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QUESTION: I am constructing a floating dock out of pressure-treated lumber and
55-gallon empty plastic barrels and want to ensure that I provide sufficient
buoyancy. Although the dock comprises a ramp and a platform, they will be
only loosely connected (with slack rope and eye bolts), and the buoyancy
calculations for the platform are easy; it's the ramp that is giving me
headaches. Specifically, one end of the ramp will be sitting on the shore
and the other end will be supported by the plastic barrels. The ramp is
T-shaped, with the lakeside end wider to accommodate the barrels. Please
give me some guidance as to how I might analyze a T-shaped structure with
the narrow end on shore and the wide end kept afloat by barrels, so that a
weight of w placed at any point on the ramp will not submerge the barrels
more than 45% (because, although I think I did a good job sealing off the
barrel caps, I'd rather not have to find out). Although I've meant for this
question to be somewhat general to allow for design modifications, I will
mention that I've built part of the ramp already, with the walkway being 3'
wide and 8' long, and the cross of the T being 5' wide and 3' long and
covering two empty barrels; I'm willing to add another section with more
barrels if needed. Any help in analyzing buoyancy of a T-shaped ramp with
the narrow end resting on land would be greatly appreciated!
I will not be able to do any quantitative calculations without knowing the
weights of the ramp and dock. Also, will the level of the water remain
I must mention that I modified the
design to make the walkway eight feet longer, so that it is now 16 feet
long. Also, I had another empty 55-gallon barrel on hand, and placed it
lengthwise under the walkway at the end where the walkway joins the
cross of the tee. The salient data for the components are as follows:
T-shaped ramp (total weight 648
Walkway (463 lb) is 3 feet wide
and 16 feet long; a 55-gallon drum is placed under the walkway on
the lake end.
Cross of tee (185 lb): 5 feet
wide and 3 feet long, covering two 55-gallon drums.
The dock platform weighs 515 lb
and is 8'x8'. Four 55-gallon drums support it.
My dock will be located in a
tributary of the Potomac River. As such, the water is tidal, varying in
depth between 2 and 5 feet. (This is the reason I did not join the ramp
and the platform rigidly, as I anticipate that the angle of surface of
the ramp relative to that of the platform will fluctuate, given that one
end of the ramp rests on the shore.)
In order to help visualize the
situation, I am attaching three JPG images which I created in Sketchup.
Please note that these images do not include the latest modification,
whereby I lengthened the walkway and placed a barrel under its far end.
Also, I recognize that you will probably make certain simplifications in
order to expedite the analysis. That is fine with me; I only wish to
obtain a rough idea of the limits of motion of the ramp and dock as I
walk upon them.
At high tide the walkway and
platform are approximately horizontal. At low tide the water level is
about 2-5 feet down.
The volume of each barrel is 55
the density of water is 64.2 lb/ft3;
each barrel is 45% submerged, so
each barrel provides a buoyant force of 7.35x64.2x0.45=212 lb;
each barrel has a weight of 21.5
I will first do the platform which is easiest if you assume
that the load is at the center.
The weight of the platform is 515
the weight of the four barrels is
the weight of the load will be
denoted as W;
the buoyant force of the four
barrels will be 4x212=848 lb.
the load is not in the center, the total buoyant force will still have to be
848 lb but the platform will tilt so that two of the barrels will be
submerged more than 45% and the other two less than 45%. I made an estimate
of how much the platform would tilt if the load were moved over to 1 ft from
one side edge. Without going into details, the heavy side would go down by
about 2.8 inches and the other side would go up by the same distance; the
corresponding tilt would be about θ=4.50. This would make two of
the barrels 65% submerged, beyond your desired limit.
Next I will look at the walkway.
weight of the tee and two barrels under it as well as the net buoyant
force of those barrels act at 17.5 ft from the shore;
the weight and buoyant force of the
third barrel act at 14.5 ft from the shore
the maximum weight W acts
at a distance x from the shore;
there is a force F exerted
up by the shore which we will not need to know;
I have assumed that the ramp has no
interaction with the platform, since reference is made to "slack ropes".
All this is shown in my diagram. If one
now sums the torques about the point of shore contact and sets that sum
equal to zero, the product Wx can be solved for.
Wx=2488 ft⋅lb. So, for x=19
ft, the end of the ramp, W=131 lb. I am guessing that this result
does not make you happy!
I am not sure how rigidly coupled the
platform and walkway are ("slack ropes"), but suppose that they are coupled
as if, when horizontal, they were rigidly attached. So now the summed torque
equation is 0=212x14.5+424X17.5-228x17.5-21.5x14.5-Wx-463x8+848x23-601x23=8169-Wx.
So W=8169/x. So, for x=19, W=430 lb and
for x=27, W=303 lb. It would seem that it is important
that the coupling be designed such that the platform can help hold up the
walkway. I figure that the walkway will only go down a maximum of about 150
at low tide; this should not significantly alter the estimates I made for
the horizontal situation. So you should allow a fairly rigid coupling like
some kind of hinge.
QUESTION: The question relates to a medical procedure. If you fill a balloon to a
set pressure to use in a semirigid tube to dilate it, will the radial force
exerted on the tube by the balloon be different if the balloon is filled
with water as opposed to air? In otherwords, does the density of the medium
filling the balloon change the radial force exerted on the walls of a hollow
tube if pressure is kept constant? Is there a known law that describes this?
I am afraid that your question is very ambiguous. I do not get the picture
at all. What is the idea, to use the balloon to pressurize the tube or vice
versa? When you say "to use in a semirigid tube to dilate it" what does "it"
refer to? Can you give me a description of what this device does and how it
is used? Would the tube be filled with water also? Is the patient horizontal
or vertical? Is the balloon above or below the tube?
FOLLOWUP: The tube is a hollow tube with muscular walls. The patient is laying
flat. The balloon is within the tube at an area along the tube that is
narrowed compared with the rest of the tube.
Ah, one picture is worth a thousand words! Water has a density about
1000 times the density of air. Therefore, although you can ignore the
pressure differences between places at different heights for air, you can
possibly not for water. That is, the pressure will be higher at the bottom
of the balloon than at the top by an amount ρgd where
ρ is the density of the fluid, g≈10 m/s2 is the
acceleration due to gravity, and d is the diameter of your inflated
cylindrical balloon. For example, if d=1 cm=0.01 m the pressure on
the back side is 0.1 N/m2=1.5x10-5 psi larger at the
front for compared to 100 N/m2=0.015 psi for water. In other
words, the back side of the esophagus (I assume that is what this is) must
support the weight of the water creating a larger push by the balloon on the
back than the front. The density of the fluid does make a significant
difference; I have no way of judging whether the difference is enough to
make a difference in the efficacy of your device.
QUESTION: But for my work, I need to figure out one parameter. Usual passenger
vehicle. We know door velocity (for instance driver door) and I need to
convert it to Joule. I am suppose that also need a mass (27 kg trim
assembled door), distance from hinge to outside handle (approximately 1000
mm) What also? Please share a proper equation.
Joules measure energy. This would be kinetic energy of rotation. You do not
tell me what you mean by "door velocity"; I assume it is rotating about the
hinges with some angular velocity ω radians/second. Since it will
be rotating about the hinges, you need to know its moment of inertial I
about that axis; if you model the door as a uniform rectangle, I=ML2/3=27x12/3=9
kg m2. If it had an angular velocity of ω=1 rev/s=2π
s-1 then its rotation kinetic energy is K=˝Iω2=178
QUESTION: I am working in vehicle assembly plant and at the
end of a final line we measure all doors speed of closing (velocity). For
this check, DURR 5000 is used. It devise install at the edge of the rear
door when the front door is measured and at the rear body panel when rear
door is measured. For this measurement, we have upper limit borders: for
front/rear own. As I know, velocity 1.62 m/s it is about 5 J and 1.49 m/s is
4 J for our C-class model, at least I found these data in my documents. I am
puzzled that the result 178 J it is not appropriate equation (after my poor
explanation). Can you please figure out in Joules or provide equation then
is known: door mass = 27 kg, hinge to handle distance = 1000 mm, velocity =
Everything I did above was correct, but since you did not give me a velocity
in your original question, I had to make up a speed of 1 revolution per
second for the angular velocity; now that you gave me some velocities, I can
see that my guess was way too big. If the door is closing, then it is
rotating about the hinges. So specifying a velocity (1.1 m/s) does not tell
me anything because the speed depends on how far from the rotation axis you
are measuring. For example, if this is the speed of the center of the door,
the outer edge has a speed 2.2 m/s and the hinge edge has a speed of 0 m/s.
If I take the outer edge (1 m) as where the speed is measured, I find that
ω=v/R=(1.1 m/s)/(1 m)=1.1 s-1 and
therefore, K=˝Iω2=˝x(9 kg m2)x(1.1 s-1)2=5.45
J. The equation you want is K=˝Iω2=˝(ML2/3)(v/R)2
=M(vL/R)2/6 where L is the
distance from the hinge to the outer edge of the door and R is the
distance from the hinge to the point where v is measured. If
L=R, then K=Mv2/6. (Keep in mind that this
(I=ML2/3) assumes that the door's mass is uniformly
distributed in the radial direction which will certainly not be exactly
true. It would be most accurate if you could get a good measurement of the
moment of inertia about its hinges.)
THOUGHT: I looked up the use of this Durr 5000 device and saw it also
applied to sliding doors in addition to the hinged doors I have considered
in my answer. In the case of a sliding door the energy is K=˝Mv2.
QUESTION: I apologize if this question seems a bit simple. I'm setting up a small
malthouse, and I have a plastic holding tank (not insulated) that holds the
water for the malting process. I'm trying to determine if I need to heat the
tank in the winter or not. The tank holds 18928 liters. The water comes into
the tank at 9 degrees C, and the ambient temperature around the tank is 17
degrees C. My question is, how long will it take the water in the tank to
reach equilibrium with the ambient temperature.
To answer this question it is necessary to know the material from which the
tank is fabricated and its thickness and its geometry (mainly the area
exposed to the ambient temp).
FOLLOWUP: Okay, the tank is a cylinder with 102" diameter, and 154" high. The
walls are 1/4 thick, high density polyethylene.
The equation for heat transfer through a conducting barrier is
where k=0.47 W/(mˇK) is the
thermal conductivity of
high density polyethylene, s=0.25"=0.00635
m is the thickness of the barrier, A=43.15 m2 is the
area of this tank, T is the temperature of the water as it warms,
and Thigh=290 K. dQ/dt is the rate at
which energy (heat) enters the tank.
The equation involving the rate of
increase of the temperature of water to which heat is being added at a rate
where m=20.5x103 kg
is the mass of the water and C=4.19x103 J/(kgˇK) is the
specific heat of water.
Combining the two equations, I find [dT/(290-T)]=3.73x10-5dt
. Integrating and solving for T I find T=290-8ˇexp(-3.73x10-5t)=290-8ˇexp(-0.134t);
the first expression is for t in seconds, the second for t
in hours, and the temperature is in K. The final result is shown in the
graph converting the Kelvin temperatures to 0C. Note that the
temperature never (theoretically) reaches 170C, but a day to a
day and a half would probably be fine for your purposes. (By the way, the
volume which I calculated and used was 20.5 m3, a tad larger than
the volume you quoted, 18.9 m3.)
(DIFFERENT QUESTIONER): You recently worked out a heat
transfer/tank problem where you came up with: the equation "Integrating and
solving for T, I find T=290-8ˇexp(-3.73x10-5t)=290-8ˇexp(-0.134t)". Is this
e raised to (values), or is it 290-8^(-3.73x10-5 t). It's not clear about
the intermediate steps? Using Wolfram on-line integration, I get:
-log(290-T) + constant = 3.73x10^-5t. Just looking for clarity.
exp(x) is standard alternative notation for ex.
The reason that an exponential function appears in the answer is that
Wolfram's log(x) is the natural log, ln(x) in more common
notation, and x=eln(x). So, if you have an
equation of the form ln(x)=y and you want to solve for
x, x=ey. Also, the integrator you used is
for indefinite integrals and you need to do an integration here from 282 to
T. You also need to remember the properties of logarithms, viz.,
ln(a)-ln(b)=ln(a/b) and ln(a)=-ln(1/a).
QUESTION: Me and my brother are having a debate about running on a treadmill vs
running on a track. Not counting mental stuff like learning how to pace i
feel that the only real difference is wind resistance. He seems to think
inertia comes into play differently for each. I feel like once you are on
the treadmill and it starts moving that the treadmill track is your inertial
frame of reference so any change in inertia would require just as much
energy as it would on an outdoor track. He seems to think that since you
don't let the treadmill ever move you actually that you never have to work
against inertia of rest while you're accelerating or decelerating on a
treadmill. I mean if the treadmill was already moving at 10mph and then you
just jumped on it i could see his point but not if you start off standing on
the treadmill while its at rest and accelerate along with it.
In terms of simple introductory physics, you are correct. However, there are
often subtle differences between simple physics and the real world when
applied to very complicated systems like the human body. Your brother is
also wrong because he is just trying to explain any differences in terms of
simple physics also. In fact, there are many differences between running on
a track and running on a treadmill due to biomechanics. A good article to
read is a post on the
RunnersConnect blog. Or just google treadmill vs. track.
QUESTION: Is there a formula for calculating the side-ways deflection wind has on
a lawn bowl(over and above the bias deflection ) running at 12 s, the time a
bowl takes from delivery to stop over a 26 m distance over bowling green
Once again, doing Ask the Physicist has led me to learn something new. I
never really knew anything about lawn bowls other than it is done on grass
and rolling balls are involved. For the benefit of others who are ignorant
of the game, let me summarize by describing the ball. (A good article on the
physics of lawn bowls balls can be found
here.) The ball is not a sphere but rather an oblate spheroid which
makes it sort of like a door knob but not so extremely flattened; but it is
slightly more flattened on one side of the ball than on the other which
results in a center of gravity being displaced to one side of the equatorial
plane as shown in figure (a). This results in a tendency for the ball to
curve left if it is rolling the angular velocity shown in the figures; this
motion is the "bias" referred to by the questioner which I am to ignore.
When rolling in the x direction (figure (b)), there is a frictional
drag force called, rolling friction D, which
opposes the motion (v) and eventually brings the rolling to a halt.
If there is a wind, there is a force W due to the
wind which tries to make the ball roll to the right (figure (a)) but if it
does roll, there will also be rolling friction trying to keep it from
rolling. In order for the wind to have any effect at all, it is clear that
we must have W>D; if this is not the case, there will only be
static friction in the y direction which will be equal and opposite
to W. A lawn bowls ball has a mass of about m=1.5
kg and a radius of about R=6 cm=0.06 m.
To get the equations of motion for the
x and y motions, we first need expressions for D
and W. The rolling friction may be expressed as D=-μmg
where μ is the coefficient of rolling friction and mg is
the weight of the ball. The force due to the wind may be approximated as
W≈źAV2 where A=πR2 is the cross
sectional area of the ball and V is the speed of the wind; this
approximation is only correct if SI units are used. The equations of motion
in the x-direction are
Here t is the time and v0
is the speed of the ball at t=0. If the ball is rolling in
the y-direction because of the wind, the equations of motion are:
It should be noted that if (źAV2/m)<μg,
these equations imply that the ball will accelerate opposite the direction
of the wind, obviously not correct; hence the wind will have no effect on
the ball if V<√(4μmg/A). In that case, ay=vy=y=0.
So, having found the general solutions,
let us now apply the solutions to the specific case from the questioner. We
are told that when t=12 s, vx=0 and x=26
m. With that information you can solve the x-equations to get v0=4.32
m/s and μ=0.037, reasonable values compared to numbers in the
article I read. The area is 3.14x0.062=0.0113 m2.
The first question we should ask is what is the minimum speed of the wind to
have any effect at all: Vmin=√(4x0.037x1.5x9.8/0.0113)=13.9
m/s=31 mph=50 km/hr; this is a pretty stiff wind, so the wind probably has
no effect on bowling under normal conditions. So, just to complete the
problem, consider V=15 m/s=34 mph=54 km/hr.
Thetrajectory during the 12
seconds is shown in the graph below; after 12 seconds the ball will continue
accelerating in the y direction.
So the bottom line is that unless you are playing in
a gale-force wind, the wind has no effect on the ball if the wind has no
component along the original direction of the ball (which I have called the
x-axis). You can tell if wind makes a difference by simply setting
the ball on the groundunless the wind blows the ball away, you need not
worry about its effect. If the wind is blowing in the +x or -x
direction, that is a whole different thing, but the questioner asked for the
THOUGHTS: This question continues to intrigue me and I have carried
my investigation further. The question originally stipulated "over and above
the bias deflection" so my whole discussion totally ignored the fact that
the ball, owing to its off-center center of mass, will curve. At the very
end of my answer I noted that if the wind is not perpendicular to the path
of the ball, it would be a different story; indeed for a spherically
symmetric ball I showed that, except for very strong winds, a wind
perpendicular to the path has no effect at all. However, for an actual lawn
bowls ball, the path curves to where a wind in the y-direction
might have a significant component along the path. I have calculated
(graphed below) the x and y positions of a realistic path
with no wind using equations (10) and (11) of the
article referred to above. To do these I used all the numbers used above
(R, m, A, v0,μ); I used the moment of inertia for a solid sphere ( I0=Icm+mR2=(7/5mR2))
and chose the COM off-center distance to be d=1 mm. As you can see,
the curving is substantial, carrying the ball about 4 m from its original
direction. You can see that now a wind of any magnitude can have an effect
on the trajectory. The angle φ which the tangent to the
trajectory is given in the article as φ=(2/p)ln(v0/(v0-μgt))
where p is a constant also given in the article. As can be seen,
once the trajectory leaves the x-axis the wind contributes with the
component of its force along the trajectory; this has the effect of reducing
the effect of the frictional force causing the ball to slow down less
rapidly. However, this is now like having a time dependent force of friction
which, I believe, will lead to equations of motion which will not have an
analytical solution but would have to be solved numerically.
QUESTION: For a snow plow that is very heavy, is there an advantage to having the
connection point of the winch line up high so that there is less weight to
pull or is there no difference? I can send a picture for clarification.
Yes send me a picture. You are talking about a winch which is used for what?
Pulling a stuck vehicle? Lifting and lowering the plow? What?
QUESTION: Yes, lowering and lifting a plow. The problem is
that the plow is out very far out and that my winch is mounted pretty low on
my machine. So instead of the winch simply lifting the plow up, right now it
is mostly pulling backwards and then that is making the plow come up. I
attached a picture of what the manufacturer recommends but I haven't had too
good of luck with them in the past. The last two pictures are of my machine.
You can see how far out the plow is and how low my winch is. Would that
pulley and cable system helped at all?
You may not want to get the full physics explanation here, so I will first
give you a qualitative explanation. The tension T
in the strap is what is lifting plow and any part of it which is horizontal
(TH) is wasted. Your gut feeling is right, " it is
mostly pulling backwards " Anything which you can do to increase the
vertical part (TV) will make the lift easier, and moving
the winch up is a good way to do this but it might be easier to have a
pulley higher up which then brings the strap back down to the winch.
The figure shows all the forces on the
plow assembly: the weight W which acts at the
center of gravity (yellow X), the tension T in the
strap (where I have shown vertical (TV) and horizontal
pieces (TH)), and the force the truck exerts on the
support which I have represented as its vertical and horizontal parts,
V and H respectively. (Note that the
various forces are not drawn to scale since T has to be much larger
than W to lift the plow.) Suppose T is
just right that the plow is just about to lift. Then the sum of all the
forces must add to zero, or V+TV-W=0 and H-TH=0.
The sum or torques must also add to zero; summing torques about the point of
attachment to the truck (light blue X), WD+THs-TVd=0.
Note that the TV is trying to lift the plow but TH
is trying to push it down. Now, in order to get a final answer for the
unknowns (which are T, V, and H) we note that
TV=Tsinθ and TH=Tcosθ
where θ is the angle which the strap to the winch makes with the
horizontal. The final answers I get are:
I put in some reasonable numbers just
to get an idea of the answers, W=500 lb, θ=200,
D=2m, d=1.8 m, and s=0.1 m. Then T=1920
lb, H=1800 lb, and V=-157 lb. The negative value for V
means that V is down, not up. In this scenario,
the pulling force has to be nearly four times greater than than the weight
we need to look at whether the manufacturer's suggestion will be better than
a straight shot to the winch. Now there are two forces pulling up, the
tension T from the pull point to the winch and the
tension P from the pull point to some anchor
higher up. Of course the magnitudes of these two tensions are the same,
P=T. The picture shows only the pulling forces, the rest are the same
as in the picture above. There are still three unknowns, T, V,
and H. I will call the angle that P makes
with the horizontal φ. I will not show the details, just give the
As a numerical example, I will use the
same numbers as above and add φ=400. Then T=624
lb, H=1070 lb, and V=-115 lb. It is definitely
advantageous to use the manufacturer's suggestion here which, in my
numerical example, reduced the force the winch needed to exert by a factor
of about 3.
QUESTION: What would a chart of an 8g impact look like for an 8g impact with a 100
ms duration? I can send a jpg of what I am told fits the requirement. But I
seriously doubt that it does fit the requirement. But then again I am barely
understanding the test. I see spikes to 20-30g. I am having a hard time
finding anyone that can answer this question. (I asked for more
QUESTION: These are the charts provide by the AF to me when they
tested a pallet that we built.
The requirement is: Ultimate load. When uniformly loaded to 10,000 pounds,
the load being restrained to the pallet by chains, the pallet installed
between restraining rails locked to the rails by 2 locks through each rail
and engaging 2 lock notches on each side of the pallet, and resting on 4
rows of conveyor as specified (see 18.104.22.168), the loaded pallet shall
withstand a dynamic load of 3 times the force of gravity (g's) for a period
of 0.1 second. The pallet shall be serviceable after undergoing the test. In
addition, the pallet shall withstand a dynamic load of 8 g's for a period
not less than 0.1 second. The pallet need not be serviceable after
undergoing such a load; however, the pallet shall remain in one piece.
These are the result of those tests. These charts just do not look right to
me in their conclusions. I am not a math major, but I do understand some
stuff and this does not look correct. ie. "Area under the curve." or correct
averaging. If you can set me straight or put me on another path it would be
So, I must admit that I do not understand all the details of the
requirements (locks, chains, notches, etc.) What I can do, though,
is estimate the area under the curve which seems to be one of your main
concerns. I drew in a rough fit (green) to the data (red) and calculated the
area under the green curves (keeping in mind that the area of the triangular
segment below the axis is negative). As you can see, the area I got as a
rough estimate is 0.37 compared to the actual area of 0.35. The average
would simply be the area/time=3.5. My guess is that your pallets did not
fulfil the requirements since the precipitous drop before the 0.1 s had
passed would seem to indicate a collapse of the pallet. Both tests show this
behavior, with the 8g test having the "collapse" occur earlier as would be
expected. However, since I do not understand the details of the test well
enough, I would not take my guess as gospel.
QUESTION: I teach middle school science, however I've spent more time advancing my
understanding in several of the other science disciplines I teach so my
ability to work through this problem is above my current skill set. However
I would really like to be able to walk my students through this scenario, in
part because I'm not good with subtlety. I would like to know what other
variables I need to measure/know in order to calculate the force a
hypothetical student's head would experience at the moment it hit the floor
after they tipped their chair back too far (despite numerous warnings not
to). I imagine I'll need the Mass of the hypothetical student + the mass of
the chair. The Time from when the hypothetical student's center of gravity
passes over the point of rotation for the chair, the Height of the
hypothetical student's head is above the floor. With Gravity being constant,
I "should" have all the necessary information to calculate this (no chance
at all will I let my students try to measure the force in real life) but I
still have the niggling sense that I'm missing some variable.
Let's first do the simplest estimate: imagine dropping the head from
about h=1 m above the floor. The speed the head would hit the floor
could be found from energy conservation, ˝mv2=mgh
or v=√(2gh)=√(2x9.8x1)=4.4 m/s. Now, you either need to
know the distance the head stops in or the time it takes to stop. There is
not much give in either the head or the floor, so let's just guess that the
distance it takes to stop is 3 mm. If you do the kinematics, the
acceleration of something moving 4.4 m/s and stopping in 3 mm is about 3000
m/s2, about 300 gs! If you take the mass of the head to be about
4.5 kg, the force is about F=ma=13,500 N=3000 lb. The time over
which this force acts is very short, about 4.4/3000=0.0015 s.
To be fancier, you would have to do the
rotational problem. The student plus the chair have a certain moment of
inertia I and when they rotate about the rear chair legs they
acquire a rotational velocity ω when the center of mass has fallen
a distance h. In that case ˝Iω2=mgh
and so ω=√(2mgh/I); now you can get the speed the
head hits the ground by writing v=Lω where L is the
distance of the head from the rear legs of the chair. But to calculate the
moment of inertia of the student/chair would be very complicated and, in the
end, the speed of the head would not be all that different from the simple
calculation above. And you still have to make approximations over how far it
would travel to stop. Just use the simple head dropping calculation to
convince the students that the order of magnitude of the force would be a
few thousand pounds.
QUESTION: I am working on a science fiction book and have a bombardment of a
planet happening. I would like to get an idea of the force of impact of 2
kinetic kill vehicles 1 is a 10 pound Depleted Uranium ball. 2 is a
100 pound Depleted Uranium dart. They are traveling at 1/10th light
speed. They impact an Earth like planet. I only need a ballpark
value. If either or both are some how destroyed interacting with the
atmosphere or cause some other effect, that would be good to know too.
I am not sure what the material being depleted uranium has to do
with anything. Also, "force of impact" cannot be calculated unless you know
the details of the collision, in particular how long the collision lasted.
You could estimate the kinetic energy classically for your projectiles
because the speed is much less than the speed of light. 1 lb is about 4.5
kg, so K≈˝mv2=˝x4.5x(3x107)2=2x1015
J=2000 TJ (terajoules); the Hiroshima bomb had an energy of about 63 TJ.
If the collision lasted 10 s, this would correspond to a power of 200 TW
(terawatts). To give you a feeling for the magnitude of this power, the
total output of all power sources on earth is about 15 TW. The 100 lb
projectiles would have 10 times the energy and power as the 10 lb
projectiles. So those things would have quite a punch. Here is the problem
that most sci-fi writers never think about, though. Whoever is firing these
projectiles has to give them this energywhere are you going to get 200 TW
in the middle of empty space?
Now, the second
part of your question. These things have a speed (3x104 km/s)
which is much faster than the fastest meteor (72 km/s) and you know what
happens to themthey burn up and break up. The recent (2013)
meteor exploded at an altitude of about 20 mi and most of its energy
(about 1500 TJ) was absorbed by the atmosphere. It had a much smaller speed
than your projectiles (about 20 km/s compared to about 30,000 km/s) but a
much larger mass (about 107 kg compared to 4.5 kg), so the
energies were quite comparable (1500 TJ compared to 2000 TJ). So I would
guess that your projectiles would not do much damage to the objects on the
surface of the planet.
QUESTION: i'm traveling at the speed of sound and a gunshot is fired at the exact
moment I am passing the gun, what is the resulting sound that I hear and for
The figure shows the plane at three times:
just after the
gun has been shot;
at the time
when the sound reaches where the plane had been at time 1;
at a time twice as long as 2.
Also shown are the corresponding
spherical wave fronts of the sound from the shot. As you can see, the wave
fronts never catch up with the plane. You will never hear the gun.
QUESTION: I am a nurse at a long term care facility. My back hurts every time I
get finished pushing a medication cart for the 8 hour shift. My question
is...If the medication cart weighs 220 pounds when assembled, how much
weight am I pushing given the fact it is on wheels?
You may assume that the wheels almost remove the frictional force of
moving forward; in other words, it takes very little force to keep it moving
once it is up to speed. The times you need to exert a significant force on
the cart would be when it speeds up or slows down. The more quickly you
bring it up to speed or bring it to rest, the larger force you need to
exert. So plan ahead and speed it up or slow it down gradually. Here is an
example: if you sped the cart up to a speed of about 6 ft/s in 1 s, you
would need to exert a force of about 40 lb, whereas if you took 2 s, the
needed force would be about 20 lb. Also, when you turn a corner, go slowly
since it takes a force to turn the cart also and the faster you take the
corner, the greater the required force.
QUESTION: I have a small (15' diameter) swimming pool. I have a round leaf netk,
about 17' in diameter, which I suspend over the pool as follows: I took 5
10' lenghts of 1" PVC electrical conduit, and joined them end-end to form a
hoop/circle ca. 16' in diameter. The round leaf net has a drawstring, so I
placed the 17' diameter leaf net on the ground, then placed the 16' diameter
on top of the net. Then I folded the excess leaf net up above the conduit
hoop and tightned the drawstring. So far so good -- net below hoop, edges
wrap around hoop, drawstring holds the net on the hoop. I suspend this
hoop-supported net from a single point in the middle of the circle formed by
the hoop/net. From this central point, I have 10 "spokes" of cord (parachute
cord) going to 10 points along the hoop. The pieces of cord forming these
"spokes" were measured carefully to all be the same length. Here's my
question: When I suspend this hoop/net from this central point and the 10
lengths of cord (spokes), the hoop/net does not want to stay in a plane.
Instead, 2 opposite sides (e.g. at 12 and 6 on a clock face) lift
and the other opposite sides (e.g. at 3 and 9 in the clock face) drop down.
I know in German, when a bycicle wheel gets stressed too much and deforms,
this is referred to as an "eight" -- I suppose because from the edge it may
resemble an eight. I suspect there is something similar going on with my
hoop/net, but don't really understand enough to know for sure. As a followup
question, I would be grateful for any thoughts as to how I might get this
hoop/net to remain in a plane. Many people suggest just tightening up
(shortening) some of the "spokes" but I have tried this and it does not
The surface defined by your deformed loop is called a saddle point.
Because the pvc is pretty flexible (you were able to easily bend it into a
hoop), unless the weight is distributed very symmetrically, it is quite
possible for this kind of warping to take place. To minimize this uneven
distribution of weight, your spokes should be connected to the points where
you have used straight through connectors to connect the pvc pipes and the
points opposite as shown in blue in my drawing. But I suspect that there
will still be an asymmetric distribution of mass and the circle will deform
again. If so, then you will need to add crossbars as shown in red; this
should keep the hoop pretty resistant to deformation. You will probably only
need 3 spokes now and if it does not hang so the plane of the hoop is
horizontal, you can just adjust the lengths. It would also be good to have a
cord coming straight up from the center, maybe another piece of pvc to which
the spokes could be attached or just another cord. You could get some idea
of the balance by just hanging it from the center.
QUESTION: I was fishing and the water was rough. Waves crashing all over. Then I
look over and see a perfect about 20M circle of calm water while the whole
lake is thrashing about. It lasted for about 5mins. The water was about
20feet deep and there was not large rock anything different where the circle
appeared. The calm circle did not appear to be like a vortex or whatever of
water. It was calm and not rotating. Can you explain the physics behind this
or at least give me a guess of what you think caused this. (It almost looked
like an object that was not visible to the naked eye caused it but I know
thats not true)
I was just about to reply that I have no idea but just googled
calm water in a storm. It turns out that there is an old sailor's trick
to calm water in choppy conditions: just pour a little oil on the surface.
There seems also to be
understanding of this phenomenon. For a good overview, see the
Physics Stack Exchange. Perhaps there was some oil in your calm
QUESTION: How could I measure the viscosity of pizza sauce (and other materials)
using "at home" equipment? I want to determine the viscosity of a sauce,
then take the pizza sauce and place it on a turntable whose speed can be
controlled and see what speed is required to make the sauce flow from the
center to the edge.
Then, I want to alter the sauce and make it more and/or less watery
(changing its viscosity) and measure that new sauces viscosity.
Then, I'll take the new sauce and re-measure the turntable speed necessary
to make the new sauce flow from the center to the edge.
Last, I want to replicate the tests enough times so I can create an equation
that would allow me know what turntable speed would be necessary to
correctly flow the sauce based upon the viscosity of the substance.
Why not just experiment with sauces until you find the right
thickness to achieve what you want? Measuring the viscosity (not an easy
task) is just an unnecessary step in the process.
QUESTION: The reason is that I want to be able to alter the
sauce, test the viscosity, then know how much to alter the speed of the
rotating table consistently. Knowing, for example, a 1% increase in
viscosity requires a 5% increase in rotation speed allows me to continuously
alter the sauces and know with certainty the speed the table must rotate.
This must be a science fair project or something because it will
certainly not be of any help in making pizzas in the real world. In a pizza
you want to spread the sauce uniformly over the whole area, right? What
causes the sauce to move out on the rotating turntable? There is a
(fictitious) force F, the centrifugal force, which pushes the sauce
out, F=mv2/R for a mass m with speed
v when at a distance R from the center. But, the speed
depends on the distance from the center, v=Rω where
ω is the angular velocity; so F=mRω2 and an
ounce of sauce experiences a bigger force as it moves out. In other words
the sauce will tend to all be pushed out the the rim of the pizza regardless
of its viscosity. If the viscosity of the sauce too large, the centrifugal
force might be too small to move the sauce at all, so there would be the
tendency for the sauce to stay in the center. In any case, I cannot imagine
that it is possible to use rotation to get a uniform spread of sauce.
If you still want to pursue this, I
found the description of a straightforward
experiment to measure viscosity μ. You get a tall cylindrical
container and fill to a depth d with the fluid (density ρf).
Drop a sphere (density ρs radius R) into the
fluid and measure the time t it takes to reach the bottom; then the
velocity of the falling sphere was v=d/t. The viscosity is
then μ=(4R2g( ρs-ρf))/(9v).
It will be a little tricky since the sauce is not transparent. Also, it is
important that the sauce be homogeneous, no chuncks in it.
QUESTION: What happens to the atmosphere inside a cupping vessel when a flame is
introduced and causes a partial vacuum (negative pressure) - which allows
the therapeutic vessel to adhere to the skin surface? What are the physics
that explain this happening?
Cupping vessels have been in use for thousands of years. Ancient
Greek and Roman physicians used them to assist in
Chinese medicine uses them to
treat a variety
of maladies; cupping therapy is generally considered pseudoscience by modern
medicine. The idea is to provide suction on the surface of the skin and is
achieved by first heating the cup and air inside and then placing it on the
skin. As the air inside cools, the pressure decreases. The physics of this
pressure decrease can be understood by examining the ideal gas equation
which relates pressure P, temperature T, volume V,
and amount of gas N: PV=NRT where R is a constant
which depends on the units you use. So, as you can see, keeping the volume
and amount of gas constant, if the temperature decreases the pressure must
decrease. The fact that P∝T is sometimes called
A similar thing
happens when canning food in glass jars. The canning is done with the
contents very hot and a lid which is slightly domed is affixed. As the
contents cool, the pressure decreases causing the dome to pop inward toward
the contents, signaling that a good seal has been achieved.
QUESTION: I am a hot air balloon pilot. I am trying to develop a mathematical
formula, for calculating where my Scoring baggy will land, when I
participate in a Ballooning event. I am given a location in a Pilots
briefing, as to where the Scoring X's are located. I attempt to fly to that
particular location, drop my Scoring Baggy on the X, and earn points
depending on how close I am to the center of the X. (Each Leg of the X is
approximately 100' to 300' long, depending on the Ballooning event.)
The Scoring Baggy (6 ounces) has a
My Balloon's Speed, as I approach
the Target, is a variable. the Scoring Baggy will share that speed once
I release it, Correct ?
And, the height I release the
Scoring Baggy from, at the time I release it, is also a variable,
question is, is there a Mathematical formula that will calculate, how many
feet from where I release the Scoring Baggy, it will land ? And how long it
will take ?
This is not a simple question. Although it would be simple if air
drag were neglected, I suspect that it is not negligible for the bag
weighing only 6 oz. I will do the calculation without air drag here. I will
not include the details, just the final results. The time t in
seconds it takes for the bag to hit the ground is t=√(2h/g)=ź√h
where h is the height in feet from which you drop it and g=32
ft/s2 is the acceleration due to gravity. For example if you drop
it from h=144 ft, t=3 s. The distance x in feet
it will travel horizontally in this time is x=vt where vx
is the horizontal speed of the balloon in ft/s when you drop it; so if vx=20
ft/s (about 13.6 mph) and you drop it from 144 ft, x=60 ft. The
equation for x can be written as x=źvx√h
which is handy if you do not care about the time. Note that the weight
of the bag does not come into this at all.
this is pretty mathematical and probably not a computation you would want to
do in the heat of a competition! For my own interest, I want to estimate how
much error is introduced by neglecting air drag. It is much more complicated
if you include air drag. For this case, I need to do the calculation in SI
units rather than English units. The reason I need to use SI units is that I
will estimate the air drag force as Fd≈ź(vy2/A)
where vy is the vertical velocity of the bag and A
is the area it presents to the onrushing air; this estimate is correct only
for SI units because it has things like the density of the air at sea level
built into it. Now, it is my understanding that the speed of a hot air
balloon moving horizontally is the same as the speed of the wind; in other
words, from the perspective of a person riding in the balloon, he is in
perfectly still air. This greatly simplifies the problem because the bag
will drop straight down as seen from the balloon, i.e. it should
strike the ground directly below the balloon. So, the bag sees two forces,
its own weight mg down and air drag up; Newton's second law becomes
This is a first-order differential equation has a solution vy=-[√(g/c)]tanh([√(gc)]t)
where c=A/(4m). This solution is also a differential
equation since vy=dy/dt where y
is the distance above the ground; solving this differential equation, y=h-(1/c)ln(cosh([√(gc)]t).
A graph of this function compared to the case above for dropping from 144 ft
(note that 144 ft=44 m) is shown. (I approximated the area of the bag to be
0.01 m2≈16 in2=4x4 inches.) The time for the bag to
hit the ground is now 3.33 s rather than 3 s, approximately 10% longer
meaning that you should drop it when you are 66 ft from the target rather
than 60 ft. If you are lower the correction is smaller, if you are higher
the correction is larger. Given the circumstances under which you must act,
I would expect the inclusion of air drag to be unnecessary unless you are at
a very high altitude and that you should just drop it when you are about
x=źvx√h from the target.
COMMENT: It occurs to me that I have assumed that the wind speed and
direction are the same at all altitudes. This will not be true in the real
world and I am told that taking advantage of this is how hot air balloons
can get some control over direction of travel. Obviously, trying to do a
calculation including this would be impossible other than for a particular
set of wind velocities as a function of altitude.
QUESTION: The ideal gas law says that pv=nrt. If I have a balloon filled with an
inert gas and put heat into the balloon while holding the volume constant
the pressure will increase on the left side of the equation and the
temperature will increase on the right side of the equation. If I remove the
heat source and allow the balloon to expand the volume will increase and due
to the gas laws the pressure will decrease by the inverse of the volume
keeping the value for pv constant. However, on the right side of the
equation with a constant number of molecules the temperature will decrease.
How can pv remain equal to nrt with the temperature decreasing and pv
remaining a constant value?
What makes you think that PV will remain constant? This is
called an adiabatic expansion, a process where no heat enters or leaves the
system. What remains constant is PVγ
where γ is a constant which depends on the gas. For example, for a
monotonic gas γ=5/3 and for a diatomic gas γ=7/5. Once you
know what the new P and V are you can get the new T:
QUESTION: So while I was a little kid due to numerous headaches I had, I was
scanned in an MRI machine. I was feeling a little anxious when the machine
was put on and my mom came to me to ease my anxiety. She had her wallet with
her and pretty much all her cards went dead. What exactly causes magnetism
to destroy payment/membership cards?
The magnetic strip is just like magnetic recording tape. There is a
layer of very fine particles which are magnetizable. Data is written on the
tape by using an electromagnet called the recording head; when the magnet is
turned on the particles become magnetized. So the data is written in stripes
in a code, sort of like the UBS labels used to scan products at the cash
register. In a magnetic strip, the card moves by a tiny coil in which a
current is caused to flow when the magnetic stripe goes past it. Since
magnetic fields are used to create the magnetized particles, magnetic fields
can be used to destroy them. Even a relatively weak field, if present for a
long enough time, can mess up the data on a magnetic strip. An MRI machine
has a huge field and it would easily demagnetize the strip.
QUESTION: I would like to perform a calculation of a man descending a tower using
cords and subjected to the action of the winds. I have been trying to find
some equations but it is somewhat difficult due to the drag force. My main
objective is to calculate the maximum horizontal distance x that the man
could reach due to the wind action against the technician descending the
tower using cords. Tower height: 78 m (please consider up tower as a
zero reference). wind speed: 20 m/s Mass of man: 70 kg Man
descending with constant speed and slowly. So, please what is the
maximum distance when the man is at 66 m from the top of the tower ?
As shown in the figure, there are three forces on the man, his
weight mg, the tension in the cords T,
and the force of the wind F. The equations of
equilibrium are F-Tsinθ=0 and mg-Tcosθ=0.
Solving, tanθ=x/y=F/(mg), so
x=ytanθ=Fy/(mg). Now, how can we get F?
There is a very good approximation to the drag force by air at sea level
moving with speed v: F≈źAv2 where
A is the area of the object presents to the onrushing air (and which
works only for SI units). Finally, x=Av2y/(4mg).
If I approximate g≈10 m/s2 and A≈1 m2
and use your numbers, x≈9 m.
COMMENTS: I should have emphasized that air drag calculations are
only rough calculations, probably accurate to maybe ą20%. Also, for
your situation, the general approximation for x as a function of
y is x≈y/7. Also, if the horizontal displacement seems too
large, keep in mind that 20 m/s is a very strong wind, about 45 mph which is
QUESTION: Please explain me that why if a thin layer of water is spilled on a
rough surface like plastered floor and we place our finger in it then why
water move away from the point of contact of our finger on that surface and
it appear to be dried around.
This is probably akin to the similar phenomenon of a foot pressing
down on wet sand and the area close to the foot is visibly dried. I found an
Physics Forums which seems to be correct: "The phenomenon being
described is called 'Dilatancy' and was discovered by Reynolds about 100
years ago. It works only when you have well compacted sand that contains
just enough water to cover all the individual grains. When you stand on the
sand you create a stress / force which causes the sand to move. In order for
the sand to move / flow individual grains have to be able to move past one
another. Imagine a bunch of oranges stacked as you might see them at a
grocers. The first layer has them all tightly arranged and then the second
layer sits down into the gaps between the oranges on the first and third
layer. Now imagine trying to move one of the oranges in the second layer. In
order to move it the oranges on the first and third layer mut move down and
up respectively to enable the orange to move. Effectively the volume of the
pile of oranges or grains of sand increases with bigger gaps in between. So
when you put your foot down on the sand it shoves sand out the way but in
doing so the volume in between grains has to increase temporarily to allow
the grains to move relative to one another. Consequently all the fluid at
the surface is sucked by surface tension into the extra gaps made by the
rearrangement of the sand. Since there is no longer any fluid at the surface
the grains of sand are now dry." If you go to the original Physics
Forums question, ignore all the early answers which are wrong.
QUESTION: Say I fill an airtight barrel with water and have a valve at the bottom
and a feeder hose at the top. If this barrel is uphill and I have the feeder
hose down lower say in a pond will the draining of the barrel through the
lower valve create enough vacuum to pull the water uphill creating a siphon?
First of all, I would call what you are proposing a pump, not a
siphon. You are trying to "suck" water uphill using a vacuum. The first
thing that comes to mind is that there will be a limit on how high the hill
is above the water level below. Even if you have a perfect vacuum, the
highest you can lift water this way is 10.3 m=33.9 ft. But you start off
with a hose full of air, so you will never get a vacuum, so you will be
limited further in the height to which you can pull the water from below.
For example, if the volume of the air in the hose were 1/10 of the volume of
the barrel, you could only lift the water 9.3 m. Or, if the volume of the
air in the hose were equal to the volume of the barrel, you could only lift
the water 5.1 m. So there is no simple answer to your question, but this is
probably not a very workable way to lift water.
QUESTION: How much heat would it take to heat 1 gallon of water to 600 deg F in a
pressurized system, from 70 deg F to 600 F in 1 hour. Not counting the ss
vessel. Also since the water is not allowed to change states are the
calculations just the Sensible heat cals or are there special calculations
needed. This is part of a R&D Application.
For my approximation to be fairly accurate, the water must remain
liquid at a constant volume. I will work in SI units so 700F=210C
and 6000F=3150C; I will convert back to Imperial units
for the final answer. I looked up
data for the specific heat of water which turns out to have a
significant temperature dependence as shown in the figure (black). I did a
quadratic fit (green) to these data and integrated over the temperature
range to get E=1356 kJ/kg. The mass of a gallon of water is about
3.8 kg so the total heat is Q=1356x3.8=5.2x103 kJ=1.44
kW⋅hr=1240 kilocalories. Keep in mind that the pressure will be very large
at 6000F, about 1800 psi.
QUESTION: How much "work" (physics definition) is actually accomplished in a gym
workout? I'm currently using F x D x reps = actual work done. The upward
lift ("D") x the amount of weight lifted ("F") x the number or
repetitions... to get actual work done. Am I in making some mistake, here?
Thanks, in advance, for your help.
By "upward lift" I assume you mean the distance lifted. So, lifting
a weight F over a distance D you would do W=FD
units of work on the weight. For example, the weight of a 2 kg mass is about
19.6 N and the work to lift it 1 m is 19.6 J. But, and here is the catch,
you use more energy than 19.6 J to lift that weight because your body is not
a simple machine like a lever or a pulley. To understand why, see the
faq page. In a nutshell,
the reason is that to just hold up a 2 kg mass, not move it up at all,
requires input of energyyou get tired trying to hold up a weight at arm's
length, right? And, what about lowering the weight back down? The work done
on the weight is negative which implies that energy is being put back into
you but know that it also takes energy for you to lower the weight at a
constant speed. A biological system is considerably more complex than
systems we talk about in elementary physics classes. I think that it is of
little use to try to analyze a workout in terms of elementary physics.
QUESTION: If the earth is curved how is it you can get a laser to hit a target at same height at sea level more then 8 km away?
How is it that it's bent around the earth?
First of all, light is not bent around the earth; it travels in
a perfectly straight line and therefore, because the earth is curved,
there is a maximum distance away for a target at the same altitude. What
that distance is depends on the altitude of the laser. You say that the
laser is exactly at sea level by which I presume you mean the surface of
the earth; at this altitude you could not hit any target also at sea
level. In the figure I have drawn the earth, radius R, a point
a distance h above the earth's surface (laser location), and
another point a distance h above the earth's surface (target
location). The distance between them is 2d. Focus your
attention on one of the triangles with hypotenuse (R+h).
From the Pythagorean theorem, d=√[(R+h)2-R2]=√[2Rh+h2];
if h<<R, d≈√(2Rh). For example, if
h=10 km, about the height a commercial jet flies, 2d≈714 km is the most distant target at the same altitude which you could hit.
I have always wondered how much energy do you do with if you let a kettle at 1800 W be running for two minutes? What is the approximate cost for this?
this is not a homework question.. just a question i wonder =)
Watt is one Joule per second, 1 W=1 J/s. Energy consumed by an 1800 W
kettle in 2 min is 1800x120=216,000 J. But, we are more used to
measuring electrical energy in kilowatt hours, (1 kW·hr)(1000
W/1 kW)(3600 s/1 hr)=360,000 J. So the energy used by the kettle is
(216,000 J)/(360,000 J/kW·hr)=0.6 kW·hr. A kW·hr
costs on the order of 5¢-15¢, so the cost would be between 3¢
This is odd, but My family has just moved into a huge house with little outdoor space. We live in a climate that is cold in the winter, and I want my children to get some exercise on a daily basis. We own a trampoline, and have space for it indoors on the Second floor of our house. The ceilings are 12 feet high, so there would be no problem with the kids hitting their heads on the ceiling. My question is whether or not the house would stand up to the force generated by the trampoline. The walls of the house are made of concrete (you can't nail into it.) I am assuming the floors are quite solid as well, as they must support the weight of the house. They are concrete as well.
My Youngest child is quite large (6 ft, 260 lbs)--he is only twelve. We need the activity.
First, a disclaimer: I can give you an idea of how much force the floor
will experience. I cannot predict whether this will cause your floor to
fail because I have no information about your floor other than that it
might be concrete. I have watched some videos and it seems that the
jumper never goes as high as h=2 m and the trampoline never
goes down as far as s=1 m. So I will just do my calculations
with those to get an upper limit on what force might be expected. Your
son's mass is about m=120 kg. An object falling from h=2
m will hit the trampoline with a speed of about v=√(2gh)≈√(2x10x2)=6.3
m/s. I will treat the trampoline as a simple spring so that I can write ½mv2=½ks2-mgswhere k is the spring constant. Putting in m,
v, and s and solving for k I find k=7200
N/m; since the force exerted by a spring is F=ks,
the largest force the trampoline exerts on your son is about 7200 N=1600 lb;
Newton's third law tells you that this is also the force your son exerts
down on the trampoline. Therefore, the trampoline exerts a force down on
the floor of 1600+W where W is the weight of the
trampoline. This is a little more than the weight of a grand piano.Keep
in mind that this is the greatest force and just for an instant; the
average force over the collision time would be half this. This is a
little more than the weight of a grand piano.
hi, can and are earthquakes be caused by celestial alignments ie planets?
Let's take a simple example. As seen from earth, Mars and Jupiter are
aligned. I estimated the force on a 1 kg object which is sitting, let's
say, on the San Andreas fault: F=3x10-11 N; the
weight of that 1 kg object is about 10 N. I would say that putting a 1
kg object on the ground is a great deal more likely to cause an
earthquake than those planets, wouldn't you?
So there's a powerline outside my bedroom window, and I thought, huh. Turns out I'm sleeping with my head in a 6mG AC magnetic field (according to two meters). Help me use physics to stop caring.
How do I estimate/calculate which puts more force on the charged particles (calcium, potassium, sodium) in my brain: a) An aqueous solution at 98.6 degrees Fahrenheit or b) a magnetic field acting on charged particles moving at some estimated speed in said aqueous solution.
My hope here is that the force of (b) is like an order of magnitude or two, or something, below the "noise floor" of (a) and then I can stop caring forever.
How about this: the earth's magnetic field is about 0.6 G, two orders of
magnitude bigger than the field due to the power line, and you are
exposed to it 24 hours a day. It is also possible that there is some
other source of field closer by than the power line which, though a much
smaller current, would produce a much bigger field. For example, if
there were a wire in the wall carrying a typical household current of 1
A, the field 2 m away would be 1 mG. There is no good scientific
evidence that any magnetic fields you are likely to encounter have any
effect, good or bad, on the functioning of your body.
Several years ago, I was caught in a massive windstorm in a skyscraper. I was on the 54th floor (approx. 756 feet from street level, full building height is 909 feet) , pulling cable, and I stopped for a break. I left a cable pulling string hanging from the ceiling (48 inches free hanging length) in the office, with a 1/4 lb weight attached, and when the storm hit, the weight began swinging like a pendulum. The arc was 16 inches (eyeballing it), and traversing the length of the arc took about 1 second. How can I calculate how far (full arc) the skyscraper was moving by observing what the pendulum in the building was doing?
A 48" pendulum has a period of about 2.2 s, the time to swing over the
arc and back. Since you were estimating, the pendulum was swinging with
about the period it would if the building were not moving at all. I
would conclude that either the pendulum got swinging somehow and the
building was not perceptibly moving or that the period of the building's
motion was about the same. If the building was swinging with a period
significantly different from 2 s, the pendulum would be swinging with
that same period; that is called a driven oscillator.
QUESTION: How strong is 0.01 newton meters of
torque? I want to build a motorized
camera slider that will pull a Canon 5d mk 2 up a 45 degree slope. It weighs
1.5 kg. How much torque do I need?
is the motor I am thinking of buying:
Rated power 0.01 W Rated speed:20 rpm Rated torque:0.01 N·m
I will work it out in general and then we will see if this motor can do
the job. There are some things you have not told me, in particular the
speed 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 is the
known quantity, FR=τ=mgR(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
N·m; the required power would be P=14.7x0.01/√2=0.1
W. I am afraid that your proposed motor is inadequate.
Unlike a submarine traveling in a horizontal line at 5 fathoms compared to 50 fathoms (the pressure in the vertical direction would be the same....) But in a vertical line I imagine from 50 fathoms it would rise slower in the first 5 fathoms than it would in its final 5 fathoms.... I.e in its first 5 fathoms of accent it had the external pressure of 50 fathoms descending (dropping to 49 fathoms pressure then 48 etc) compared to 5 fathoms pressure descending to 0 fathoms pressure as it ascends.
Is this correct....
As long as the water is considered incompressible (which it is for all
intents and purposes), the net force on the submarine due to water
pressure (called the buoyant force) is the same at every depth. Even
though the pressure in the water is enormously bigger at great depths,
the pressure difference between bottom and top of the submarine is the
same. Therefore, if drag forces are neglected there will be a net force
up on the submarine which is constant and equal to the buoyant force
minus the weight of the submarine. However, the drag force is not
negligible and is approximately proportional to the speed. So as the
rising submarine speeds up the drag force down becomes bigger and bigger
until it is eventually equal to the buoyant force minus the weight;
hence the net force is zero and from that depth up to the surface the
speed is constant. You might also be interested in a related
earlier question about
If you took two 1000 mile long metal bars, laid one horizontally on the
ground and stood the other vertically, would they weight the same and if
By weight we mean the force of attraction between the earth and the
object. Therefore, they would not weigh the same because the vertical
bar has most of its mass farther from the center of the earth than the
horizontal rod does. The weight of a point mass on the surface of the
earth is W1=mMG/R2
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 W1=mg.
But, if the point mass is a distance H above the surface,
its weight is smaller, W2=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 W3=mg/(1+(L/R)).
In your example L=1000 mi is not small compared to R≈4000
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.
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
according to the formula of variation of g above the surface of earth
g' = g(1+2h/R) and the radius of earth is 6400 km .If we put R/2=3200 in the previous formula,did it mean that after 3200 km space start ?
First, let's see where your "formula" comes from. The acceleration due
to gravity g may be written as g=MG/R2
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.
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.
Is a board suspended between two points and evenly weighted across it's
length (by books, for instance), less likely to bend if the ends are clamped
or nailed so they don't move? This is in contrast to if the ends simply
rest on end supports.
Yes. Keep in mind, though, that the board will be exerting a horizontal force
on the support. Be sure the end supports are strongly attached to
If the end points are dado slots (the boards are shelves for media storage), and the shelves and dados are cut so that the shelves fit the dado slots very precisely on all three sides, then horizontal travel and twisting could be eliminated, which would obviate the need to attach to the wall. Correct?
For the board to bend, the ends are pulled toward the center and lifted
up (see figure). I would anchor the ends.
One more question though, and a more challenging one I hope.
If one wanted to screw a square block of wood to a wall, where would the optimal places be to put the screws? Assume that the block of wood is 12" on a side, two edges are parallel to the floor and only two screws are used. The wood will be weight-bearing and the weight will be evenly distributed along the top edge of the block. Also, assume that the thickness of the wood is small relative to the other dimensions and that the screws can find equal purchase in the wall in back of the block of wood being fastened.
My gut tells me that (a) the screws both should be put along a horizontal line parallel to the floor, and, (b) it doesn't matter where that line is, high or low, since the weight is evenly distributed. What I can't intuit, is if it's better to have the screws be 3" or 4" or some other distance from a vertical side, as long as they are symmetrically placed.
In the general case, if N randomly placed screws are used, what is the math that tells me what percentage of the weight each screw bears according to its geometric placement?
I really do not think it is worth worrying about. The reason is that it,
paradoxically, is likely not the screws which keep the block from
falling, it is the static friction between the wall and the block; the
function of the screws is more to press the block against the wall than
to hold up the weight. Therefore, regardless of where you put the
screws, tighten them as tight as you can get them since the amount of
frictional force you can get depends on the normal force, the force
pressing the wall and block together.
I have measured the thickness of steel pipes with a thickness meter (pocketmike) with a sound velocity of 5813 m/s.
No I need to change my thickness values with a sound velocity of 5920 m/s.
This because this value is more accurate for this steel pipes I measured...
Which formula I need to use?
I would assume that the device measures a time. In that case, the
thickness 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
QUESTION: What ideas are the basis of the special theory of relativity and what conservation laws does it preserve?
This isn't from homework, but a class discussion that I didnt understand
This could be called an unfocused question, but the answer is pretty
brief so I will carry on. Most textbooks will tell you that you need to
postulate that the laws of physics are the same in all inertial frames
of reference and accept that the speed of light (electromagnetic
radiation) is independent of the motion of the observer or of the
source. My take on this is that only the first postulate is needed
because Maxwell's equations are laws of physics and they predict that
the speed of light depends only on two physical constants, not on the
motion of the source or observer. All important conservation laws—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 ½mv2 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.
This came up in a discussion about the use electro-magnetic rails guns whose power is contained in the speed of a given mass when it slams into a stationary object as there is no explosive material involved, the energy produces is converting the kinetic energy of the projectile into heat on impact.
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?
You have many misconceptions, so I have edited your question(s) by
itemizing the parts of your question. I will answer by parts.
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=½mv2
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=Rωcos(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=6.4x106x7.3x10-5=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
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
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.
the projectile stikes something, like the moon in your question, all
that matters is what its velocity is with respect to the moon.
Due to the ideal gas law/approximation the speed of sound does not change with altitude or barometric pressure. Yet ask any long range shooter and they will tell you, from practical experience, that the lower the barometric pressure or the higher the altitude, the less drag there is on the bullet and therefore the bullet takes longer to slow resulting in less time for gravity to affect the bullet leading to less "drop" for the bullet over long ranges. Why then doesn't the ideal gas law apply to bullet trajectory?
I do not know about the speed of sound, but I would certainly not assume
that anything about sound could be applied to bullets. The drag force
Fd of a bullet of speed v in air is well
represented by Fd=½Cρv2
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.
Would it be possible for an 'average' human to throw a 500 g weight 40 ft but
with only a 2 ft high corridor of travel without the projectile touching any
of the corridor's sides? This isn't homework, it's a really ridiculous question that we can't physically try at work! We work in a theatre that has two people positioned 40 ft apart and roughly 30 ft up in the air near the ceiling. Ages ago someone said that they could throw a spanner from one person to the other person. Some people agreed, some said that it wasn't possible due to the arc needed. This question comes up every few months and I just wondered if you might have an answer.
So the problem is, with what speed would you need to throw the spanner
in order that it reach a maximum height of 2 ft above its starting
height when it has gone 20 ft? This is a straightforward kinematics
problem. You are probably not interested in the details, so I will just
give you the final result. The necessary speed would be 57.7 ft/s=39.4
mph. A major league baseball pitcher can throw a fastball with a speed
of around 100 mph, so there is probably somebody in your theater company
who could throw the spanner with the necessary speed, but I would not
want a 1.1 lb wrench coming at me with a speed of nearly 40 mph!
In some of the more realistic space combat science fiction there is a concept called a 'BFR' (Big Fast Rock) which, mined from a dead world or asteroid, melted to molten and then reformed to a near-perfect density distribution with collars of ferrous metal impressed in it, is shot at some fraction of light speed from a large EML cannon running down the long axis of mile long ships. I would like to know how to calculate the impact force release for a 2,000 lb BFR moving at .10, .15 and .30 C. I'm assuming it's going to be in the high Megaton range and I don't know what the translative math per ton equivalence in TNT.
What you want is the energy your projectile has when it hits. The energy
in Joules is E=mγc2 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
I have a question related to determining the force at impact.
Here's the question in two parts:
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).
The "force at impact" depends, essentially, on the time to stop. If she
has a speed 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 is the acceleration due to
gravity. So, yes, if you increase the mass you increase the force
proportionally; I guess I would toss that extra 45 lb over before I
jumped! Regarding your second question, call the height of one floor
h. Jumping from the second floor would result in a speed of v2=√(2gh);
jumping from the third floor would result in a
speed of v3=2√(gh) which is
about 1.4 times larger than v2. In general, jumping
from the nth floor would result in a speed vn=√[2(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 v2=27.7
ft/s=18.9 mph and v3=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 t2=2x3/27.7=0.22 s and t3=2x3/39.2=0.15
s. Finally, the average forces during impact are F2=160+5x27.7/0.22=790
lb and F3=160+5x39.2/0.15=1467 lb. These are
approximately 5 and 9 times her weight (g-forces of about 5g
I want to ask about the Mandela effect theory. I do not want detailed answers concerning blackholes and cosmology,instead i want to know if it's real. I personally do not believe in time travel and all the "clues" people have been providing don't seem credible enough. So please if it is true what is the exact physics background that proves it? And how can i have further information and dig deeper into this theory?
I never heard of the
Mandela effect. For readers who, like me, are similarly ignorant it
is a special case of a psychiatric phenomenon called
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
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: As surface tension means force acting on surface then why not its unit if force per unit area i.e N/m2.
If it is N/m then which length will we have to take here.
Think of the surface as an elastic sheet. What you want to call the
surface tension is how hard you need to pull on this sheet in order to
make it rip. The way this is usually measured is to create a film which
is actually a very thin volume with surfaces on both sides; a simple
device to do this is shown in the figure. Now, pull on the sliding side
to stretch the film. When the film breaks, measure 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.
I just came across an ad saying that
the earth's magnetic field keeps us healthy and strong, and saying that if we live in the world with no magnetic field, all life on earth will end (assume that sunlight / water / oxygen exist / and for some weird reason gravity still works)
So my question is, is this valid? If there is no magnetic field, even though we have sunlight, water and air, we would perish?
magnetic field has no direct effect on you. There are many magnetic
products touted as having miraculous beneficial effects on your health;
do not buy them, they are a scam. There are important indirect effects
of the magnetic field which help provide a healthy environment at the
surface of the earth, though. Probably the most important is that the
field extends far into space and protects the earth from the solar wind
which is a stream of charged particles (mainly electrons and protons)
which are deflected around the earth by the field. If there were no
field, the effect of this "wind" would be to strip the upper atmosphere,
including the ozone layer which protects you from intense ultraviolet
radiation from the sun. But I do not think that buying a magnetic
matress pad, for example, would help that issue! Lack of a magnetic
field cannot result in our perishing, though, since it is well known
that the field reverses direction periodically; the reversal would not
be instaneous which would mean there would have been a period at least
hundreds of years long when the field was negligibly small. The most
recent reversal was about 780 million years ago when humans existed and
we survived somehow. One thing about your question is strange: "…for
some weird reason gravity still works…" The earth's
gravity has nothing to do with its magnetic field.
QUESTION: If you tow a boat from a tow path which is the best point to tie the rope onto the boat?
The rope will exert a force on the boat, obviously. This force will tend
to do three things: exert a torque which will tend to rotate the boat
about a vertical axis through the center of gravity (you don't want
this), have a component along the bank which will pull it along the
canal (that is what you want), and have a component which will pull the
boat toward the shore (you don't want this). To minimize the tendency of
the boat to turn, attach the rope close to the center of gravity of the
boat. To minimize the tendency to drift (not turn) to the bank, make the
rope as long as possible so that most of its force will be exerted along
the bank. Some tiller will be needed to make small corrections, but
those should be minimal.
Does water slow down gradually as it rises vertically out of a fountain? It would seem that it should, but that would mean the water behind would catch up and create a constantly wider jet. However it seems that the jet stays relatively consistent in width (subject to some splashing) until it reaches the top. At which point it splays and falls. This would suggest it is at a consistent speed all the way up until it suddenly decelerates and falls.
Any "piece" of the water has a downward acceleration (slows down on the
way up, speeds up on the way down) at all times; ignoring friction
effects, each drop of the water follows the same trajectory as a thrown
ball would follow. So, if one piece is slowing down, why doesn't the
piece behind it catch up? Because it is slowing down too at exactly the
same rate. If you threw two balls up, one right after the other, the
second would never catch up. Of course, if the stream of water were
going vertically up, the water falling from the top would collide with
the water rising and, as you say, splay out.
It's a question about R value, and directed to you partly because it appears you are in New Zealand. I'm looking at a video from an inventor in New Zealand who is recycling plastics into building blocks. I cannot find him on the net to ask what is the R value of these blocks. I know it is a long shot to ask you, but I wondered if there is a way to calculate R value, based on assumptions about physical properties of plastics such as milk jugs, trapped air and density in these blocks.
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
Heavy water is Deuterium oxide. Water is H2O. 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?
Deuterium (D) is a naturally occuring isotope of hydrogen; approximately
0.015% of hydrogen atoms are deuterium, which is about 1:6,670.
Therefore about 1:3,330 molecules of water would be HDO and
approximately 1:44,400,000 (6,6702)would be D2O.
The term "heavy water" refers to D2O 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 D20 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% H20,
and 25% D2O. Where did all that D2O 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% D2O
are routinely achieved.
Is it possible for a skydiver to not be strapped in to a parachute, just holding onto it,
or even wrapping his arm into the straps. When he pops the schute, is it possible to hold on to it? Would his arm withstand the sudden deceleration or could it be torn from the shoulder socket?
There is, of course, no simple answer to this question. It depends on
how fast he is falling and the parachute design. It also depends on
"luck" as the graph shows since the two graphs are for identical
parachutes, weights, and speeds. The drag force (the force the parachute
exerts up on him) as a function of time is shown in gs. One
g would correspond to the weight of the parachutist. Both graphs
end up after about 7 s at gforce=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!
If I have a .25 inch airtube in
a 2 ft deep fish tank, how can I calculate the force necessary to expel the first bubble of air at the bottom of the tank if the other end of the tube is at the surface?
The weight density of water is 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
QUESTION: I am an avid sports fan and I have often wondered if the experts may be wrong about the myth of the rising fastball in the game of baseball. I played baseball for over 20 years and I can tell you that the ball does appear to rise when certain pitchers throw hard put a heavy backspin on the ball. I have been told that experts say it is nothing more than a visual trick your eyes play on you because a rising fastball is considered to be physically impossible. I can tell you first hand that a softball pitcher I know can throw a ball that rises after being thrown on a straight trajectory. I suspect the Magnus effect may have something to do with the anit-gravitational behavior of the ball. Do you think this could be what causes the ball to appear to rise as it travels, or is it just our perception?
There is such a thing as a rising fastball, but it does not actually
rise; it simply falls more slowly than a nonspinning ball does. An
experienced hitter knows intuitively what a normal fastball does and
when presented with a rising fastball he will swear that it rose because
it actually fell less. Incidentally, by rise or fall, I am talking about
the direction of the acceleration. So a ball which is thrown at an angle
above the horizontal is obviously rising but it rises at a
rate of rise until it reaches the peak of its trajectory and then begins
back down; the rising fastball will actually rise farther. Then why do
we say it is a myth? It is easiest to understand by looking at a ball
thrown purely horizontally. Can spin cause the ball to actually go
upwards? To answer this, you need to think about all the forces on a
pitched ball. There is the weight, FG, 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 FD 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 FM 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.
I'm learning about ideal gases in class and learning about the ideal gas law, but we also (very briefly) mentioned that at low temperatures and high pressures the ideal gas law no longer holds very well. We didn't go much pass that but I did a bit of research and an explanation included another set of constants calculated experimentally, but again only really explained very briefly that the size of the molecules become much greater and the inter molecular force between the molecules is no longer insignificant (as is a factor of the kinetic molecular theory of gases). I guess my question is, what is so substantial to the aspects of low temperatures and high pressures of ideal gases that causes them to follow a different set of parameters?
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
pairs because N is very large. So, finally,
where c is a proportionality constant which depends on what the
particular gas is. This is sometimes called the van der Waal equation.
When I had solar panels installed on my house the old fashioned meter ran backwards when the Sun was bright. They have now fitted a digital meter which can sense when energy is being sent into the grid. I can see how this could be done with DC, but how does it work with AC? In AC the current is switching direction at 50 Hz. How can the meter sense the 'direction' of the energy flow?
You are right, the average current is zero. However, the current is not
the power—the power is the product of the current and the
voltage. Both current and voltage are sinusoidal functions of time,
i(t)=Isin(ωt) and v(t)=Vsin(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.
Thank you for your reply to my question about the power direction with energy generated by solar panels. I now understand that phase angles are crucial.
The inverter in the loft takes DC from the panels and converts it into AC. To feed energy back into the grid does the inverter have to deliberately adjust the phase or does this occur naturally when there is an excess of energy?
All that matters is what the phase is at the meter. If energy is flowing
into your house the phase will be one way, if flowing out it will be the
Why don't liquid pipelines that run downhill rupture from the weight of the liquid in them? For instance, a 14 mile long section of 10" ID pipe carrying crude oil that runs at a 22.5 degree angle has about 301,592.89 gallons of crude in it. That Crude weighs 91,328,360.39 pounds. I calculate that the static weight at the bottom of that run should be about 22,832,090.10 pounds of oil. That exceeds the tensile strength of the pipe wall by a factor of 50 to 100. At first I thought it was because a closed pipe would have sort of a vacuum at the top, but then I realized that would make the pipe crush from atmospheric pressure. I'm missing something simple I am sure, but darned if I can figure out what it is...
There is something seriously wrong with your numbers. They imply that
the weight of one gallon of crude is about 9x107/3x105=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+½ρv2=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, Ptop=PA and ytop=1000
m and at the bottom Pbottom=P+PA
and ybottom=0; here, PA is the
atmospheric pressure. Therefore, PA+900x9.8x1000=P+PA+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.
Suppose I have a 10 tons weight hanging 5 meters up in the air. I want to get electricity by lowering the weight against a dynamo (for example).
How much energy do I get?
A 100 W light bulb needs 100 W of power when it's ON. So, if it stays on for 10 hours it will consume 1 KW, am I right?...
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:
200,000kg * 9.81m/s2 * 100m =
196.2MW/3600 = 54,500KW/h Is the formula right? If yes, how do I apply it? Because I get extremely small numbers if I change his 200,000 with my 10,000, and his 100 meters with my 5 meters.
Plus, I don't know what KW/h means. All I'm interested is knowing how many Watts are available at any given moment while the weight is falling.
The first thing we need to get straight is what a watt is. The unit of
energy in SI units is the Joule (J); 1 J is 1 N·m where a Newton
(N) is the unit of force and the meter (m) is the unit length. A Joule
is the kinetic energy which a 2 kg mass moving with a speed of 1 m/s
has; or, it is the work you need to do to lift a 1 kg mass to a height
of 1/9.81 m. A Watt (W) is the rate at which energy is delivered or
consumed, 1 W=1 J/s. Therefore, your 100 W bulb consumes 100 J of energy
every second. Incidentally, if you look at your electricity bill, you
will be billed for how many kW·hr you have consumed; a
kW·hr is a unit of energy, 1
kW·hr=1000x3600 J=3,600,000 J.
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 104x9.81x5=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.
If there's less gravity on the moon, then why do astronauts move slower? Why don't they have more spring if gravity's pull is weaker? I'm thinking that this should be a matter of resistance, but obviously there's a factor I'm not accounting for.
Sure, with less gravity (about 1/6 of earth) an astronaut can jump much
higher, but it will take longer to get to the top and back down than on
earth; that would look like slow motion, right? Or, just think about
taking a step. Your center of gravity is forward of your trailing foot
and so you rotate about that foot. The rate of rotational acceleration
is only 1/6 that on earth, again like slow motion.
Engineers in the Bay of Fundy are trying to harness tidal power to generate electricity. It's claimed that they can generate enough electricity to power the entire Atlantic Provinces of Canada.
This got me to thinking, if that much power can be harnessed, where has that energy been going? I'm assuming it goes to heat and warms the water. If that's the case, would harnessing the power therefore cool the water and potentially harm aquatic ecosystems?
My research shows that all the power projects on the Bay of Fundy will
generate about 20 MW. These will be powered by underwater turbines. I
did a very rough estimate of the total power available by the falling
water: The average rise in sea level is 15 m, area of the bay is 13,000
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.
i was studying something on the internet where i need to know one question i.e If an object Travelling at about 2500 km per hour or may be higher collides with other object in the ocean around 90 meters below the ocean level is it possible that the debris of that object will go around 2 km approx ahead in the water itself?
Extremely unlikely. Let me show you a very rough calculation which will
demonstrate this for an extreme case. If one object collides elastically
with another mass whose mass is very small compared to the incident
mass, the collided-with mass will recoil with a speed twice the incident
speed, in your case that would be 2x2500 km/hr=5000 km/hr≈1400
m/s. The drag force on an object with speed v in water may be
approximated as Fdrag=½CρAv2;
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=v0/(1+kt)
s-1and v0=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!
Would my weight be the same on the surface of earth and one mile underground? How about one mile in the atmosphere?
Above the surface of the earth the gravitational force falls off like 1/r2
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=R2/r2=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
Helium. Where do we get it from if it is lighter than air and doesn't react with any other elements in the normal human tolerant environment?
Good question. Even though it is the second most abundant
element in the known universe, there is virtually none in the atmosphere
(because it is so light that its average speed is greater than escape
velocity and it shoots off into space) and is not tied up in rocks,
water, or other chemicals (because it is inert) like hydrogen is, for
example. This element was not even discovered until 1868 as a spectral
line in the sun (where untold zillions of tons are being produced every
second from nuclear fusion) and not found on earth until 1895 when trace
amounts were found coming from uranium ore; the source was as a nuclear
decay product in α-decay. The first large amounts were
discovered in 1903 as a byproduct mixed with the methane in natural gas
wells; today large scale amounts come only from helium trapped
I am curious about a topic. In golf, if I hit a ball very hard and then I hit one very softly, is the one
hit very softly more likely to move or sway in its straight path?
You refer to "its straight path". No golf ball goes in a
straight path, so I presume you mean that it does not curve left or
right; such a ball, if not curving, would have a projected path on the
ground (like the path of its shadow) which is straight. For a
right-handed golfer, a ball which curves right is called a 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;
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
We had a phenomenon happen recently about 8:30 PM that is inexplicable to us, but there must be some explanation. My wife and I were in separate rooms when we both heard an extremely loud noise from the living room! The noise sounded like a large glass that just hit a hard tile floor, but loudness was magnified. As it turned out we came into the living room to find a glass platter that we had sitting on the coffee table for about a year just shattered. âIt broke completely by itself as there was no one in the room.
Do you know how this may have shattered/blew apart all by itself?We used the platter to put 3 little oil lamps on.
Inside our house the next morning at 6 AM we heard thunder outside so thought it might have had something to do with the barometric pressure.Very low barometric pressure and the type of glass it was made up combined just right to explode it like that? The temperature was a constant 68 degrees as it was for the months that was on the table.
If you have any idea about this, we would appreciate it.
Glass, as you know, is manufactured at very high temperatures.
It has a quite large coefficient of theremal expansion (a large change
in size for a small change in temperature) and is a poor conductor of
heat. This means that as it cools it does not all cool at the same time.
This can result in very large stresses being "frozen in" at some
locations. What causes it to spontaneously break is usually difficult to
determine; most likely it had recently been bumped or your oil lamps
might have caused hot spots on the glass. Such things could have caused
a tiny fracture to begin and the final shattering could easily come at
some unpredictable later time. Unusual but not unexpected.
On a skate board going down a .5 mile hill at 45 degrees slopes if I weigh
187 lbs how many mph would I be going by the bottom of the slope.
This isn't homework I'm a 35 year old heavy equipment operator and my son had a accident on skate board and we are curious how fast he was going when he wiped out.
I just wasn't sure how accurate I was when I said about 30-35mph. Please if you don't know just tell me so I can find someone who does—we got bets on it now amongst the family.
Wow, 45º is pretty darn steep! A half mile would
correspond to his having dropped by about 0.35 miles≈560 m. If
there were no friction at all his speed would have been 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 Fdrag≈¼Av2
where A is the area his body presents to the onrushing wind. When Fdrag
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 ¼Av2=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!)
I am curious about generating power in space. Why do they always use solar instead of the windmill type of generation?
A coil/magnet rotating. It seems to me, once the rotation is started, it would continue forever? Thus if you used a rocket to start the rotating part of the generator, and it kept spinning, could you use the magnetic field to protect say, an astronaut inside the generator? If it was big enough. Would you get perpetual energy if you used the electricity created in say, a microwave rocket engine or electromagnet. Or does the magnetic force alone cause the spin to lose momentum?
So, you start something rotating in a vacuum and it never stops
because there is no air drag. You could even imagine making extremely
low-friction bearings so you could mount this on the side of your
spacecraft and it would at least spin for a very long time before
slowing down. But, the minute you hook it up to a generator you are
asking it for energy so it immediately begins to slow down, giving its
kinetic energy to you to power a light bulb, maybe. There is no free
lunch in this universe, and if you want energy you need something to
give it to you and the sun is the most convenient source in our
If the planet earth was perfectly smooth and spherical will the water cease to flow?
Not if everything else stayed the same. If the earth were
completely isolated, not rotating, and without atmosphere, water would
flow until it formed a uniform layer over the earth; eventually any
currents would damp out due to the viscosity of the water. The fact that
the earth is rotating and heated by the sun and has an atmosphere would
mean that the water would try to distribute itself mostly uniformly but
with an equatorial bulge; however heating and cooling of the atmosphere
would cause weather patterns and the resulting winds would move the
water around just like what happens today. Also, the moon causes tides
which are, by definition, motions of the water. You probably could think
of many more reasons the water would not become totally static.
We're trying to design a vessel for working in the vacuum of space. We have a vacuum chamber that can pull a 0.01 atm partial vacuum, so the question is : How does the force difference compare on the walls of a vessel, with 14 psi inside to outside chamber or space, i.e. force comparison between 0.01 atm in the chamber and the 10-14 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.
Ok, the 14 psi inside your chamber is about 0.953 atm and the
pressure outside is 0.01 atm making the net pressure difference 0.943
atm. The pressure outside in space is, for all intents and purposes,
zero, so the net pressure difference would be 0.953 atm. So, the percent
difference is 100x(0.953-0.943)/0.953=1.05%, about what you guessed. In
other words, the force on any part of the walls of your chamber is
about 1% smaller than it would be in space.
Not sure if this is physics or not but how can you record silent sound ? I found the patent for it on google it's
and I wanna know how you can record it because a regular microphone doesn't pick it up.
The idea here is essentially the same as AM radio where the
high-frequency radio wave is modulated by the audible message. For this
invention the radio wave is replaced with sound of a frequency larger
than is audible but modulated by an audible signal. You could certainly
make a detector (call it a microphone if you like) to detect these
high-frequency sound waves; ultrasound imaging in medicine does just
that. Then you would need some electronics to extract the audible signal
from the carrier, just like you need a radio receiver to extract the
audible signal from the AM radio carrier.
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: I hope you don't think that the rope would spontaneously rise up if
it were longer than the circumference of the earth; you would have a
slightly slack rope laying on the ground. You are specifying the difference
in radii between one circle with a circumference C and another of
is not really physics. But, it is easy enough to do. If C is the
circumference of the earth, then C=2πRwhere R
is the radius of the earth and
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!
What will happen if we fill water in the tyres of our cars instead of air?Does it have any effect?.
Three important issues I can think of. First, it would add a lot
of weight to the vehicle which would hurt your gas economy. Second, the
moment of inertial of a wheel would be larger requiring a larger torque
having to be exerted for either acceleration or braking. Third, air is
compressible and water is not and so the wheel would not have a
cushioning effect on the ride.
I just saw a
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?
This video was shot in Russia in a reduced-gravity jet provided
by S7 Airlines. Weightlessness is not achieved by falling but by
following the parabolic path which a projectile would follow. Imagine
that someone shot you from a cannon with a speed of 500 mph (the typical
speed of a commercial jet). If there were no air drag, you would follow
a parabolic path. The plane which contains you now follows that exact
path and that is how you appear to be weightless. An alternative way
would be for the plane to simply go straight down with an acceleration
of 9.8 m/s2, but I think you can see that this would not be a
very safe situation. Anyhow, back to you 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: Have scientists done experiment on what is the value of gravity below the earth surface as depth increases? if done pl. provide chart g vs depth.
deepest hole ever drilled is only about 12 km deep. I could
not find any reference to attempts to measure g at various depths
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
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.
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?
"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+½at2,
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=6x1010v/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
along the same line as yours.
I am writing a blog and had my friends to ask of what they would want to talk about.. I get this one. I have no clue with physics, so, I am asking. Explain why a photon particle- which is a very small bundle of energy and travels at the speed of light- seems to defy the laws of physics by never losing speed or velocity?
There is no law of physics which says an object naturally loses velocity. To
change the velocity of something you have to apply a force to it, it must
interact with something. If you had a bowling ball which had no forces on it
it would continue going with a constant speed in a straight line forever.
This is just Newton's first law. However, Newtonian mechanics is not valid
for a photon, but it behaves like any other particle when it experiences no
interactions which would change it. There is one important difference—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.
Gasoline contains 40 megajoules of energy per kilogram and gasoline trucks have around a hundred tons of it. So how does a gasoline truck exploding not produce an explosion similar to a small nuke?
Since it is always a little ambigous what is meant by a ton, I did my own
calculation using the volume of a tanker truck of about 10,000 gallons and
the density of gasoline of about 2.7 kg/gallon. I got the total energy
content of about 1012 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:
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.
If a man could travel with a near speed of light, what would happen to him if he'd run though an elephant. Would that man be demolished or could he survive that impact?
Are you kidding? A man going 200 mph, far below the speed of light, would be
instantly killed in this scenario. A 100 kg man going at 80% the speed of
light would bring in about 5x1018 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.
Bodies of water bend with the curvature of the planet. How large would a body of water have to be in order to measure a difference of 1 inch from one end to the other.
often get questions like this. I have used this figure many times before.
Here 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≈1-½θ2+…
Finally we can write that θ=s/R. If you now solve
for s you will find s≈554 m.
I'm a science geek and an IT person who just found out that there are sixteen types of water ice. I've been googling the phenomena, and can only find discussions of the physical make-up at the atomic level. Can you help me with some discussion of the various forms of ice at a macro level? Like, what does it look like, how does it act, etc. Everything I've found is in science speak, which I really can't envision myself.
Refer to the figure above which came from the Wikepedia article on
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.
If atoms at room temperature move around at roughly the speed of a jet airplane, how fast do atoms that make up a jet airplane move when said airplane is moving (at room temperature)?
I assume you are talking about molecules in a gas. It is much more
complicated to talk about molecular speeds in solids. So let's talk about
the air in the airplane cabin. The speeds are distributed from very low to
very high, but the most probable speed would be around 1000 mph, about twice
the speed of a commercial jet. But, that does not mean that the most
probable speed of molecules in a jet flying by would be around 1500 mph
because the molecules measured inside the airplane would be moving in all
directions so the result of adding 500 mph to all the molecules going 1000
mph would yield anything from 500 mph to 1500 mph. Measured inside the
airplane, the average velocity of all the molecules would be zero because
for every molecule going in one direction with some speed 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.
In what medium would sound travel the fastest?
I have not done an exhaustive search, but the largest I found was beryllium
with a speed of 12,890 m/s. Diamond is a close second with a speed of 12,000
if a cat fell from 100 feet how fast was the cat in mph falling when it hit the ground
The terminal velocity of a cat is about 60 mph and that speed is achieved
after falling about 50 ft. Therefore falling from 100 ft the speed would be
about 60 mph. See earlier answers (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.
When I was very young, I used a hand-held fish scale to try and measure my
own weight by attaching the hook (normally placed in the gills of the fish
just caught) to my belt and then pulled up on it. I'm interested in the
physics behind my folly and, expanding it further, if the scale was attached
to a light (not heavy) bar (that would support more than my weight) and the
scale could measure a fish weighing more than me, and I was strong enough to
lift more than my weight above my head, how close could I expect to get to
seeing my weight on the scale (attached to the bar I'm pushing up, and a
belt-device around my waist)?
The way you always solve a statics problem like this is to first choose a
body, then apply Newton's first law which states that the sum of all forces
on the body must equal zero; I choose the scale as the body.
What are the forces on the scale itself? Your hand pulls up with a force
Fhand, your belt pulls down with a force Fbelt,
and the earth pulls down on the scale (its weight) with a force Wscale.
These three have to add to zero which means that Fhand=Fbelt+Wscale.
Now, the scale is calibrated so that it reads zero when Fbelt=0,
so the scale will read whatever force your hand exerts up. Note that when
you do this analysis, the desired quantity, 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.
Thank you! So it sounds like it wasn't a folly after all (except for the fact that the scale I used wouldn't support my weight at that age).
Of course it 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.
Hi, I was wondering what are the chances of survival from falling from the ninth floor of a building, going over the science of that how does surface affect the fall, body weight and trajectory. What is the difference from falling from a third story window as opposed to a higher up one?
Someone else also asked this question; apparently it refers to a recent actual
incident of a student falling out a dorm room window about 85 ft≈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)=26-½gt2. 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=26-5t2 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=mΔv/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
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.
A rough estimate including air drag would have his speed at the ground be
about 21 m/s rather than 23 m/s as above. Given the rough estimates in all
these calculations, this 10% difference is indeed negligible.
When in the early universe did life become possible? I'm assuming it would have had to have been at some point when the temperature had cooled down enough so that the protons and neutrons of basic elements of life like carbon and oxygen could bond and come together. Also, what do you believe is the purpose for intelligent life in the Universe?
I usually do not answer questions about astronomy/astrophysics/cosmology as
stated on the site. I can give you a little information here, though. The
universe just after the big bang was almost exclusively hydrogen. No
elements which could be used to form planets where life might develop or the
material for life itself. The first stars started forming about 100 million
years after the big bang, basically pure hydrogen stars. These had to be
much larger than the sun (perhaps 300-1,000 times heavier) because of the
lack of heavier elements and consequently their lifetimes were too short (a
few million years) for life to evolve (it has taken about 5 billion years to
reach our stage of evolution, about half the sun's lifetime). Inside these
stars nuclear fusion caused heavier and heavier elements up to about iron to
form; then when those stars used up all the fuel, they exploded (super
novae) scattering the heavier elements to eventually be part of clouds of
dust and hydrogen which subsequently formed new stars with the possibility
of planets. (As you can see, it is not "cooling down" which is responsible
for heavier elements which are created in stars which are super hot.) So, we
are now at least several billion years into the age of the universe. More
detail on early stars can be read in a
Scientific American article. The "purpose" for anything is not within
the purview of physics.
What is the purpose of utilizing a percentage of body weight to determine how much weight to bench press/push/whatever? I know this seems like a fitness question and not a physics question, but what I am interested in is WHY weight would be used to determine how much one could (or should be able to) lift/push?
For example: A gym teacher wants to grade his students on their strength. He decides to use abililty to push a weighted sled across the floor as the measure. He wants to make the task equally difficult for every student in order to make the grading fair. So, he decides that each student will push 2x his/her body weight for 5 minutes and the grade will be based on how FAR the student is able to push.
So, Student A weighs 170 pounds and pushes 340 pounds (including the weight of the sled) for a total of 160 yards. Student B weighs 240 pounds and pushes 480 pounds (again, including the weight of the sled) for 80 yards. Student A pushed farther and gets a better grade, but Student B complains that he had to push much more weight so he should not get a worse grade.
Does Student B have a legitimate complaint or does his heavier weight contribute somehow to his ability to push that doesn't have anything to do with his strength? As in, does his weight help push the sled in some way?
Sorry, I don't know enough about physics to ask this question using proper physics terms like force, mass, etc. I hope you will still answer my question!
I cannot comment on the rationale for correlating weight to strength. I can
certainly comment on the physics of your particular example of sled pushing.
I would first of all comment that this example is certainly not one solely
of strength because, since it is a timed activity, endurance as much as
strength is being tested; if one student, for example, were a heavy smoker,
he would likely become exhausted more easily. As a physicist, I would equate
"strength" with force. The specific example you give, though, seems to me to
be more related to energy (work done by the student) or power (rate of
energy delivered) than strength; purely in terms of strength, the heavier
student exerts more force. The force F which each student must exert
depends on the weight w he is pushing and the coefficient of friction
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
t , so WA/WB=PA/PB=dAwA/dBw
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)
The demonstration wherein one pours water from a clear vessel and one shines a laser pointer (a red laser pointer in my case) through the water and it appears the laser beam bends with the water pouring out of the clear vessel;
What causes the beam to bend with the water? Would this same type of effect work with other fluids than water, such as air or glass?
It is caused by
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.
Why simple harmonic motion is called simple?
Take a look at the figure above. The red curves are all pure sine waves, just
having different frequencies. These are all called simple harmonic motion
because they may be expressed as a single sine wave. Now, take all four of
these and add them together. The result is the black curve. This curve is
still harmonic (which means that it is periodic, it repeats itself after
some time, called the period, has elapsed) but it is not "simple" because to
describe it you must use four different sine waves.
Hi ...here is a link to Felix Baumgartner's freefall jump from space -
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?
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.
Will the drag coefficient be the same in air and water?
There is no simple answer to this question. The drag coefficient CD
is a constant characterized by the geometry of an object used to
calculate the drag force FD if it is proportional to the
square of the velocity v: FD= CD ρAv 2
where A is the area
the object presents to the fluid flow and
is the density of the fluid.
Whether or not this equation is true depends on a quantity called the
Reynolds number Re=Lρv/η
where L is a length along the direction of flow and
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,
FD≈CD ρAv 2+kv.
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
ρ water ≈1000
Pa·s. Then vwater/vair≈(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, CD=0.47,
A=7.9x10-3 m2, L=0.1 m, vair=1000x1.8x10-5/(0.1x1)=0.18
m/s and vwater=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.
As I said above, for 1<Re<1000, FD ≈av+bv2
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 FD≈av≡½CD ρAv 2
so CD=2a/( ρAv )=2aL/(A ηRe )≡c/Re
where c is a constant. If FD is proportional to
v, CD is inversely proportional to Re; this is
called Stokes' law. As we saw above, Re>1000 leads to CD=constant.
Ferguson and Church have derived an analytical expression which very
neatly reproduces data for the transition from Stokes' law (1/Re) to
constant CD. Calculations for a golf ball are shown at the
left (constant CD, the green line, is called turbulent in
the legend). Note, as we have stated, that the transition occurs in the
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