressure at the bottom
would equal to the weight of the tube of water divided by the area of the
tube (+P atm ). Also, I have
previously answered your
question about the air in a tube in the atmosphere; it would not suck out all
the air in the atmosphere for the identical reason your water tube would not
create a perpetual geyser. Since you do not know any physics, let me
convince you another way; remember physics, at its heart, is an experimental
science, so let's do an experiment. Your idea should work equally well if
the ocean were only 5000 m deep, maybe not just so robust a geyser, right?
That is, your idea should scale. So, try the following experiment. Take a
straw and put a little cork to plug the bottom and push it down into a glass
of water; pop that cork out some way and see what happens. Does the water
come up the straw and continually squirt out the top (tiny geyser) or does
it come up to the level of the water surface and stop? Maybe it will rise a
bit above the surface but it will settle back to the level of the water in
the glass.
QUESTION:
ANSWER:
QUESTION:
ANSWER: x',y',z' )
shown to the right. The Coriolis force is given by 2mv' xω v' m and
ω is the angular
velocity of the earth ( ≈ 7.3x10-5
s-1 ). Now, for a body dropped from some height h , the
direction of the velocity is in the negative z' direction, so the
direction of
v' ω x' =[( ω· λ )/3]√(8h 3 /g )
where λ is the latitude. For example, at the equator where λ =900 ,
and you drop it from 100 m, the deflection would be x' =2.2x10-2
m=2.2 cm.
QUESTION:
ANSWER: F to a mass m ?
Newton's second law tells you that you cause the particle to accelerate,
that is if you exert the force over some time t the velocity will
change from v 1 to v 2 . Writing Newton's
second law, F=ma=m (v 2 -v 1 )/t .
Rearranging this equation, Ft =m (v 2 -v 1 );
so you can see that the amount which the product mv (which is called
the linear momentum p ) changes is numerically equal to the product
Ft which is called the linear impulse J . If the force is
constant, the time need not be short; t could be an hour and J
would still tell you the change in p . I hope you can appriciate that
J =p 2 -p 1 is simply an alternative
way to write Newton's second law. If F is not constant, you can still
calculate J but it is just harder. In essense what you do is
calculate Ft for many vanishingly small values of t all along
the path the particle takes and sum them to get the total J . (If you
know integral calculus, this is just integrating F dt over the
time interval.) Of course, Newton's second law is a vector equation, so J
and p are vector quantities, J p 2 -p 1 .
If you have a system of particles which interact only with each other, the
net force on the whole system is zero because of Newton's third law, so
J =0 and p p
QUESTION:
ANSWER: creates a static electric field. A charged particle
moving with constant velocity (constant speed in a straight line) creates
both an electric field and a magnetic field and both change with time. But
neither of these situations are said to radiate electromagnetic
fields. Electromagnetic radiation propogates through space as waves; visible
light, for instance, is an electromagnetic wave. Electromagnetic fields can
be created when a charged particle accelerates. E.g., an antenna radiates
radio waves when electrons in the antenna are made to oscillate back and
forth (accelerating). The derivation of how accelerating charges radiate is
a topic in an intermediate E&M course and beyond the scope of this site.
QUESTION:
ANSWER: invisibility
cloaks ".
QUESTION:
ANSWER:
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
unhuh...
ANSWER: look this up in an elementary physics
text —I
am not a tutor. The reason your argument is invalid is that dt is
infinitesimal so if a t is perpendicular to the
velocity vector v Limit (as
dt —> 0){ √[v 2 +(a dt )2 ]}=v .) Just use
your common sense: a car going in a circular path has a constant magnitude
of velocity, right? And nonconstant direction or velocity, right?
QUESTION:
ANSWER: only forces on that body
affect its motion. So, if you are interested in what body B will do, you
look only at that body. The force which B exerts on A is not a force on B.
So, if B has a mass of, say, 2 kg and there are no other forces on B, it
will have an acceleration of a B =10/2=5 m/s2 .
Since the 10 N force continues, A and B remain in contact, so a A
must also be 5 m/s2 . One force on A is the 10 N force from B
which is in the opposite direction as the acceleration. But the net force on
A must surely be in the direction of the acceleration, so there must be some
other force (probably due to you pushing on A) on A which is bigger than 10
N and points in the direction of the acceleration. For example, if the mass
of A is 4 kg, F -10=4x5=20 N, so F =30 N. Finally, you could
look at A and B together as the body. In that case the forces they exert on
each other do cancel, the acceleration is 5 m/s2 , and the total
mass is 6 N; therefore, there must be some external force (you again)
causing this 6 kg to have an acceleration of 5 m/s2 : F=ma =6x5=30
N. All three ways of looking at this problem are consistent with each other.
QUESTION:
ANSWER: 0 would work. The point is for the incident
light to be reflected in exactly the opposite direction from the direction
it came in. The only way I know to achieve this is with
corner reflector
which is three mirrors connected as in the inside corner of a cube.
QUESTION:
ANSWER:
Q
ANSWER:
QUESTION:
ANSWER: L = ∞.
But,
that is only if your proton and your electron are the only things in the
universe! You can calculate the ionization energy, the energy necessary to
move an electron from the ground state of hydrogen to infinity; it is 13.6
eV=2.18x10-18 J. You can also calculate the energy necessary to
move it to any other distance. For example, the energy to move it to about
0.5 μ=5x10-7 m (which is about 100 times the radius of the
hydrogen atom) is 13.599 eV, essentially the same as to infinity. So you can
see that most of the energy is supplied in close to the proton because that
is where the field (and therefore the force) is strongest.
QUESTION:
ANSWER: π + is
composed of an up quark (charge +2e /3) and a down antiquark (charge +e /3)
and would have a charge +e and be able to form an "atom" which you might
call pionium.
ADDED
NOTE: π + and one
π - . I cannot find any reference to this pion+electron
system, so I guess you can call it whatever you like.
QUESTION:
ANSWER:
QUESTION: https://www.youtube.com/watch?v=vvbN-cWe0A0
135,890 feet - or, 41.42 km (25.74 mi)
In the video you see speed at which he is supposedly travelling, 729 mph (1173km/hr), 46 seconds after he jumped. The footage is cut then seconds later speed strangely falls to 629 mph...anyway after 4 mins and 18 seconds of free falling he releases parachute. My questions are What speed would he be going at a 4 mins and 18 seconds into freefall? Also how do you explain the deacceleration when it was at 729 mph then down to 629 mph?
ANSWER:
earlier answer .) There is a better video at
https://www.youtube.com/watch?v=raiFrxbHxV0 ; this video shows data
acquired by instruments on Baumgartner. Here some clips from that video:
The first three are the times also
showing speeds; these are roughly the speeds you refer to in your question.
The second three show speeds and altitudes. The third graph shows the whole
history of altitude, speed, and mach speed. The speeds are all larger than
indicated in your video. The speed at 4:20, 113 mph, answers your first
question. The speed at 0:50 is about the maximum, 847 mph or mach1.25. The
speed at 1:01 has dropped to 732 (about the same change as your video); I
suspect this deceleration is due to two factors: there is getting to be much
more air resulting in higher drag and, probably after he broke the record he
oriented his body to increase drag and slow down. The official maximum speed
after everything was calibrated was 843.6 mph.
QUESTION:
ANSWER: C D
is a constant characterized by the geometry of an object used to
calculate the drag force F D if it is proportional to the
square of the velocity v : F D = C D ρAv 2
where A is the area
the object presents to the fluid flow and
ρ
is the density of the fluid.
Whether or not this equation is true depends on a quantity called the
Reynolds number Re =L ρv /η
where L is a length along the direction of flow and
η
is the viscosity of the fluid. It turns out that only if Re >1000 is
the velocity dependence approximately quadratic. If Re <1 the drag
force is approximately proportional to v . Anywhere between these
extremes the drag force is a combination,
F D ≈C D ρAv 2 +kv .
C D ≈constant
only for the case Re >1000; the drag coefficient would then be the
same for identically shaped objects. It is interesting, though, to get a
feeling for what the relative speeds in air and water corresponding to Re =1000
are. The necessary data at room
temperature are
ρ air ≈1
kg/m3 ,
ρ water 3 ,
η air ≈1.8x10-5
Pa·s, and
η water ≈1.0x10-3
Pa·s. Then v water /v air ≈(1.0x10-3 /1000)/(1.8x10-5 /1)=0.056
(which is true for any Re ). Taking a sphere of diameter 0.1 m as a
specific example for the critical Re =1000, C D =0.47,
A =7.9x10-3 m2 , L =0.1 m, v air =1000x1.8x10-5 /(0.1x1)=0.18
m/s and v water =1000x1.0x10-3 /(0.1x1000)=0.01
m/s. Since the speeds are relatively low, you can conclude that for many
cases of interest the quadratic drag equation holds for both water and air.
Therefore, for many situations you can approximate that the drag coefficient
in air and water are about the same. For a specific case you should check
the Reynolds numbers to be sure that they are >1000. The bottom line is that
if the drag force depends quadratically on velocity, the drag coefficient
approximately depends only on geometry, not the properties of the fluid.
Keep in mind that all calculations of fluid drag are only approximations.
ADDED
COMMENT: Re <1000, F D ≈av+bv 2
where a and b depend on Re ; e.g., a ≈0 for
Re >1000 and b ≈0 for Re <1. So, for Re <1, a
is just a constant and F D ≈av≡½C D ρAv 2
so C D =2a /( ρAv )=2aL /(A ηRe )≡c /Re
where c is a constant. If F D is proportional to
v , C D is inversely proportional to Re ; this is
called Stokes' law. As we saw above, Re >1000 leads to C D =constant.
Ferguson and Church have derived an analytical expression which very
neatly reproduces data for the transition from Stokes' law (1/Re ) to
constant C D . Calculations for a golf ball are shown at the
left (constant C D , the green line, is called turbulent in
the legend). Note, as we have stated, that the transition occurs in the
region
1<Re <1000. Note the sudden drop in the data around Re =500,000.
I believe that this must be due to the
dimples on the golf ball which
reduce drag.
QUESTION:
ANSWER:
QUESTION:
ANSWER: per
se . When something is turning it has rotational kinetic energy and
therefore a spinning planet has more energy than an otherwise identical
planet not spinning. In general relativity you usually hear about gravity
being caused by mass warping spacetime. However, mass is just the most
obvious source of gravity and what it is which really warps spacetime
is energy density and mass has a lot of energy (E=mc 2 ). The energy due to the turning
is infinitesmal compared to the mass energy of the object so you would never
be able to distinguish the difference due to the turning by looking at the
gravity.
QUESTION: at the same speed [in the same time] no matter the charge pushing it forward. More powder and it moves faster but vertical speed is the same. I imagine two equal weight gliders of 20/1 and 30/1 lift/drag ratios would also express the same vertical speed?
My question is, wouldn't this gravity effect also hold true underwater? An underwater glider expressing 30/1 L/D would travel 50% further, faster than a 20/1 L/D in a given vertical?
ANSWER:
QUESTION:
ANSWER: earlier answer .
QUESTION:
ANSWER: hindsight
is 20-20 "? It is unrealistic to expect that when something new is
discovered that we should somehow know all the effects that might have on
anything else. At the time radioactivity was discovered, nobody even knew
what atoms were composed of or what their structure was. And it was found to
be very tiny bits of matter (e.g. what we know as electrons today)
and who would have thought that getting hit by something trillions of times
lighter than a speck of dust could be harmful? It took experience before it
was appreciated how dangerous it could be. One of the best known examples of
such experience is the case of
radium watch dials .
Radium, mixed with a phosphor, was painted by women onto watch dials and it
would glow in the dark. The workers were encouraged to point their brushes
with their lips to make the fine lines required. Subsequently many became
ill with cancer and other radiation sickness. It seems stupid now, but the
dangers were not known until they were discovered. Also many of the health
effects were long-term effects, the effects not appearing until years or
decades after exposure. One example I know about from personal experience is
the effect of radiation treatment for acne which was popular in the 1950s. I
very much wanted to have this done but my father forbade it, making me
furious with him. Years later people who had had this treatment started
coming down with cancer thanks, Dad! Yet another example is the
shoe-fitting x-ray machine. When I was a kid it was really fun to buy new
shoes because you could look down at an x-ray of all the bones in your foot
and how well the shoe fitted you; again, when more was learned about
radiation, these machines were all sent to the scrap heap. When it began to
be appreciated that there were dangers, studies began to be done to try to
set allowable dosage levels. But, it is unethical to do such experiments on
people, so lab animals had to be used which always presents problems with
scaling and other variables. Much of what we know today was learned
after-the-fact by doing long-term statistical analyses over decades of
medical records.
QUESTION:
ANSWER: Φ M ≡∫∫B ·A Φ M =BA
, is the flux through an area A . Then, if you interpret
Φ M
as a number, you would simply say, for example, there were 10 lines through
an area of 1 m2 if the magnetic field is 10 T. You could ask the
same question about electric flux,
Φ E ≡∫∫E· A E=Q /(4 πε 0 r 2 )=1/(4x3.14x8.85x10-12 x12 )=9.27x109
N/C. The flux at a distance of 1 m from the charge would therefore be Φ E
=EA = 9.27x109 x4x3.14x 12 =1.17x1011
Nm2 /C. Incidentally, this is the flux regardless of where you
measure it because the area of the sphere
(4 πr 2 )
surrounding the point charge appears both in the denominator of the field
and in the numerator of the flux and cancels. Hence, you could say that
there were 1.17x1011 lines of
electric field emanating from a 1 C point charge. You can see this from
Gauss's law,
Φ E =∫∫E · A Q /ε 0
if the integration is over a closed surface enclosing the charge Q .
QUESTION:
ANSWER:
QUESTION: 50 The Physicist ] light years away. It seems that for him, he will have traveled 20 light years in one year. That is faster than the speed of light. What don't I understand?
ANSWER: v =0.999c . The
earth observer would see an elapsed time of 100x0.999=99.9 years (your 100
years). The traveler will see the distance shrunk to D' =
10 √(1-.9992 )=0.447
light years, so the time of travel is t =2x0.447/0.999=0.895 years
(your 1 year). You should read the earlier answer on the
twin
paradox .
QUESTION:
ANSWER: W of all the water, so P 2
would be W /A bottom where A bottom
is the area of the bottom of the container. This assumes that atmospheric
pressure is the same everywhere in the vicinity of the container; in other
words, I have ignored the buoyant force due to the air on the whole
container because it will surely be much smaller than W .
FOLLOWUP
QUESTION:
ANSWER: P 2 =P atm +W /A bottom .
There is no problem that P 2 ≠ P 1
because the force which the container exerts on the table is not P 1 A bottom .
Think about it—the
sides of your container exert a downward force on the bottom of the
container.
QUESTION:
ANSWER:
palindrome composed of the allowed letters, e.g. AHA, OTTO, WOW,
YAY, MOM, TIT, TUT, … .
But, wait, it is somewhat more difficult than that! Look at the pictures of
the Camel cigarette package and its reflection, in particular, the word
CHOICE on the side. So words composed of letters with symmetry upon
reflection about a horizontal axis will also reflect the same: B, C, D, E,
H, I, K, O, X will have a reflection which is the same as the original, but
only if turned upside down. There is no restriction for these words to be
palindromes; besides CHOICE, some others would be BED, HEX, BOX, BIKE, HIKE,
CHIDE,
QUESTION:
ANSWER: appears to be coming from
somewhere but is not actually coming from there. When you look into a
mirror, it appears that the light from the image of what you are looking at
is coming from behind the mirror, but it is actually coming from the mirror
itself.
QUESTION:
ANSWER: d vs . t 2 , the slope of the line
will be ∝a .
You know that d =½at 2 , so suppose that I call t 2 =u .
Then clearly the slope of the line d =½ au
is
½ a . Your
equation d /t 2 =V /t is clearly
wrong: d /t 2 =½ at 2 /t 2 =½ a
and
½ a≠v /t.
QUESTION:
ANSWER: g ≈10
m/s2 .)
We can agree that the acceleration is a =250 m/s2 and that
is undoubtedly 25g . Now, we need to write Newton's second law for the
person, -mg+F=ma =-500+F =12,500, so F =13,000 N. This is
the average force by the superhero on the person as she is stopped, so the answer that the
average force is 12,500 N is wrong. When one expresses a force as "g s of
force", this is a comparison of the force F to the weight of the
object mg , F (in g s)=F (in N)/mg =13,000/500=26
g s; this simply means that the force on the object is 26 times the
object's weight. So neither of you is completely right, but if there is any
money riding on this, your friend should be the winner because the only
error he made was to forget about the contribution of the weight to the
calculation of the force. I am hoping that superman knows enough physics to
make the time be at least 0.3 s so that Lois does not get badly hurt!
QUESTION:
ANSWER:
QUESTION:
ANSWER: jet streams . The
prevailing direction of these winds is easterly and their speeds are as
large as 100 mph. Therefore when you travel east (as from Dubai to Mumbai)
you have a tail wind, and when you travel west (as from Mumbai to Dubai) you
have a head wind. Since the greatest speed an airplane can fly, say 600 mph
for commercial jets, is airspeed, that is relative to the air, the ground
speed with a 100 mph wind would be 700 mph going east and 500 mph going
west.
QUESTION:
ANSWER:
right-hand rule such that counterclockwise is a positive vector on the
axis.
QUESTION:
ANSWER:
QUESTION: 24 kilograms and radius of 6371 kilometers. So how do I put these together in a formula to get the 9.81m/s gravity?
ANSWER: F which each feels is
F=GMm /r 2 where G is the universal gravitational
constant, G =6.67x10-11 N·m2 /kg2 ,
M and m are the masses of the two masses (take M as the
mass of the earth), and r is the separation between M and m .
This also works if the masses are spherically symmetric. But, of course, you
also know in your case that F=mg . So, if you solve for g, g=MG /r 2 ≈ 6.7x10-11 x6.0x1024 /(6.4x106
)2 =9.8
m/s2 .
QUESTION:
ANSWER: O E1 . Then, you
have to look at what is called the transition matrix element for this
operator, <Ψ |O E1 |Ψ' >=∫Ψ ·O E1 ·Ψ' dτ
which is an integral over all space and Ψ' and Ψ are the
excited and ground states, respectively. If this is nonzero, then the decay
will decay with some half life determined by the value of the matrix
element. If it is zero, the state will still probably decay to the ground
state but via a different kind of transition.
QUESTION:
ANSWER: μmg
(where μ is the coefficient of static friction) so the greatest
acceleration the coin could have is μg ; for example, if μ= 0.3,
and you caused the acceleration of the cardboard to be any greater than
about 3 m/s2 , the cardboard would slip under the coin. But, if
you gave the cardboard an acceleration of 0.2 m/s2 the coin would
move with it.
QUESTION:
ANSWER: early universe were much more massive than the sun; this
resulted from the fact that the early universe was essentially all hydrogen and a bit of helium.
Stars began forming only 100-200 million years after the big bang. Very large stars burn much hotter than smaller stars and therefore have much shorter lifetimes, a few million years rather than
a few billion years for less massive stars (e.g . the sun with an
expected lifetime of about 10 billion years) which would explain why many
black holes are very old. The rate of expansion of the universe is not
inconsistent with the model of a big bang about 14 billion years ago (in
fact, the rate is one of the measurements used to estimate the age of the
universe) even though we do not understand all the details (e.g . dark
energy and accelerating expansion).
QUESTION:
ANSWER: net work done on each person was the same.
Where the energy came from is different. On the stairs, the energy is
supplied by what the guy ate for lunch. In the elevator, the energy is
supplied by the motor lifting the elevator. Regardless where it came from,
mgh of energy must be somehow supplied.
QUESTION:
ANSWER:
QUESTION:
ANSWER:
QUESTION:
ANSWER: earlier
answer about
gravity of a cylinder which shows how the tendency is toward a spherical
shape.
QUESTION:
ANSWER:
QUESTION:
ANSWER: p is mass m times velocity v must be
wrong; the relativistically
correct expression for momentum is p=mv / √[1-(v /c )2 ]
where c is the speed of light. For a photon, this is a little tricky
because p =0/0, but you can also write that E 2 =p 2 c 2 +(mc 2 )2
where E is the energy of the a particle of mass m , so p=E /c
if m =0; since E=hf for a photon, where h is Planck's
constant and f is the frequency, the linear momentum of a photon is
p=hf /c . Therefore the frequency of the photon must be
uncertain according to the uncertainty principle. Since the energy of a photon is
hf , there will be an uncertainty in the energy of the photon.
QUESTION:
ANSWER: definition of the
Hall coefficient R H , R H
≡ V H t /(IB )
where V H is the Hall voltage, t is the thickness of
the sample, I is the current, and B is the magnetic field. The
Hall voltage may be
shown to be V H =-IB /(nte ) where n
is the charge carrier density (units [m-3 ]) and e is the
charge (units [C]]) of a charge carrier. Therefore R H =-1/(ne )
so the units of R H are m3 /C.
