QUESTION: Years
ago, I used to listen to the NPR
Saturday radio program "Car Talk"
They always had a puzzler question
that they asked (new puzzle) ant
then answered the prior week's
puzzel later in the program. A
puzzler was asked that I never heard
the answer to. I'll try to repeat as
much of the details as I remember. A
guy is floating in outter space and
is "buck naked". He is within reach
of two iron bars. One is magnitized
and the other is not. How can he
tell which bar is magnitized? They
didn't say it, but I presume that
the non-magnitized bar would be
attracted to the magnitized bar, and
vice versa. That's as much as I
remember. I have wondered for years
what the answer was but searches of
NPR "Car Talk" website only turn up
re-plays of the radio programs and I
don't know the dates of broadcast
where I might be able to listen to
the next-week's program to hear the
answer.
ANSWER:
Other
mentions of car talk are at
carstuck ,
pickup ,
saltwater .
QUESTION: I would
like know what affect solar flares
have on the earth's orbit. Since we
talk about solar sails as a means of
propulsion, wouldn't that mean the
solar wind pushes against the earth
as well?
ANSWER:
QUESTION: I
recently thought of this question
that I could not confidently answer
myself, and I have googled as well
to no avail. I apologise if this
question seems trivial, but I am
hoping to be blessed with some of
your knowledge. The question is: How
does visual and tactile nerve input
match in our perception? For
example; if I tap the back of my
hand, I see it happening instantly,
and feel it happening at the same
time. The pathway of course as you
know is light hitting the back of my
retina, then having that nerve
impulse travel through the optic
nerve towards the brain, while the
nerve input from the back of my hand
does the same as well. And
apparently, nerve impulse conduction
velocity does not even come close to
the speed of light. Is it because of
the differing lengths of the nerves
involved allowing them to somehow
match each impulse upon arrival in
the brain? I thought of this, but it
feels so simple and incomplete.
ANSWER:
QUESTION: I was
watching a film starring Arnold
Schwarznegger and when explaining
the futuristic weapons (90s film),
an actress said the bad guys were
using railguns as rifles. When I
looked it up on google, it said that
the concept was not viable. If it
can be done and built as a rifle;
what challenges do we need to solve
and if not what are the unsolvable
problems ?
ANSWER:
here . There are also links to
several more questions "about sci-fi
and video games guns" on the faq
page.
QUESTION: The
fabric of spacetime itself is
expanding exponentially. This would
imply that eventually there will be
an amount of spacetime with
absolutely nothing within its
volume. Eventually this area would
be a complete void being too distant
from any interaction with anything
whatsoever. Not a neutron, not a
photon or anything. Absolutely
nothing. What kind of events would
occur in such an environment?
ANSWER: E Δt ≤ℏ
for example, if an M =5 kg
bowling ball popped into existence
that would be ok as long as it
disappeared before a time Δt =ℏ/ΔE
where ΔE =Mc 2 =5x(3x108 )=4.5x1017
J and ℏ=1.05x10-34
J·s, so Δt =4.7x10-51
s. So a big mass would exist an
impossibly short time. What normally
happens is that a
particle-antiparticle pair, say an
electron-positron pair, pops into
and promptly out of existence. This
is what is called virtual pair
production and you cannot observe it
directly. However, if you do some
electrostatic experiment in a
perfect vacuum you will not get the
correct answer unless you account
for the virtual pair production
processes. This is why this is
sometimes called vacuum
polariztion .
QUESTION: Someone
originally posited the question: "Is
it possible for two different balls
of identical mass and velocity to
have different kinetic energies?" At
first I agreed with most others that
it was impossible and used the KE
equation as the basis for my
opinion. After some thought however,
I came up with a scenario that makes
me question this as follows: Imagine
two spheres of identical masses, one
with an internal void space of some
amount. Impart to them both a spin
and a velocity, both of which being
equal. Now if you were to measure
the KE of the homogenous spere as it
moves, it would be a constant as one
would expect. With the spere with an
internal void spinning as it moves
along its path, would this not
create a kinetic energy level whose
vale plotted on a graph would look
like a sin wave whose frequency
would match the spheres rotation and
the amplitude would be a
relationship between the total
volume of the sphere and the amount
in the void space. I think I belive
that thte kinetic energy problems I
have done all assumed the center of
the item I was calculating for as an
easement of solution. I think that
if the ball is psinning under these
circumstanes you would have t
consider the sphee having a 'moent'
that you have to calcullate its
position for insead of the center
and thus the sin wave effect. Am I
close to the solution or is it truly
just the KE equation regardless of
what is going on iwthin the mass?
ANSWER:
etc . You simply say that one
sphere is spinning as it also moves
with speed v . Even a ball
not having a translational speed at
all but spinning has kinetic energy.
The total kinetic energy of rotation
is K rotation =½Iω 2
where I is the moment of
inertia of the rotating object and
ω is the angular velocity.
So if one of the balls is spinning
it has more kinetic energy than the
other. If both balls are spinning
with the same angular velocity, the
kinetic energies could still be
different because of the moment of
inertia. For example, the moment of
inertia of a uniform solid sphere is
2MR 2 /5 and of a
hollow spherical shell 2MR 2 /3.
QUESTION: My
question is about how torque and
power each contributes to the
performance, and acceleration of a
car. The acceleration should be
equal to the force the wheels pushes
the road with, devided by the mass
of the car. The force is determined
by the torque of the engine if i’m
not mistaken. On the other hand,
acceleration of a car can be
described as the accumulation of
kinetic energy, given constant mass
of course. The rate of that is
determined by the power delivered to
the wheels. My hope is that you can
give me a better understanding of
«torque vs power».
ANSWER:
F times the distance from r
from the axis of rotation, T=rF ;
if you double the torque in a
particular situation, you will
double the rate at which the
rotational speed (RPM) increases.
Torque for a combustion engine
usually envisions the drive shaft as
the agent for delivering torque from
the engine. But you really cannot
just say that a particular engine
has a particular torque because the
torque depends on many variables
such as the RPM of the engine and
also on the gearing between the
drive shaft and the wheels. Now, the
engine delivers energy to the car at
some rate. If something is
accelerating in a straight line, the
force F times the velocity
v is the power P ,
P=Fv . For example, suppose
you want to see what force is
required to deliver P =200
horsepower (hp)=1491 watts (W) at
the speed v =50 mph=22 m/s;
F=P /v =1491/22=68
N=15 lb. Now the same thing can be
done for delivering energy to a
rotating object with a torque: The
torque required to deliver power
P to something with a
rotational velocity ω is
given by the equation P=Tω ;
in this equation P is
in N·m/s, T is in N·m, and
ω is in s-1 .
I presume you would like to
work in RPM and horsepower, so
making those conversions,
P =(2π (ω /60)T )/746,
with ω in RPM, P
in hp, and T in N·m.
T =7114(P /ω ).
So if you
want to know how many N·m of torque
you need to generate 200 hp at 1000
RPM, it is 7114x200/1000=1423
N·m=1050 ft·lb.
QUESTION: If light
slows down ( I appreciate it is by
very little ) in water what speed
does it travel at after leaving the
water and entering a vacuum again?
If it returns to c then where does
the extra energy come from to speed
up again? If it does not return to c
then does that not mean that the
only light in the universe
travelling at c is light that has
never been slowed down by anything
else? This came up in a conversation
with some friends ( we have weird
conversations ) I have a sneaky
feeling that it will be some special
case because of the nature of light
( being massless? ) that means
conservation of energy does not
apply. Could you explain in terms a
55yo with a degree in maths ( not
physics ) can understand?
ANSWER: Let's
start with a fundemental fact about
waves: the speed of a wave v
equals its wavelength λ
times its frequency f ,
v =λf . I assume you
know what wavelength and frequency
of a wave are. Now imagine a very
long string which we are wiggling at
one end with some frequency some
frequency f . So there will
be waves of frequency f
traveling down the string. The speed
of waves on a string depends on the
mass density of the string (grams
per meter, for example) and the
tension in the string, how taut it
is. Now, suddenly the mass density
gets bigger at some particular
location on the string, so at that
point the wave on the new part of
the string, while still vibrating
with the same frequency as the
lighter part of the string, moves
more slowly. As a result, the
wavelength gets shorter. Note that
no energy leaves or enters the wave
when the speed of the wave changes.
If the string becomes lighter again,
the waves will speed back up again.
Light behaves
the same way, slowing down in the
medium and speeding back up at the
interfaces with the vacuum. The
difference is the mechanism for
slowing the waves down or speeding
them up; v =λf
still holds. There is another way
you can look at light: You may know
that light waves of some given
frequency may be thought of as a
swarm of photons, each having an
energy E=hf where h
is Planck's constant. Since f
is the same in and out of the
medium, the photons always have the
same energy whatever the wavelength
of the corresponding waves.
QUESTION: (the
video holds my try to solve a
textbook question with my intuition,
it's not about how to solve a
textbook question but rather it is
that whyy my intuition is incorrect.
Although it's not necessary to watch
a video of me struggling to solve a
problem and in the end get the wrong
answer.) Sir it's not about the
textbook question.. but I am asking
that why is equivalent mass of a
pulley system (the mass of a block
by which we can substitute the whole
pulley) different than, the (m1+m2)
where m1 is mass at the left side
and m2 is mass at the right side of
the massless, frictionless pulley.
Whyy is it not m1+m2 because I only
see that much mass there on the
pulley And in the equivalent mass
formula, 4m1m2/(m1+m2) Where does
some extra mass come from??? Well,
if we measure the weight of the
whole pulley system, it should be
(m1+m2)g !! And not 4gm1m2/(m1+m2)
right? see some say that the tension
in the upward string that is holding
down the whole pulley system will
depend on the accelerations of
masses happening inside the pulley,
But Those are internal
accelerations, and if one block goes
up, the other one goes downand
therefore balancing the
accelerations. Can you please solve
the confusion on the equivalent
weight of the whole pulley being
different then the actual weight of
the pulley??? Why is it different?
ANSWER: m 1 is
at rest. The questioner then was
correct in his reasoning that the
tension in the thread connected to
m 1 must be m 1 g
and it will be the same on both
sides of the upper pulley. He then
seems to think that the whole
lower pulley sistem can be thought
of as an object of mass m 2 +m 3 .
This is only true if those two
masses are equal. If they are not,
then one accelerates down and the
other up and all bets are off, you
cannot do the problem that way.
(Whoops, I just did what I said I
wouldn't—tell him what he was doing
wrong!) The first thing to do is to
look at the lower pulley. There is
an upward force m 1 g
and two downward forces which must
be equal in magnitude which I will
call T ; the pulley must be
in equilibrium, so T =m 1 g /2.
Next I will look at m 2 .
I will choose the +y
direction to be vertically up and so
m 1 g /2-m 2 g =m 2 a 2
where
a 2 is the
acceleration of m 2 .
Next I will look at m 3 .
I will choose the +y
direction to be vertically down and
so
-m 1 g /2+m 3 g =m 3 a 3
where
a 3 is the
acceleration of m 3 .
But the magnitude of the two
accelerations must be the same so
a 3 =±a 2 .
Since I have chosen +y to
be in opposite directions, a 3 =a 2 =a .
If you now solve the two equations
(eliminating a ) you will
find [4/m 1 ]=[1/m 2 ]+[1/m 3 ].
QUESTION: I'm a
recent physics graduate, and I've
realized that most of what I've done
while studying physics is very
theoretical, with the exception of
some basic kinematics, circuits, and
optics labs. However, if at all
possible, I would like to perform an
experiment to show Relativity (or at
least the underlying postulates) in
the real world for myself. After
all, it's different enough from our
everyday experience to be difficult
to accept on an intuitive level,
even if you learn the math and such.
I figured I'd ask around and see if
you had any ideas for such an
experiment using either materials
you can find at home or basic
undergraduate lab equipment.
ANSWER: Well, I
couldn't think of a table-top
experiment which would meet your
criteria. So I did a quick Google
search and found a question
basically the same as yours on
Physics Forums . I think you will
find the ensuing discussion
interesting but probably nothing
which would fit your wishes.
QUESTION: Hello,
I'm another 11th grader and I'm
trying to understand how EM waves
work. The electric and magnetic
waves are in the same phase.
However, I was told that during
induction the changing magnetic
field induces the electrice one, and
vice versa. From this I'd assume
that B is shifted by compared to E,
supposedly being its derivative, but
that's not the case. I looked some
things up on ChatGPT and the
internet and found the following
Maxwell equation: After getting a
rough understanding of gradients and
partial differentiation (took a long
time), I concluded that the rate of
change in the electric field equals
the negative rate of change of the
magnetic field (divided by the
change in time). Which still doesn't
make sense! If E and B change
alongside each other (beside some
constant, or directional
difference), then charged objects
would gain a constant magnetic field
and freshly magnetized pieces of
iron would gain an electric field,
which they could keep, until their
magnetic field declines.
Besides the
former, I'm struggling at
understanding how energy is stored
within photons, if E and B oscillate
at the same time. Due to how
inductors and capacitors can be
charged, I am of the opinion that
electric and magnetic fields contain
energy, like in this equation about
energy density: However, if E and B
reach zero at the same time, that
would mean the photon has no energy,
which it clearly has, when E and B
are at their maximum. This goes
against the conservation of energy,
which means I'm really missing
something. Can you help me in this
regard?
ANSWER: [This
question had equations and a figure
which I have edited out because I do
not do LaTex symbol language. Also,
because the questioner is not really
prepared to understand
electromagnetism at this level.]
I admire
your quest to understand
electromagnetism. You are already in
a good place in understanding
qualitatively. Until you have
studied vector calculus and
differential equations, however, be
satisfied that you have already
achieved way beyond your years. I
will briefly give you some more
information to help with your
pursuit of this difficult subjece.
The first equation you show is ∇xE B t .
This is not a gradient of
E , rather a
curl of E E B
The other
equation you quote (corrected here)
is energy density of an
electromagnetic wave,
U =½(ε0 E 2 +B 2 /μ 0 )
in empty
space. What you are missing is that
this equation has been
time-averaged. And before it is time
averaged any spot in space would be
oscillating between 0 and some
maximum value; there is no violation
of energy conservation as you
suggest. Regarding the photons, this
is not what the energy density is
for them. If the frequency of the
wave is f then the energy
of one photon is hf where
h is Planck's constant.
Hence, for photons U photons = Nhf
where N is the number of photons per
unit volume. In both cases U
has units J/m3 .
QUESTION: For the
past several months, I (an aerospace
engineer with an interest in quantum
mechanics) have been trying
understand Heisenberg's frame of
mind when he first came up with his
crucial insights in 1925. Everywhere
I look I see essentially the same
statement: "His leading idea was
that only those quantities that are
in principle observable should play
a role in the theory, and that all
attempts to form a picture of what
goes on inside the atom should be
avoided." What none of the sources
has conveyed to me is the following:
granted that at the time Heisenberg
developed his theory there was not a
microscope powerful enough to see
individual particles within the
atom, but how did he conclude that
seeing the positions and motions of
the particles was "in principle"
impossible? This was two years
before he derived the Uncertainty
Principle, which in any case is
predicated on his original
derivation of quantum mechanics. So
without any of that "knowledge," he
just arbitrarily decided that some
quantities which had not yet been
observed were in principle
unobservable? I don't understand
that. Can you help me?
ANSWER: Your
question is strange. What does
"Heisenberg's
frame of mind" even mean? Anyhow,
that isn't physics, that's history
of physics or maybe psychology of
physics. In 1925 Heisenberg
constructed the first mathematically
rigorous theory of quantum
mechanics; it was based on matrix
algebra. Prior to that there was
only the Bohr-Sommerfeld
semiclassical description of what I
refer to as quantum physics. Within
a year an alternate theory, based of
differential equations, the root
being the Schrödinger equation which
is a wave equation. The latter was
generally more accessible to
physicists and is the favored theory
to introduce to students for that
reason. Still, Heisenberg's quantum
mechanics is very elegant but more
difficult to calculate with. But the
two were subsequently shown to be
100% identical in rigor and both
describe the exact same physics.
What is different is the mathematics
used by each.
The
uncertainly principle does not say
some quantities are unobservable. It
says that some pairs variables are
'conjugate variables' and therefore
you cannot measure both to arbitray
good accuracy, you cannot know both
of them exactly. It is actually
built right into quantum mechanics,
so no 'arbitrary' decisions needed
to be made.
I think you
should give priorty to learning
quantum mechanics rather than
Heisenberg's state of mind.
QUESTION: Is it
possible for a person to jump/drop
onto an impact force measuring
device and have the initial impact
force be less than body weight?
Reference/context: Science world had
a kinesiology exhibit, and one of
the displays was about impact force
measurement. The person gets weighed
and then climbs up to a platform
approximately 18-24 inches high. The
person then jumps/drops down onto an
impact force measurement device, and
a display shows impact force by body
weight. After a couple attempts I
achieved an impact force reading of
1.2 times my body weight, and I feel
I could have done better. Could that
number have been below 1.0?
ANSWER: I need to
know what the device actually
measures. "Impact force" is not a
quantity in physics. A force is a
force, but an impact is the product
of a force times a time. For
example, if you have a mass m ,
hit the device with a speed v
and come to a stop in a time t ,
your average impulse over that
interval is I=Ft=mv . where
F is the average force you
feel during the collision time. So
you could write F=mv /t.
I suspect that F is
what this device measures. Assuming
m is some fixed number, how could
you control how big or small F
is? If you increase v or
decrease t , F will
be bigger; if you decrease v
or increase t , F
will be smaller. An everyday example
is consistent with this: If you fall
and hit your head on a cement
sidewalk, you might get badly hurt
because your head stops in a very
short time; if instead you fall on a
thick foam mat, you would probably
not get hurt because it takes a
pretty long time to stop.
So let's see
if there is a way to get an F
which is less than your weight.
Being a scientist, I must use SI
units, but I will try to convert to
Imperial units when it seems
helpful. Mass is measured in
kilograms (kg), length in meters
(m), and time in seconds (s). So
suppose that your mass is 100 kg,
the height is 0.5 m, and the time is
t . ; acceleration due to
gravity is about g =10 m/s2
so your weight is W =1000 N
(a newton is mass times g
and is 1 N=0.224 lb). So, your
weight would be about 224 lb. since
v/t is your acceleration, I finally
come up with F /W=a /g
where a is your average
acceleration. So you just need to
have the rate at which you are
slowing down less than the
acceleration due to gravity. If you
drop from 0.5 m your speed when you
hit the device is about v =3.16
m/s. So if you can make v /t =10,
F /W= 1.0, so t =0.316
s. If you make the time greater than
about a third of a second you will
get below 1.0; you can do that by
watching what parachutists do when
they land—go into a crouch as slowly
as you can. Of course, I do not know
how the measuring device works so my
numbers may be off, but anything you
can do to increase the time from
when you hit and when you stop will
lessen the average force
you feel.
QUESTION: When
Hiroshima and Nagasaki where bombed
with nuclear weapons in World War 2
there was total devastation. However
within a relatively short period of
time people were able to move back
in and re-build the cities. However
when there is a nuclear power plant
meltdown like Chernoble or Fukushima
the area around the power plant
remains contaminated with
radioactivity for thousands of
years. Why is that?
ANSWER: I have
previously
answered this question.
QUESTION: I'm
having trouble with Hawking
radiation. It seems to me that if
you separate a pair produced at the
"event" horizon, with one half of
the pair falling into the black hole
and the other half escaping, you
have added mass\energy to the black
hole rather than subtracted from it.
The problem for me here is the
assumption that the mass\energy to
produce the pair in the first place
is somehow subtracted from the black
hole?! To me this seems to
presuppose that the mechanism for
the pair production is well
understood and the reservoir for the
mass\energy for this pair production
can communicate across the "event"
horizon.
ANSWER: AI have
answered this question before.
QUESTION: Red
shift/blue shift—light traveling
toward us (Andromeda Galaxy as an
example of blue shift) has a shorter
wavelength than stars/galaxies that
are moving away from us. These light
waves are part of the visible part
of the electromagnetic spectrum
(right?), but what about the parts
that are invisible to the
eye—ultra-violet, infrared, radio
waves, etc? Do these waves change
frequency in the above examples? (I
know about Doppler radar and how it
works, so is the same principle at
play here?)
ANSWER: All frequencies
of electromagnetic waves are Doppler
shifted.
QUESTION: Black
holes because of their mass and
gravitational 'pull' suck things in
and there is no escape (at least as
I understand it), and this includes
light. But light has no mass, so
what is it about light that makes it
susceptible to the gravity of a
black hole?
ANSWER: Our best theory
of gravity is general relativity. In
general relativity a photon, the
quantum of the electromagnetic
field, has energy. (Of course it
does because the electromagnetic
field itself has energy.) The energy
E the photon has is
E=hf where h is
Planck's constant and f is
the frequency of the field. If you
throw a ball straight up in the air
with some initial speed v ,
it initially has energy E =½mv 2 ,
but the higher it goes the smaller
v becomes until it is zero;
so when it runs out of energy it
stops and, like any mass, will fall
back to the ground. The photon,
however, still loses its energy but
in another way: Because of the red
shift of light moving away from a
heavy object, the photon's frequency
will get smaller as it rises. For
most objects, a star for example,
the photons will end up with a
pretty small shift in frequency. For
a black hole, however, the photon
loses all its energy before it can
escape, that is, it simply
disappears. If you are wondering
where the energy of the photon went,
it went to increasing the mass of
the black hole by an amount m ,
mc 2 =hf
or the mass of the black hole has
increased by m=hf /c 2 .
Incidentally, There are two kinds of
red/blue shifts. One is the one you
are probably familiar with,
generally called "Doppler shift" and
caused by motion of the source or
observer. The other, as discussed in
this answer, is referred to as
gravitational red/blue shift.
QUESTION: I've
just generally been thinking about
this for a while, and it's difficult
to get an answer from Google. I
think if I understand things
correctly, it's actually vital for
our universe that the speed of light
is constant, and this constant
effectively creates the
electromagnetic spectrum we observe.
Light can't go faster. It can only
increase in frequency as more energy
is added. If there was no constant
speed of light, it might be the case
that all energy in the universe
would evaporate instantly? Maybe? I
don't know. I'm just trying to
conceptualize what's going on. But
what would be the physical mechanism
by which the speed of light is
constant? It must be a property of
space? Do physicists have a solid
understanding of what causes this?
ANSWER: Go to the faq
page. You will find "Why
is the speed of light constant to
all observers? "
QUESTION: I am an
11th Grade Student, and my teacher
currently teaching us newton's laws
of motion, in the above image there
is a pulley arrangement, with all
surfaces smooth! my teacher was
discussing about this in the last
few minutes of school and i couldnt
understand it. He said that the big
10kg block will move towards the
left! what? why would it move
towards left? it would need a force
in the left direction right? but
here, i dont see any force on the
10kg block which is on left
direction... [followed by 364
unnecessary rambling words!]
ANSWER: Once
again I am faced with a question
which violates site ground rules
requiring "...single, concise,
well-focused questions." Oh well, I
have just edited it down and, rather
than doing a complete solution, I
shall only address the student's
primary concern: when released, what
happens to the 10 kg block? I always
told my students to "choose a body"
when starting a Newton's second law
problem. I choose the pulley which,
although not stated explicitly, must
be frictionless and massless. The
next step is to find all the forces
on that body . Clearly the
string is pulling on the pulley to
the left by T 1
and down by T 2 ;
because this is a massless pulley we
can assume T 1 =T 2 =T.
But, wait, is there anything
else touching the pulley which might
exert a force on it. Yes! This setup
would not work if the pulley were
not attached to the 10 kg block. The
block must therefore exert some
force on the pulley to hold it on
the corner of the block,
F in my diagram.
If the block exerts a force on the
pulley, the pulley exerts an equal
and opposite force on the block, and
that force has components to the
left and vertically down. One more
thing: The questioner indicates
that, when the system is released
and the two 5 kg masses start
accelerating, the tension is the
string is 5g . This is
wrong.
QUESTION: My
question is about the way steam and
heat works when food is cooking. For
instance, if you cook food too long
in a steamer, the food generally
gets mushy if cooked too long.
However if you heat up food in an
airtight container in the microwave,
the food does not get mushy if
cooked for too long. In fact it will
burn and get crunchy instead,
despite having steam and lots of
moisture in the container. Is there
a physics-related reason that the
two steaming methods vary so
drastically in outcome if
overcooked?
ANSWER: In a
steamer there is a reservoir of
water continually adding stream to
the inside of the pot. The cooking
happens from the outside in. If you
cook too long it will be overcooked.
The same happens simply boiling the
food. When you microwave you heat up
the whole thing uniformly. And any
steam that is in there came from the
food itself, not nearly as much as
in a steamer. But if you do that too
long you will boil the water out of
the food and it will certainly
become "crunchy".
QUESTION: I need
to move an 80 pound speaker 28.5
inches above the floor using a dolly
on a 7 foot long plank. How much
force will I need to apply? (A very
long question heavily edited by me!)
ANSWER:
QUESTION: You have
4 hot hard-boiled eggs you want to
cool down quickly. They are in a
pot. You throw in a handful of ice
and then add cold water. Do you just
cover the eggs with the water or
fill the pot? Which is faster?
ANSWER: I want to
start out by noting this is an
academic question. The way cooks
normally cool hot foods quickly to
halt the cooking is to plunge the
hot food into an ice bath, water to
which enough ice has been added to
bring the temperature of the water
down to 0°C (ice water).
This can
get to be a very complicated system.
I will simplify it to include only
what is going on with the water,
ice, and eggs. In other words,
imagine that these three are
enclosed in an insulating box
through which heat cannot flow. This
is probably a good approximation
because cooling the food happens
quickly enough that the interaction
with the bowl and air in the room is
minimal. What happens is:
ICE:
There
are two ways ice can absorb
heat: (a) if the temperature is
less than 0°C, it can warm up
until it gets to 0°C, and (b) at
0°C it will absorb heat but not
increase in temperature, instead
melt. Most home freezers have a
temperature of 0°F=-18°C, so
initially, the way the ice will
absorb the energy is by warming
up; when the ice reaches 0°C it
will absorb energy by melting.
WATER:
The question seems to ask,
mainly, how much water should we
add. I will look at several
amounts of water. I will start
the water at 15°C, a typical
temperature for tap water.
EGGS: The
mass of an egg is about 50 grams
and we have four of them. They
have just come out of boiling
water at 100°C.
I struggled
with answering this question. The
main problem is that the speed at
which the temperatures change
depends sensitively on things like
how heat diffuses from the centers
of the eggs into the water and the
size of the pieces of the "handful
of ice"; crushed ice will absorb
energy much faster than a big block
of ice. So I cannot answer the
question, which presumes that only
the amount of water determines the
speed to come to equilibrium—there
are many more factors which play. I
think it likely that it would take
longer times for large amounts of
water to come to equilibrium, so I
could glibly suggest that the answer
is the smaller amount of water is
faster; but the problem has too many
variables to know for sure. What I
have decided to do is demonstrate
the physics of how the final
temperature T is achieved
and work that out for one example: I
will find the minimum amount of ice
we must add to 1 liter of water so
that the equilibrium of the
ice/water/eggs system will be T =0°C.
Heat is just
the flow of energy. If an object has
a mass M and a temperature
T 1 and an amount
of heat Q is added its
temperature increases to T 2 .
Then the change of temperature is
proportional to Q , Q =MC (T 2 -T 1 ),
where C is a
proportionality constant called the
specific heat of the material
absorbing energy. If the object is
instead having energy taken from it,
T 2 -T 1 <0
so Q <0. If a solid is
melting it is absorbing energy but
not getting warmer; in that case one
needs a constant called latent heat
of fusion, L which is the energy
absorbed per kilogram of material
melted, Q=ML. So, the
following is the computation.
and the
latent heat of
fusion
for:
ice:
L =3.33x105
J/kg.
I will
calculate the amount of ice which is
needed for the final temperature to
be T =0°C starting with 1 kg
(a liter) of water.
M w =1
kg of water at T w =15°C,
M e =0.2
kg of eggs at T e =100°C,
and
an
unknown mass M i
of ice at T i =-18°C.
So, the
equation (which is simply energy
conservation) to solve is:
Note that
this is a linear equation for M i
as a function of M w .;
this is shown in the figure. So, 1
kg will about cover the eggs and
about a third of a kilogram of ice
will totally melt when T
reaches 0°C. If you start with more
ice, there will simply be ice still
unmelted in the pot when equilibrium
is reached.
QUESTION: I am
currently studying for my physics
exam. A student in the group chat
asked us all a question and it got
us stunned. Can you help? Please be
aware that this is NOT a homework
question, we all just share
different opinions on the answer.
Question: A
container with very thick/sturdy
walls is on a frictionless surface.
Inside the container, on the left
side, there is a cannon with
cannonballs. The cannon fires all
its cannonballs against the right
wall, where they fall directly to
the ground upon contact (they do not
bounce back). What is the effect on
the container?
a) The
container stays in place
b) The
container moves to the right
c) The
container moves to the left
d) Not
enough information
I personally
think the answer is A, but a lot of
people are saying C, and others also
say D. I'm lost.
ANSWER: See the
figure. I assume that the box and
cannonball are rigidly attached and
have a combined mass of M ;
the cannonball has mass m .
Before the cannonball is fired, the
whole system is at rest and so its
linear momentum is zero. Afterwards
the cannon ball is moving (relative
to the ground which is where the
momentum was previously zero). But
the total momentum must still be
zero (conservation of momentum)
because there are no horizontal
external forces on the system. So,
mv-MV =0 or V=mv /M ;
when the cannonball hits the wall
its horizontal linear momentum
becomes zero. There are still no
horizontal external forces so the
motion of the box must also stop.
Clearly the answer is c).
There is
another way you could look the
problem. Since after the the cannon
is fired the cannonball, previously
at rest, now is moving. So the
cannon/box must have exerted a force
F -F
on the cannon/box
etc .
QUESTION: I am a
high school senior who writing
software for the avionics of our
school's model rocket for the TARC
rocketry competition. I have taken
AP mechanics (Calculus based) and
had a difficult time figuring out
the projectile path of a rocket.
After burnout the rocket will coast
to a target apogee (around 800 ft)
at subsonic speeds. The rocket has
onboard an accelerometer (plus
gyro), a really accurate clock, and
an altimeter (with barometer). I was
wondering if there was a way to
predict the exact apogee of the
rocket based soley on the
accelerometer readings and the clock
(while knowing all the constants:
mass, drag coefficient, etc.)? My
team and I have tried several ways
to do this but we get stuck in
dependency loops. Drag is based on
velocity while velocity is based
subsequently on drag. We tried using
energy, momentum, even modified
kinematics (for changing
acceleration), none of which worked.
What we need is a single function
that takes in current vertical
acceleration and the constants and
predicts a delta height which we can
added to the current height to
predict apogee. Also, since air
pressure is part of drag and that
changes based on height according to
a function, it would be nice if we
could put that into the predictive
function. I really don't want to
make this question too long but if I
could add one more layer of
complexity. We are using air brakes
to control the altitude to achieve a
target apogee. We planned to run the
above mentioned predictive function
in a loop and reverse engineering it
to create another function to find
out how much more drag area we would
need to achieve the target altitude
since the airbrakes are radial flaps
that just increase our cross
sectional area.
ANSWER: First of
all, you have not given me nearly
enough information about what you
are trying to do. Is the time the
rocket is burning part of your
calculation? It is very different
(and harder) than the trajectory
after burnout for several reasons.
Is the trajectory one-dimensional,
i.e . straight up and back
down? Or does the motion of the
rocket also have a horizontal
component? Does the accelerometer
give you the direction of the
acceleration vector? And, if not, it
is not very useful. Do you know the
mass of the rocket before and after
burnout?
All the above
is necessary for me to give you any
guidance. But the way you are
approaching the problem, common
among newcomers to physics, is all
wrong. One of the most important
things to consider when solving a
problem in the real world (not the
ideal models of on-the-blackboard
problems in a physics classroom) is
to determine what is important and
what can be approximated without
serious error. Except in textbook
problems, you should never expect to
be able to get the appropriate
equation in just one step, you need
to make sure you understand things
step by step. You should be starting
out by imagining you are doing a
textbook problem—ignore drag and
ignore the change of mass during the
burn. Then you might put in drag.
Here is a little tip: A fairly good
approximation for air drag D
is D ≈¼Av 2 .
Here A is the area
presented to onrushing air and v
is the speed; the vector
D
must be
opposite the vector v D will come
out in Newtons. The estimate for D
does not look dimensionally correct;
that is because lots of stuff is
included in that little factor ¼
like the fluid is air, the density
is atmospheric pressure at sea
level, the shape is a generic
sphere, etc . Including
D in your calculations will
tell you if it is actually very
important for your problem.
That's
probably all I should do at this
point. If you answer my questions
about what you are trying to
accomplish, maybe I can give you
some other instructions.
By the way,
if you know the acceleration as a
function of time you should be able
to determine the position and
velocity as functions of time if you
know the initial position and
velocity. It would require numerical
integrations because a will not be
constant.
REPLY:
I am only
simulating the rocket after
burnout. As such, the problem is
more like finding the trajectory
of a cannonball mid-flight.
Like a
cannonball, the projectile will
have velocities in the
horizontal and vertical
direction, however only the
vertical component is needed as
I am trying to predict apogee.
The problem
is that at an arbitrary time
during the ascent of the
cannonball, we will be given
vertical acceleration (drag and
gravity with direction), current
altitude, and time (if that is
needed though it might not be),
and we must determine the apogee
time and altitude.
Constants
like mass (after burnout), cross
sectional area, drag
coefficient, etc. are known.
Drag is
quadratic (v^2) Without drag,
the apogee is simply found with
kinematics (constant
acceleration), but since drag is
a changing force that imparts a
changing acceleration unlike the
constant 9.81 of gravity, no one
on the rocket team could figure
it out.
Also, I have
found other equations online that
include drag and give altitude as a
function of time, but the equation
we need should just give out a
number for max altitude based on the
given values stated above. That's
why we tried using energy formulas
to solve the problem since at the
top there is no kinetic energy, only
potential (again only vertical
components).
ANSWER: There is
a good reason why you are having so
much trouble: You are high school
students who, at most, have had
introductory differential and
integral calculus. As I will show,
this problem will require
differential equations. And not just
simple linear differential
equations, but coupled second-order
nonlinear differentials. Such
equations can only be solved
numerically, almost never
analytically. I have given a little
thought and have thought of a way
that just might work for you. It
won't be easy. One thing before we
get started: You said "only
the vertical component [of the
velocity] is needed ."
This is not correct because to get
the behavior of the vertical
component you must also know the
horizontal velocity.
One thing
which I am not clear on is what you
are given. It is imperative that you
know the initial velocity, both
magnitude and direction, to solve
this problem. When you say
"we will be
given vertical acceleration" I think
you must mean "we will be given
acceleration" because the
acceleration is not vertical. When
you say "drag and gravity with
direction" I assume that the drag
force, both vertical and horizontal
components are given. The drag force
is usually written -C|v |v ,
so Dx =-C √[vx 2 +vy 2 ]vx
D y =-C √[vx 2 +vy 2 ]vy v =√(D /C ) andvy /vx =Dy /Dx =tanθ
θ= tan-1 (Dy /Dx )
is the angle v makes with
the horizontal. From the initial
drag you can now get the two
components of v vx =√(D /C )cosθ
and vy =√(D /C )sin θ .t =0) values of the
components of the velocity. I will
choose that the initial values of
x and y are both
zero, i.e. I choose the
origin of my coordinate system to be
at the initial position of the
rocket.
So now we are
ready to do some physics. What's key
in classical mechanics? Newton's
second law, F =ma m (d2 r t 2 )=-C |v |v -mg 1 y 1 y m is the
mass, r t ,
and g is the acceleration
due to gravity. I realize that
vector calculus is something you
have not yet studied, but it's not
so bad because we just write it in
component form:m (d2 x /dt 2 )=-C|v |vx =-C √[vx 2 +vy 2 ]vx m (d2 y/dt 2 )=-C|v |v y -mg=-C √[vx 2 +vy 2 ]v y -mg. These are second-order (second
derivative), non-linear (square
root), coupled (vx
and vy both
appear in both equations) equations.
But we can make them much more
tractable by noting that (d2 x /dt 2 )=(dvx /dt )
and (d2 y /dt 2 )=(dv y /dt ).
Finally
we have vx /dt )=-β √[vx 2 +vy 2 ]vx
andv y /dt )=-β √[vx 2 +vy 2 ]v y -g β=C/m . Now we can't
solve these analytically but we can
do it numerically (that's where you
come in!). This will just involve
arithmetic. vx ={-β √[vx 2 +vy 2 ]vx }Δt
andvy ={-β √[vx 2 +vy 2 ]vx -g }Δt. So here is your computer
programming problem. You know β ,
g , vx ,
and vy at t= 0,
they all are just numbers. Choose
some Δt , say Δt =1
s and then you can calculate the
changes in the velocity components.
Now you know the (approximate)
values of the component velocities
at t =1 s. Now you calculate
with those and get the values of the
velocities at t =2 s.
Continue until you find the time
when vy ≈0; going
past this t vy
will become negative. You will
probably need to try several values
of deltat until it is small enough
that the final answer doesn't change
(you are, after all, simulating what
you would get with infinitesimal Δt).
After you find the total time
to the top, you still haven't
answered where it is. What you need
to do is to integrate into your
program calculation and tabulation
of Δx=vx Δt
and Δy=vy Δt
at each interval, and calculate
running sums. The final sum of Δy
is your answer.
I know that
this is a lot for a high schooler to
digest. But I have tried to simplify
as much as I could. As I warned you
at the beginning, when you have air
drag to contend with, you have a
much more complicated problem than
with constant forces. If you are
able to pull it off I would love to
see your results.
ADDED
THOUGHT: To check
that I have not made an algebra
error somewhere I have written a
PYTHON code to do the calculations I
suggested above. I randomly chose
reasonable boundary conditions,
β =0.001, v x t =0)=100
m/s, v y t =0)=200
m/s, g =9.8, and Δt =0.25
s.
y as a function of x .
QUESTION: We have
read for years that the universe is
composed of certain percentages of
visible matter, dark matter, and
dark energy. I can understand that
there are ways of reasonably
estimating the amounts of visible
and dark matter. They are described
for us in published articles. How
does anybody estimate the amount of
dark energy? And what is it anyway?
Is dark energy expected to have all
the characteristics types that
regular energy has - kinetic,
thermal, etc.? All of the articles
that I have seen simply present a
percentage associated with dark
energy. There don't seem even to be
some typical characteristics of dark
energy that would allow us in the
public to form a concept of what it
is. Falling back on the Classical
Physics model, dark energy has to be
the ability to do work on dark
matter, if that helps! Regardless of
what dark energy actually, is one
still wonders: how is it estimated?
ANSWER: Site
ground rules clearly state that I do
not do
astronomy/astrophysics/cosmology. I
can tell you though that it is not
clearly understood what either dark
matter or dark energy actually are.
FOLLOWUP
QUESTION:
But as a last
thought, they tell us that dark
matter can be evidenced by its
observed effect on real matter. And
then they quote actual percentage
estimates for the amounts of dark
matter and energy. And yet dark
energy, I think, has not even been
detected. There must be a process
used to arrive at these numbers.
That's what I am really after. Can
you direct me to anyone or any
organization that I can communicate
with to find out just how it is
done? My thought is that, if I can
understand the process, I can more
clearly Understand what "they" think
these things are. Maybe not.
ANSWER: Dark
matter is thought by most
astrophysicists to be some kind of
particle which interacts only via
gravity. It is very hard to
"observe" something if you cannot
interact with it using
electromagnetism. There are numerous
anomalies in the universe which seem
to not be described by our current
"best" theory of gravity, general
relativity. You may make a model
of hard-to-detect particles with
mass but which only interact with
the rest of the universe via
gravity, but if that model works it
does not mean that you have
"detected" dark matter, you have
only fitted the data, not
necessarily one which tells you what
dark matter is ; at this
level it is empiricism. It may be
that dark matter is not stuff, but
an indication that we don't
understand gravity as well as we
think we do.
Dark energy
is a different kind of thing. You
say that it "…has
not even been detected…" That is not
correct, it has been "detected" just
as surely as dark matter has. The
evidence for it is that very distant
galaxies are observed to be not just
moving away from us but
accelerating. In other words you can
assume that there must be some
universal repulsive force pushing
the universe apart. In my opinion,
dark energy is in just as detected
as dark matter is.
There is an
interesting back story about dark
energy. Circa 1918, when
Einstein was putting the finishing
touches on his theory of general
relativity, it had not yet been
discovered that the universe is
actually expanding; it was assumed
to be static. That presents a big
problem since gravity is a
long-range attractive force which
should have been causing the
universe to be collapsing. So he
added a term to general relativity
to hold everything in place and it
became called the cosmoligical
constant λ . Much later in
his life, when it was discovered
that the universe is not static, he
said that λ was his
"biggest blunder". In 1998 when
acceleration of distant galaxies was
observed, new life was given to
λ ! If you are interested in
delving into this more deeply, check
out this
link.
QUESTION: I was
wondering if a feather and a bowling
ball actually pull towards the earth
at the same rate? Why would the
gravitational fields produced by the
feather and the bowling ball not
increase the rate of their
individual attraction towards the
earth?
ANSWER: First we
need to specify that this notion
requires that there be no forces
(like air drag in the real world)
other than gravity acting, the earth
must be assumed to be enormously
more massive than the feather and
bowling ball (which is true), and
the gravitational force which the
earth exerts on the objects is
uniform (which is true only close to
the surface of the earth). Also, you
seem to be a bit muddled as to what
is going on. The reason that the two
masses accelerate is that each feels
a force due to the gravitational
field of the earth, it has nothing
to do with the gravitational fields
of the objects themselves. You seem
to be thinking that the bowling ball
feels a bigger force and that is
true; the force anything feels
(which is called the weight of that
thing) is proportional to (depends
on) its mass. But the effectiveness
of that force, what acceleration
(the rate at which its speed will
change while falling) it falls with
is inversely proportional to its
mass. So, if the mass of the ball is
1000 times bigger than the feather's
mass, the force on it will be 1000
times bigger. But, the more massive
an object is, the more force
required to give it a certain
acceleration is. It just so happens
(because of Newton's second law that
the force necessary is proportional
to the mass. Therefore although the
ball has 1000 times the mass, it
takes 1000 times as much force to
accelerate it as it takes to
accelerate the feather. The
acceleration is g =9.8 m/s2 .
If you want a more mathematical
discussion, see an
earlier answer .
QUESTION: One of
the issues YouTube physicists
discuss regarding interstellar
travel is that the high speeds we
would achieve would render the
spaceship more vulnerable to
constant bombardment by particles,
gases, and dust, requiring a thicker
hull in the direction of travel.
Assuming interstellar travel is
possible with current or near-future
tech (not a given), is it true that
more particles and gases would
impact in the opposite direction of
travel? That would seem to assume
that most particles are more or less
at rest in the frame of reference we
are coming from with Earth. Wouldn't
they all be going at random
velocities up to near the speed of
light relative to us? Within the
solar system, are particles mostly
at rest relative to Earth's frame of
reference? Within the
gravitationally-bound Local Group?
ANSWER:
Unfortunately, the space between
planets, stars galaxies are not true
vacuums as we often think of them.
So, inevitably, atoms and molecules
will be present in varying
densities, sometimes just moving
around in relative vacuum, sometimes
congregated in clouds. And you are
right, they will be moving in random
directions with a distribution of
speeds, but the speeds are all very
small compared to the speed of
light, c . But a ship
traveling to distant stars would
have to have speeds which are not so
tiny compared to c . And to
someone on that ship, all the
particles will be moving mainly at
speeds close to the speed which the
ship has. When you say "mostly at
rest" you do not define what that
means. What is true is that their
speeds are very slow compared to c;
if that's what you mean by "mostly
at rest", then yes most particles
which you encounter will end up
coming toward you with a speed very
close to the speed you are
traveling.
FOLLOWUP
QUESTION:
Thank you for
answering my question!! I have
refined it to clarify the point you
brought up.
We are
measuring the velocity of the
spacecraft relative to Earth's frame
of reference. Therefore, we are
measuring the velocities of the
particles it might encounter from
the same frame of reference -
Earth's. You can't compare any
particular velocity to the speed of
light without a frame of reference.
You must have an observer with its
own frame of reference to measure
the speed. Why should we assume that
particles outside the Solar System
or Local Group travel at very low
speeds compared to the speed of
light from our frame of reference?
Shouldn't we see the same number of
high-speed particles bombard the
ship from all directions, regardless
of its velocity relative to Earth?
Does the idea that the spacecraft
will encounter more particles in the
direction of relativistic travel
rely on the fact that all particles
in the Local Group are
gravitationally bound together?
ANSWER: My bad! I
should have specified my reference
frames. Let's call the earth our
"rest" frame. First some numbers:
Typical
atom/molecule densities most
places in our galaxy is 10-100
particles/cm3
Typical
atom/molecule densities in
intergalactic space is 1
particle/m3
Typical
temperatures of most places in
our galaxy is 10-100 K
Typical
RMS speed of molecules is slow,
e.g. , CO2 at
10 K is about 300 m/s
So if we
say that probably particles with
speed 10 v RMS
are the fastest with any
appreciable number, those have
speeds 3000 m/s=10-5 c .
Since the
biggest speeds relative to the earth
are so tiny compared to the speed of
light, they look like they might as
well all have been at rest compared
with the motion of the ship.
Therefore, on the ship with some
velocity U every
atom or molecule will be seen moving
with a velocity very close to
-U .
Suppose that
you are driving in a forest at 50
mph. Every tree from your frame of
reference is moving backwards at 50
mph, not just trees in front of you.
QUESTION: Since
the discovery of the Higgs Boson,
which is said to be what makes
matter essentially makes things be
physical. Then wouldn't it make
sense that there is indeed a
graviton? In where the particle is
in a wave function until check and
collapsed into a particle? Not
unlike the double slit. Couldn't
this be the case? And it is my
opinion that this could fit nicely
into an current theory.
ANSWER: I don't
understand hows Higgs or double
slits have anything to do with
whether "…there is indeed a
graviton…". I will assume that the
main issue is you want to know is
whether "…there is indeed a
graviton…". As you may know (see the
next question) general relativity is
generally depicted as a geometric
model, deformation of spacetime by
mass. So gravity is not a force,
merely geometry? If that is the
case, why seek a quantum theory of
gravity where the quanta (gravitons)
would be "messengers of the force"
when there is no force, only
geometry? But a few years ago I came
across a short
essay which made the whole thing
clear to me. A theory does not have
to have only one framework of
characterizing it. For example, when
quantum mechanics was in its
infancy, two competing mathematical
frameworks were used, both of which
seemed to come to the same result
when calculating: Matrix mechanics
was formulated by Heisenberg, Born,
and Jordan; differential equation
formulation of quantum mechanics was
introduced by Schrodinger. The two
were eventually shown to be
identical. Similarly general
relativity has the properties of a
field theory and can be interpreted
as having forces and being
quantized. Hence, there should be a
theory of quantum gravity. If you
want to read more see my
earlier answer and be sure to
look at the
essay linked there.
QUESTION: Hi. I’ve
been pondering gravity lately. Why
is it that -- although Einstein's
general relativity best explains
what we call "gravity"-- one still
sees statements such as "all masses
attract each other – you and your
pet, the earth and the moon . . ."
etc., etc. What magical "attraction"
are they talking about? Also,
could you recommend a good
illustration of what actually
happens when an object near the
earth "falls"?
ANSWER: The
theory of general relativity (GR) is
quite complicated mathematically. To
understand a complicated physical
theory requires that we have a
qualitative explanation of some
effects predicted to visualize. GR
is a theory which uses advanced
geometry formalism, a
four-dimensional spacetime (time
being the fourth dimension). It
therefore becomes a theory we think
of as a geometrical theory and
envision massive objects to deform
the spacetime around them resulting
in attraction. But, there's a
problem—since we live in a
3-dimensional world we really can't
visual a 4-dimensional world. So
what is usually done is to imagine a
2-dimensional plane which, if we put
a bowling ball in it, gets deformed.
If there is a marble nearby it will
roll toward the bowling ball; I
often refer to this as the
trampoline model and it is a cartoon
not to be taken too seriously but to
provide an example of an object
deforming some space around it. You
can also give the marble a little
shove and it will orbit the bowling
ball as shown in the figure. The
marble also deforms the space around
it but since it has so little mass
compared to the bowling ball, it is
not apparent in the figure.
QUESTION: Recently
I’ve got interested in one question.
Lets assume we have 2 balls with
same mass in space on some distance
from each other (no other objects
around). At the initial moment of
time they are staying still
(velocities = 0). Then, due to
gravitational attraction, they start
moving toward each other. At some
moment they collide and collision is
elastic — so they move away from
each other, slow down and then again
start moving toward each other (then
collide ets.) The question is — will
they bounce forever (because of
conservation of energy) or they
eventually, like, stick to each
other?
ANSWER: The
problem with your question is that
you ask for an answer in an ideal
situation, perfectly elastic
collisions and perfectly empty
space. Under those conditions, the
two balls would bounce forever. The
fact is, though, that no macroscopic
balls can collide perfectly
elastically and no vacuum is
perfect. So in the universe we live
in, the balls would lose a little
energy every time they collided and
quickly stop, "…stick[ing]
to each other…". And even if you had
perfectly elastic collisions, actual
"empty space" (intergalactic space)
contains about one molecule per
cubic meter; it might take eons, but
eventually the balls will collide
with those molecules giving some of
their energy to them and eventually
sticking together. And don't forget,
there are lots of photons zooming
around which could add or subtract
energy when they hit the balls.
QUESTION: A while
back I was reading an article about
the electromagnetic spectrum, the
visible light spectrum specifically.
And a question raced through my
mind. If I understand correctly the
colors we perceive are based on
reflective properties of objects in
our environment. Certain reflective
properties reflect different
wavelengths of light and we perceive
them as colors. My question is in
the absence of all light does
anything truly have a color? And if
so what do you think that looks
like? An infrared photo is the
closest that I can imagine.
ANSWER: If a tree
falls in the forest and there is
nobody nearby to hear it, does it
make any sound? Your question is
similar—both are about semantics,
not physics. If you say that sound
is your perception of it, the tree
makes no sound if there is nobody to
hear it; on the other hand if you
say sound is the presence of
compression waves traveling through
the air, there is sound. Color is
trickier, though, because the color
you perceive depends on how it is
illuminated. We usually think of
white light as a distribution of all
the wavelengths of sunlight. So just
saying something is red is a grossly
undetermined description because
something red in one illumination
may look pink in another. So physics
avoids this confusion and simply
uses the word color in a qualitative
sense, just like you might describe
something as warm to denote
temperature. So if you want to know
what "color" something is you
specify the distribution of
wavelengths emitted from an object
when it is illuminated by a
different distribution of
wavelengths.
QUESTION: I was
trying to bounce my Walmart ball at
the ground and have it bounce off
the ground, hit the ceiling, bounce
off the ground and then hit the
ceiling again. I noticed that no
matter how hard I threw, it would
never hit the ceiling twice. I
assume it is impossible for a
physics reason and would like to
know specifically what. The ceiling
is about 8 ft high, ball is about 42
inches around.
ANSWER: The
"physics reason" is that your arm is
incapable of giving the ball enough
energy. Each time the ball has a
collision with ceiling or floor it
loses a fraction f of its
energy, so if its incoming energy is
K , its outgoing energy is
K (1-f ); as it
travels from the floor to the
ceiling it loses energy V=mgh
where m is the mass of the
ball, g is the acceleration
due to gravity, and h is
the distance the center of the ball
travels; as it travels from the
ceiling to the floor it gains energy
V . What I did is to
calculate the energy step-by-step to
find
the
minimum initial energy K 0
needed for the final kinetic energy
to be zero, just as it reaches the
ceiling for the second time. Then I
solved the equation I got for K 0 .
The result was
K 0 =V (F 2 -F +1)/F 3
where F =1-f , the
fraction of energy retained in a
collision.
The
energy is just the kinetic energy,
K 0 =½mv 2 =mgh (F 2 -F +1)/F 3
so v =√[2gh (F 2 -F +1)/F 3 ].
So let's look
at your case, the radius of the ball
is about 6" so h is about
7'=2.1 m, g is 9.8 m/s2 ,
and I will guess that F is
about 0.5 which means half of the
energy is lost in each collision (I
dropped a similar ball and it
bounced back about halfway from
where I dropped it). Therefore,
v =15.7 m/s=35.2 mph. I doubt
that you could give it that
velocity.
QUESTION: I'm
doing my level best to finish my
first speculative fiction novel, and
this is a question to which I'm
unable to have anywhere close to
100% confidence in my answer. I'm
hoping your much more vast breadth
of knowledge might help give me a
bit more certainty with it; even if
you have to do some educated
guessing - which is more than I'd be
able to do (though I've shared my
guess below). So here goes: I would
like to have as close to an accurate
sense of how water would move on
other worlds; specifically, were all
atmospheric conditions identical to
Earth's, how would water (and most
other liquids then) move on the Moon
("The Expanse" showed liquids moving
much more slowly on the Moon) and on
Mars? My guess is, if they were
right - I assume Astronomer/producer
Naren Shankar has an pretty solid
sense of it - and atmospherics are
essentially the same, the biggest
factor is the strength of the
gravity; more than the speed of
rotation (but this is a factor I'm
uncertain about). Stronger gravity
will make liquids move more quickly,
I believe, but if they were right on
that show, then twice as fast as on
the Moon - or if rotation speed is a
bigger factor than I'm thinking,
that's something I'll have to
educate myself about.
ANSWER: The speed
v in deep water is v deep =√(gL /2π )
where L is the wavelength
of the wave and g is
gravitational acceleration; 'deep'
means h ≥½L . So the
speed is proportional to the square
root of g and to the square
root of L . For example, a
water wave on the moon would have a
speed of about v moon =√(1.6/9.8)v earth =0.4v earth .
On Mars, v Mars =√(3.7/9.8)v Mars =0.61v earth .
For shallow water, h ≤0.04L ,
the speed is approximately v shallow =√(gh );
here the speed is proportional to
root h rather than root
L . There is even an approximate
expression for the intermediate
depths, v inter =v deep tanh(kh )
where k =2π /L .
This is probably more than you need
to know, but if you want to see the
whole reference which I used, go
here .
QUESTION: I
understand that time slows down for
objects moving at relativistic
speeds. However, I don't understand
why the slow-down in time directly
affects things. Why are biological
reactions slowed down (so I age more
slowly). Or why are chemical or
nuclear reactions slowed down? Even
though they call time the 'fourth
dimension', it seems to have a much
more direct effect of things than
space does. If space where increased
or decreased by 50% in extent, I
wouldn’t become 50% bigger or
smaller. If physical objects were
directly affected by the growth or
reduction of space dimensions, how
would you ever be able to measure
that change?
ANSWER: You
misunderstand time dilation. If you
are moving with some large speed
relative to someone else, your clock
runs at some rate which is totally
the same as you would at
any speed. You do not see that clock
running at a different rate because
you were moving relative to someone
else. If your ship had no way to
look outside, you would not even
know you were moving. What is
different is the rate which you
clock runs is different than the
clock which is at rest relative to
you runs. You should read my
description of the
twin paradox .
QUESTION: So I am
having problem in understanding if
the acceleration must always be
constant when we think of Free Fall.
For example, imagine if I were about
70,000km above Earth’s surface and
the only object in my vicinity is
Earth, then I’d only feel
gravitational force of earth.
Suppose I am falling towards Earth,
then by definition what I experience
is free fall. But when I get closer
to Earth, say when I am 1,000km
above the surface, the acceleration
I’d feel would not be the same as
what I felt when I was about
70,000km above the surface. Having
said that, can I conclude that the
only reason we experience constant
acceleration when we experience free
fall on Earth is because the height
for the motion is significantly less
than the radius of Earth and
therefore, we can ignore the minute
changes in gravitational
acceleration?
ANSWER: The
gravitational force F on a
mass m at some distance
r from the center of a
spherically symmetric sphere of
radius R and mass M
is F=MmG /r 2
where G is Newton's
universal gravitational constant.
Because of Newton's second law,
a=F /m , so a=MG /r 2 .
If you set M and R
to be the mass and radius of the
earth you will find that a
is about 9.8 m/s2 . That
is the only r which has
this as the value of a ; in
other words, the acceleration is
never uniform. But, as is often
the case in science, a good
approximation can yield perfectly
adequate approximate results
provided that you understand the
limitations. It is clear from your
question that you already understand
that very well. I will give you a
quantitative way of estimating how
high h in altitude you can
go before the approximate
acceleration differs from the
ground-level acceleration:
First,
a (r=R )=g =MG /R 2 a (r=R+h )=g'=MG /(R+h )2
Next,
write
(R+h )2 =R 2 [1+(h /R )2 ]
so that we can write g'=g /[1+(h /R )2 ]
or g /g' =1+(h /R )2
Suppose
we want there to be only a 1%
error at some h ; what
is h ?
(g-g' )/g =g (1-g /g' )=0.01g =g (h /R )2
so h =√(0.01R 2 )=0.1R
QUESTION: If the
Laws of Thermodynamics and the
Conservation of Matter state thate
neither energy nor mass can be
created nor destroyed, why do
theoretical physicists fixate on the
big bang as the creation of the
universe? To me it seems logical
that both energy and mass (being
interchangeable) are eternal, and
that the big bang that caused the
present universe may just be the
latest of the likely cycle of
eternal big bangs. What do you
think, is it logical?
ANSWER: There is
no such thing as "conservation of
matter" because matter can be
created or destroyed. The total
amount of energy in the universe
cannot be changed although the
amount of mass may change. However,
I see no reason why the "eternal"
amount of energy in our universe
should preclude "fixating" on the
big bang.
QUESTION: But
energy cannot exist without matter
and matter cannot exist without
energy. As both would be stagnant in
the absence of the other. The laws
state that in an isolated system
neither mass nor energy can be
created or destroyed. Mass is a body
of matter of indefinite shape or
size. Therefore mass is matter and
cannot be created or destroyed. How
then, do you validate your
proposition that it can be?
ANSWER: This is
utter nonsense. "The laws" state
that in an isolated system the total
energy must remain constant. Mass is
just a form of energy, E=mc 2
. You are not alone in believing
that conservation of mass is a law
of physics; I often get questions
based on this assumption. So. where
did this "law" come from? Modern
chemistry was developed about 400
years ago. Experiments showed that
if you measured all the mass which
contributed to a chemical reaction
and then measured all the mass after
the reaction, the two, within the
accuracy of measurement, would be
equal. The catch is that the
accuracy of the measurement is too
poor to say that the two are
precisely equal. When special
relativity was developed, it became
clear that mass could be converted
into energy because E=mc 2 .
In an exotheric reaction, energy is
released in the form of heat and the
origin of that energy is a tiny
decrease in the mass. Indeed, even
with the most sensitive instruments
it is virtually impossible to
measure this; chemistry is a very
inefficient source of energy. But if
you look at nuclear reactions, the
mass change is easy to measure;
e.g ., the mass of a helium
nucleus (two protons and two
newtrons) is about 3% lighter than
the summed mass of two protons and
two newtrons. Stars, like the sun,
are using nuclear reactions to
release energy from mass. If the sun
were burning using chemistry, it
would have been totally reduced to
ashes millions of years ago.
QUESTION: I was
wondering how the oceans can absorb
CO2 from the atmosphere if they have
achieved equilibrium.
ANSWER: Lot's of
reasons. First, the CO2
doesn't just swim around but
iteracts chemically so that the
water dynamically reduces levels
allowing more CO2 to
replace that decrease. If the amount
of CO2 in the atmosphere
is increasing which causes the
equilibrium to be undone whether the
CO2 is decreasing or not.
So the real equilibrium is dynamic,
not static. By the way, the
chemistry causes the ocean to become
more acidic, not desireable.
QUESTION: I have
come across this paper written by a
person called John Mandlbaur. He
claims:
"My papers are
rejected because they come to a
conclusion which the editor doesn’t
believe. That is what rejected
without review means. Numerous
physicists have described my maths
as perfect and nobody can defeat
it."
Now I do not
believe this because although I am
no physicist, I know physics well
enough that papers are not rejected
without good reason so I am sure his
is rejected because the calculation
and/or reasoning is fatally flawed
but being no physicist myself I am
not qualified to comment on the
paper. If however you identified the
relevant flaws in straightforward
English, I could post your reply on
Quora and hopefully prevent others
from being taken in by these claims
of Mr Mandlbaur's. If this is not
for you, but you can recommend
someone else to explain why the
paper is refuted in plain English
please let me know.
ANSWER: I
interacted with this fellow
about 7 years ago. We had a very
lengthy back-and-forth discussion of
his "ideas" which eventually became
so outrageous that I relegated
him to the
"Off-the-Wall" Hall of Fame ". I
don't want to call names, but there
has never been a questioner more
worthy of off-the-wall status than
him. It is very sad that he has held
on to these delusions for such a
long time. I think it was Bill
Murray who said "It’s hard to win an
argument with a smart person, but
it’s damn near impossible to win an
argument with a stupid person." I am
sure that if you go over these
ramblings you will have plenty of
examples to convince yourself that
Mr Mandlbaur is
a crackpot.
QUESTION: I have
come across this paper written by a
person called John Mandlbaur. He
claims:
"My papers are
rejected because they come to a
conclusion which the editor doesn’t
believe. That is what rejected
without review means. Numerous
physicists have described my maths
as perfect and nobody can defeat it.
Now I do not believe this because
although I am no physicist, I know
physics well enough that papers are
not rejected without good reason so
I am sure his is rejected because
the calculation and/or reasoning is
fatally flawed but being no
physicist myself I am not qualified
to comment on the paper. If however
you identified the relevant flaws in
straightforward English, I could
post your reply on Quora and
hopefully prevent others from being
taken in by these claims of Mr
Mandlbaur's. If this is not for you,
but you can recommend someone else
to explain why the paper is refuted
in plain English please let me
know."
ANSWER: I
interacted with this fellow
about 7 years ago. We had a very
lengthy back-and-forth discussion of
his "ideas" which eventually became
so outrageous that I relegated
him to the
"Off-the-Wall" Hall of Fame . I
don't want to call names, but there
has never been a questioner more
worthy of off-the-wall status than
him. It is very sad that he has held
on to these delusions for such a
long time. I think it was Bill
Murray who said "It’s hard to win an
argument with a smart person, but
it’s damn near impossible to win an
argument with a stupid person."
QUESTION: I am
writing a science fiction story, and
I’d like the science to be accurate.
So: if a space ship starts from
Earth and steadily accelerates at 2
G’s, how far will it be from Each in
three months, and how fast will it
be going? Thank you so much; what
I’m writing is a love story
intertwined with the difficulties of
interstellar commerce.
ANSWER: Before
you do anything you should read the
four previous
answers which you can find in
the faq page. If you don't care to
get into the details you can just go
to
this link which will let you
know how I did the calculations for
you. You first need to understand
that acceleration is not a very
useful quantity in physics if
velocities are comparable to light
speed; this is discussed in the
previous answers I refered to above.
Since you are interested in high
speeds, we need to first determine
whether classical Newtonian physics
can be used or whether we need to
use special relativity. We first
need to specify who is measuring the
acceleration since it is not the
same as seen by everyone with high
relative speeds; I assume that you
are interested in where the ship is
and how fast it is going after three
months. The figures show
calculations I have done assuming
g =10 m/s2 , for
both relativistic (red) and
Newtonian (black) mechanics, and an
acceleration observed from the rest
frame of 2g . and
essentially velocity as a function
of time. The first figure calculates
up to about t= 23 months,
the second to 3 months. Note that
the first calculation has the
relativistic calculation approaches
c at large t
whereas the Newtonian calculation
increases linearly forever, already
at about twice light speed at about
one year. But, as shown in the
second figure, the two calculations
are approximately equal at 3 months.
Therefore at 3 months you could say
that the speed of the ship is about
half of light speed and the distance
traveled was about d =½at 2
=10x(7.78x106 )2 =6x1014
m=0.06 light years. It would be much
more complicated to calculate the
distance relativistically.
QUESTION: My
question is about spinning black
holes. Or rather spinning
singularities. Like singularities
are one dimensional objects right? A
point of infinite density. Anyway.
don't you need like...another
dimension to be capable of spinning?
Like what axis does the point spin
around? How can you rotate when
there is no geometry? I Thought it
was a point. How does a point
rotate? It wouldn't make sense
though. The thing would suddenly
stop spinning because there was no
more geometry to rotate. Anyway,
everytime I try to think about it I
sorta reach this block in my head.
ANSWER: Not one
dimension, zero dimensions. Still
your questions are pertinent. The
first thing you should accept is
that we do not know what the physics
of a black hole is. The laws of
physics are not known inside the
Schwarzschild radius. But reasonable
calculations which can be done seem
to lead to a true mathematical
singularity. I feel that most
physicists are uncomfortable with
the literal idea of infinities. That
said, there is no reason why a point
particle cannot have angular
momentum. When black holes initially
form from the gravitational collapse
of a dying star and the star
originally had angular momentum,
that property will be conserved.
From that point of view, one would
expect just about every black hole
to be "spinning". You are thinking
classically, but most elementary
particles have intrinsic angular
momentum (spin) but if you try
to interpret it as the physical
spinning of some mass distribution
about some axis you will fail.
Electrons are thought to be point
particles and they certainly have
spin.
QUESTION: I am
trying to figure out how quickly
(approz) a tangential velocity (I
think) something would have to have,
rotating around the earth at an
altitude of about 400kms, to have a
centripetal force of 1g ie 9.82
newtons. I think that's the right
terminology? Here's my thinking:
I think
escape velocity at that distance
is (very) roughly 27,360kph?
So if
you're going at just or just
under that speed you're in
constant freefall round the
earth, ie experiencing zero
gravity, in a vertically
constant orbit. (First hurdle: I
don't even know if that's
correct!)
FYI, I'm
imagining a hypothetical band
looping entirely around the
earth at ~400kms altitude, for
example, creating its own
centripetal force.
If you
stood on the inside of that
band, head towards the Earth,
and this is where my thinking
gets v v doubtful, would you
then have to spin whatever the
whole artefact (I think it comes
out at something like a
tangential velocity of 29,366kph
to create a 1g centripetal
force?) on top of the originally
calculated escape velocity in
order to create 1g's reactionary
force?
So, is the
end tangential speed the band
would need to rotate approx
56,726kph?
ANSWER: I am
sorry to tell you that most of the
things you say are nonsense. Let me
just tell you how things are rather
than picking your question apart.
First, qualitatively: A circular
orbit of radius r (r
being the distance from the center
of the earth) will have a particular
velocity associated with it; you
seem to think that you can choose
the velocity you would like to have
in that orbit. The gravitational
force on any orbiting object gets
smaller the farther away from the
center of the earth you are; as you
know, the gravitational force on an
object of mass m at r=RE
is
F (RE )=9.82m
where
RE 6
m is the radius of the earth. So the
acceleration is g (RE )=9.82
m/s2 .
But, in
general g (r )=ME G /r 2
where G= 6.67x10-11
N·m/kg2 is Newton's
universal gravitational constant and
ME = 6x1024
kg. So now you can see that every
orbit is in free fall except the
acceleration of that fall is
dependent on the radius of the orbit
and is only approximately equal to
what you think of as g at
near-earth orbits. We can also
calculate the speed at any orbital
radius r from Newton's
second law
F=mg (r )=mv 2 /r =m ME G /r 2 .
So, solving, v (r )=√(ME G /r ).
You are
interested in the altitude h =4x105
m so r =RE +h =6.41x106
m. Doing the arithmetic, g (r )=9.74
m/s2 and v (r )=7.9x103
m/s=28,440 km/hr.
ADDED
THOUGHT: Rereading
your question it appears that you
might want to be making "artificial
gravity" like the big donut-shaped
space ships which you see sometimes
in sci-fi movies like 2001 A
Space Odyssey . But I don't know
why you would want to do it around
the earth where there is already
gravity present. If you are
interested in rotating space ships
to create the appearance of gravity
because of the centrifugal force, I
have answered questions along those
lines in the past. Find links on the
faq page .
QUESTION: I am
trying to figure out how quickly
(approz) a tangential velocity (I
think) something would have to have,
rotating around the earth at an
altitude of about 400kms, to have a
centripetal force of 1g ie 9.82
newtons. I think that's the right
terminology? Here's my thinking:
I think
escape velocity at that distance
is (very) roughly 27,360kph?
So if
you're going at just or just
under that speed you're in
constant freefall round the
earth, ie experiencing zero
gravity, in a vertically
constant orbit. (First hurdle: I
don't even know if that's
correct!)
FYI, I'm
imagining a hypothetical band
looping entirely around the
earth at ~400kms altitude, for
example, creating its own
centripetal force.
If you
stood on the inside of that
band, head towards the Earth,
and this is where my thinking
gets v v doubtful, would you
then have to spin whatever the
whole artefact (I think it comes
out at something like a
tangential velocity of 29,366kph
to create a 1g centripetal
force?) on top of the originally
calculated escape velocity in
order to create 1g's reactionary
force?
So, is the
end tangential speed the band
would need to rotate approx
56,726kph?
ANSWER: I am
sorry to tell you that most of the
things you say are nonsense. Let me
just tell you how things are rather
than picking your question apart.
First, qualitatively: A circular
orbit of radius r (r
being the distance from the center
of the earth) will have a particular
velocity associated with it; you
seem to think that you can choose
the velocity you would like to have
in that orbit. The gravitational
force on any orbiting object gets
smaller the farther away from the
center of the earth you are; as you
know, the gravitational force on an
object of mass m at r=RE
is
F (RE )=9.82m
where
RE 6
m is the radius of the earth. So the
acceleration is g (RE )=9.82
m/s2 .
But, in
general g (r )=ME G /r 2
where G= 6.67x10-11
N·m/kg2 is Newton's
universal gravitational constant and
ME = 6x1024
kg. So now you can see that every
orbit is in free fall except the
acceleration of that fall is
dependent on the radius of the orbit
and is only approximately equal to
what you think of as g at
near-earth orbits. We can also
calculate the speed at any orbital
radius r from Newton's
second law
F=mg (r )=mv 2 /r =m ME G /r 2 .
So, solving, v (r )=√(ME G /r ).
You are
interested in the altitude h =4x105
m so r =RE +h =6.41x106
m. Doing the arithmetic, g (r )=9.74
m/s2 and v (r )=7.9x103
m/s=28,440 km/hr.
ADDED
THOUGHT: Rereading
your question it appears that you
might want to be making "artificial
gravity" like the big donut-shaped
space ships which you see sometimes
in sci-fi movies like 2001 A
Space Odyssey . But I don't know
why you would want to do it around
the earth where there is already
gravity present. If you are
interested in rotating space ships
to create the appearance of gravity
because of the centrifugal force, I
have answered questions along those
lines in the past. Find links on the
faq page .
QUESTION: I
suppose this is an off the wall
question, but it is quite practical,
and I’m not sure where to ask it. A
web search turns up nothing of
interest. I’ve convinced myself that
my coffee cup is interfering with
the 2.4GHz signal between my mouse
and its receiver, which is plugged
into the back of my display. The
spatial orientation of the three
objects is not ideal, but my desk
setup is a bit limited, basically
thus:
2.4GHz
wireless mouse on pull-out
keyboard tray, mouse on the
right.
My ceramic
coffee mug on the surface of the
desk, just above and beyond the
mouse, also on the right.
My display
is just above and beyond that on
the desk, with the mouse
receiver plugged into a USB port
in the middle of the back of the
display.
Ordinarily,
these three things tend
unconsciously toward an unfortunate
syzygy*. I suspect the coffee cup is
interfering with the signal, causing
severe mouse lag. So far
unexceptional. I don’t see why the
ceramic mug or its mostly water
contents might not be reflecting the
signal. But there doesn’t seem to be
any interference when the mug is
full. Only after drinking half or
more do I notice the lag. If I move
the mug way over to the left, that
seems to eliminate the lag. So my
question is, does the water level in
the mug change the overall resonant
frequency of the mug such that it
can interfere with the signal at
some levels but not others? Is it
like filling a jug to different
levels to get different tones when
you blow across the opening? And
considering that microwave ovens
work on a similar frequency by
heating water, would water be an
especially problematic interloper at
2.4GHz?
ANSWER:
Interference and blockage of
electromagnetic waves are tricky to
understand and track down. Like
sometimes when I am driving and come
to a stop there will be a specific
spot where a local radio station
will suddenly almost be totally
blocked. All the things you have
tried are reasonable: I salute you
because physics is, at its root, an
experimental science. If you could
tolerate moving your cup to the left
side, that is likely your best
solution. The 2.4 GHz seems like a
likely culprit here, particularly
since there seems to be negligible
blockage once the water is removed.
Cups of different construction would
be useful to look at, e.g. ,
a metal cup with and without water.
I doubt that it is anything to do
with interference like the blown
over bottle because sound waves are
of the same order-of-magnitude
wavelength as the resonating object
whereas the GHz waves are nowhere
close to the size of the cup. Good
luck!
*If you, are puzzled, as I was, by
the funny word
syzygy, it means when three or more
objects line up as in eclipses for
example.
QUESTION: Sir,My
doubt is in POTENTIAL ENERGY .I am
really struggling in understanding
this topic .please help me out.Sir
my question is that -"is potential
energy just a term to make sums look
easier?" Sir, taking an example - if
we throw a ball with v velocity
perpendicularly upward then at the
highest point the velocity becomes
zero therefore the kinetic energy
becomes zero but at this point we
say that the energy has been
transferred into potential energy
but acutually where did the energy
go ? And thus can we say that we
defined potential energy so that our
famous rule does not gets violated
i.e. "energy can neither be created
or destroyed it can only be
transferred from one form to another
" is it so ??? What does it mean
that energy is stored???(potential
energy)?????
ANSWER: My, you
really are struggling. It is clear
that you are just beginning your
physics education, so I will keep my
answer addressed only at that level.
In fact, I will only discuss problem
involving a uniform gravitational
field. First, suppose we know
nothing about potential energy. So,
the example you give is taking a
ball of mass m and throwing
it up vertically with some velocity
v . Then it starts with
energy E 1 =½mv 2
and rises to some height h
where all its energy is gone, E 2 =0.
So energy was not conserved; why
not? Because some force must have
done work on the mass to take its
energy. Of course, that force is
gravity which is mg .
Because the force is constant and
pointing down and the displacement
is positive is positive, the work is
negative, W=-mgh . The work
done is equal to the change in
energy W=E 2 -E 1 =-mgh =0-½mv 2 .
You could use this to find out how
high the ball will go, h=v 2 /(2g );
you have probably done such a
problem before you knew anything
about potential energy.
I next want
to generalize this problem. The ball
is thrown vertically from a height
y 1 with a
velocity v 1 and
at a later time it is at different
height y 2 and
now has a velocity v 2 .
So now the work done is -mg (y 2 -y 1 )
and the change in energy is
½mv 2 2 -½mv 1 2 .
So, rearranging, I can now write (½mv 2 2 +mgy 2 )-(½mv 1 2 +mgy 1 )=0.
If there are no forces other than
gravity which are doing work on the
ball, the quantity ½mv 2 +mgy
never changes! So now I get the
brilliant idea (just one way of
looking at this problem) to call
K =½mv 2 the
energy due to motion (kinetic
energy) and V=mgh the
energy due to position (potential
energy). So you can call
gravitational potential energy a
clever trick which allows you to
never worry about calculating the
work done by gravity. Another great
reason to use potential energy is
that energy is a scalar quantity
whereas force is a vector quantity.
There are
some things you have to be careful
of. If you are using V=mgy
for gravitational potential energy,
you must choose a coordinate system
where +y is up; e.g .,
if the top of a cliff is at y =0
m, the bottom of the cliff cannot be
at y =100 m. You cannot use
potential energy for just any old
force, it has to be a conservative
force. A conservative force is one
which does zero total work over a
closed path (ending where you
started). E.g., gravity
does negative work on the ball and
positive work on the way down. If
you push a book across a table you
must do work against friction; but
now if you push it back it will be
the same amount of work so you
cannot have a potential for
friction. Your "famous rule" is not
very useful. I think you should
think in terms of the energy
equation, E final =E initial +W ext ,
the energy you end up with is the
energy you started with plus what
you added. W ext
is the work done by all external
forces, forces for which you have
not introduced a potential energy;
forces represented by potential
energy functions are not external,
you have internalized them.
QUESTION: I saw an
older question on your site
regarding how fast Flash could run a
quarter mile. Using similar
assumptions, in Science Fantasy -
verse Warhammer 40000, there are
Space Marines who are biologically
enhanced super warriors with 19-22
implanted extra organs that make
them able to overwhelmingly
outperform a regular human. Space
Marines are roughly 7-8.5t tall
(depending on if it is earlier
variant with 19 organs or later one
with 22 organs then the height is on
the taller end of the spectrum) with
a mass of 750 pounds unarmoured.
Space Marines have been decribed
running 87 km/h taking six meter
strides. Seeing that much shorter
and less massive Flash could run
quarter mile in ten seconds giving a
velocity of 144 km/h, using the
formula, how fast would you put the
maximum velocity a Space Marine
could run given their mass of 750
pounds? Can you give one estimate
using the rubber boot sole, and one
for ceramite boot sole that their
armour is made of? And when a Space
Marine wears armour, his mass is
increased to 500-1000kg, so how does
wearing armour affect the maximum
velocity when affected by laws of
friction, gravity, etc...?
ANSWER: Oh my,
you missed the whole point of that
caluclation. The mass does not
matter because the frictional force
he can exert without slipping is
proportional to his mass. And his
acceleration is inversely
proportional to his mass; just the
same reason why the acceleration of
a falling mass (in vacuum) is
independent of mass. And strength
doesn't matter either if he can
exert the maximum frictional force
without slipping, μ static mg.
QUESTION: How an
ellipsoid object, say an American
football...ball is thrown through
the air. It shows projectile motion,
but the other interesting
observation is that the front of the
ball points in the direction of
travel. I have thought that air
resistance possibly applies unequal
torques on different sides of the
ball perhaps, but I read somewhere
that it has to do with gyroscopic
precession - that this axis is
different to the one for angular
frequency, and somehow that explains
it. I'm not sure though, any simpler
way to understand this?
ANSWER: If we
were to pass the football with
spiral in a vacuum, the spin axis
would remain pointing in the same
direction because of conservation of
angular momentum; this is the case
for the lower trajectory shown in
the figure. In air, though, the
angular momentum mainly remains
pointing in the same direction as
the velocity of the ball. I always
am surprised to find an everyday
phenomenon the physics of which is
unexplained or explained only
recently. This is such a case and
was explained only recently (2020)
in a
paper in American Journal of
Physics . A more accessible
description of the phenomenon can be
found at
this link . Your hunch that the
air has something to do with it was
spot-on.
QUESTION: so, tell
me if I'm wrong but if an object is
energy it doesn't have mass yet it
is still affected by gravity, but
you need mas to be affected by
gravity 1. why is energy still
affected by gravity 2. is there any
way whatsoever for something without
mass that isn't energy be affected
by gravity.
ANSWER: No, you
do not "need mass to be affected by
gravity". The
only object we know which has no
mass is the photon (a quantum of
light). And because it moves, it
has kinetic energy which is
well-known; you shouldn't say that a
photon is energy though,
rather it has energy. When
a photon finds itself in a strong
gravitational field it does not go
in a straight line (as you see
things) which you might have
expected, but rather it follows the
curved lines of the space where the
gravity is; in other words, in a
curved space a curved line may be
the shortest distance between two
points. Your second question refers
to something which has
neither mass nor energy; that isn't
really something , is it?
QUESTION: I hope
this question isnt stupid, also
English isnt my first language, but
here I go: People always say that
you can't travel faster than light
speed, but what is the point from
which you count the speed (0 m/s)?
Because the earth is already moving
through space, so would we need to
accelerate less when accelerating in
the same direction and more if we
accelerate in the direction it comes
from?
Follow up, may
be explained by the answer to the
first question: does the light
emitting from let's say a LED
travelling at 60mph reach a target
100 miles away faster than a
staionary LED (assuming we turn both
on the second they are next to each
other, so when the moving one passes
it)?
ANSWER: One of
the triumphs of the theory of
relativity is that its success
validates the principle of
relativity: The laws of physics are
the same in all frames of reference.
From this principle we can deduce
that the speed of light in a vacuum
is independent of the motions of the
observer or the source. So that
answers your 'followup': a bystander
would see the two light flashes
arrive at the same time. Your main
question asks which frame of
reference may a moving object not
exceed the speed of light? Since all
observers see the same speed of
light, the answer is any frame.
QUESTION: Dad I
have a (possibly) physics question:
how come tempered glass just
spontaneously combusts without any
apparent cause? I remember my old
roommate's shower fully falling
apart during the pandemic when no
one was even there
ANSWER: Hi
Hannah! I don't think you mean
"spontaneously combust" because that
would mean burning/flaming. Glass
does spontaneously shatter or break
in unexpected circumstances,
however. I have answered variations
of this question several times (
e.g. ,
1 ,
2 ,
3 );
1 is probably the most complete.
QUESTION: I have
been thinking about a question, like
how we can write the change in unit
vectors over a finite time interval,
now it
may
seem like a homework problem but
trust me it's not it's a genuine
question that I have been thinking
about. I know about the change in
unit vectors in an infinitesimally
small-time interval 'dt' and that
it's given by the magnitude of the
really small angle that the unit
vector has turned about, but what
about a change in the unit vector
over a finite time interval how can
we write that. I tried to find the
answer in many sources, but I was
unable to find any.
ANSWER: It is not
really clear to me what you are
asking. If you are talking about
Cartesian coordinates, they never
change. But if you have a coordinate
system like spherical polar
coordinates, the directions of the
unit vectors are very dependent on
where a point happens to be. I will
use a two dimnensional polar
coordinate system where the vectors
have both radial and tangential
components to keep the answer from
being too complicated. The figure
shows a red dot moving along some
path (red dotted line) which shows
the path taken by the particle
between two times t 1
and t 2 . The unit
vectors are shown in green. If you
know those two locations you can see
that the unit vectors all rotate
through an angle θ 2 -θ 1
in the clockwise direction. The path
taken by the particle will be
described by some vector function
f fr
and fθ as shown
in the figure. So the average rate
of change of direction of unit
vectors is ω avg =Δθ /Δt =(θ 2 -θ 1 )/(t 2 -t 1 ).
QUESTION: A ship
at the speed of light turns on
headlights, understand that a person
on that ship sees light go out at
the speed of light, relativity. What
does a person "see" from outside
that area of relativity? Can they
see that, is that currently
possible, what if, if not current?
Do they see a ship with no lights,
light but no ship, a spectrum of
both that equals the speed of light?
I am, and with more questions, being
as specific as possible: I find it
hard to articulate a concept from
the written word, please accept and
understand my ignorance/lack of
understanding!
ANSWER: First of
all, no ship can travel "…at the
speed of light…" and I do not answer
questions which stipulate at or
faster than the speed of light. So I
will answer your question assunming
v =0.99c , 99% the
speed of light. As you say, an
observer on the ship sees the light
in front of him having a speed of
c regardless of how fast
the ship is moving. An observer
relative to whom the ship is moving
with speed 0.99c also would
measure the speed of the light to be
c . This is one of the
postulates of the theory of special
relativity: The speed of light in a
vacuum is independent of the motion
of the source or the observer.
However he would not
"see"
the same "color" of the light
because of the doppler shift.
Suppose that the light as seen by
the ship had a wavelength of 450
nm=4.5x10-7 m, blue; for
0.99c speed of the source the
doppler shifted wavelength would be
about 4.5 nm, low-energy x-rays.
Hence the stationary observer would
see nothing but if he could measure
the speed of the x-rays coming out
the front of the ship he would see
them as having a speed of c .
QUESTION:
ANSWER: Imagine
an object to the left of a spherical
lens as seen in the figure. There is
an imaginary line drawn
perpendicular to the lens and
through its center call ed the optic
axis. There is some object
consisting of many colors. I have
chosen to look at just three points
in this object, one red, one blue,
and one green. Each rays from these
points illuminates exactly one point
where the image is formed; although
I have only drawn three rays for
each point, every ray passes through
the the same single point on the
image. Apart from size and whether
the image is inverted or not, the
image is an exact copy of the
object. Your statement that "…light
itself is not coloured…" is not
correct--the color of light is
determined by its wave length.
QUESTION: Assuming
no engineering challenges like
parasitic losses in materials etc….
Is it theoretically possible to
take: Speaker A. Play a tone. 440hz.
Digitally. Using EXACTLY 100 watts
of power. Speaker B. Play a tone.
440hz. Digitally. Using EXACTLY 100
watts of power. Have speakers A and
B facing each other head on. EXACTLY
180 degrees. The 2 speakers are
playing the same note. Same signal,
timbre, frequency amplitude pitch
etc…. Would these two “sounds”
cancel out perfectly such that “in
between” there would be zero
“sound”. I understand that small
changes in angles and frequencies
will have enormous effects on there
and how the sound will dissipate if
this system is not perfectly
balanced from the get go. Very loud.
But if the experiment were set up
perfectly, is it theoretically
possible to zero out the wave? Full
100% destructive interference?
Chaos? Feigenbaum?
ANSWER: No, the
result is a standing wave of the
same frequency. In the animation the
original waves from the left (blue)
and right (green) combine to make
the black waves which have the same
frequency and wavelength but zero
velocity. This is due to the
superposition principle which states
that if two or more waves are moving
in the same medium, the net
displacement is equal to the sum of
the individual displacements at each
time and location. Note that the
black curve is sometimes totally
zero, when the two traveling waves
are out of phase with other. If two
identical waves are traveling in the
same direction they could completely
cancel if they were 180° out of
phase. In the second (nonanimated)
gif, imagine the green and blue are
moving in the same direction. The
black would be the resultant wave.
QUESTION: I am
trying to understand heat, that is,
mostly the idea of hot and hotter.
My understanding is that heat is
basically motion, and heating
something up (cooking, let's say) is
about transferring motion from one
place to another. So that the
changes that take place in cooking
are really caused by the molecules
of stuff rubbing against each other.
So hotter really means MORE motion,
because the actual temperature at
which things burn, like natural gas,
can't be increased just because you
burn more of it. Does that make some
sort of sense to you? Clearly my
mind wanders in the kitchen, and
this has been nagging at me for a
long time.
ANSWER: I believe
that your question is answered by an
earlier answer . You refer to
motion; in my earlier answer the
kinetic energy referred to is K =½mv 2
where m is the mass of some
atom or molecule in your flame or
pot or food, and v is its
speed (motion).
QUESTION: I have a
question about fluid
velocity/acceleration and its
effects on tubular friction. Is the
relationship linear? Meaning , if
the fluid velocity is being
increased, is there a moment or a
few moments where the friction
increases more than what it would’ve
been if the relationship was linear?
ANSWER: From what
I have been able to learn while
researching your question is that
"friction" as in two solids sliding
on their surfaces is not applicable
to fluid inside a pipe. The fluid
exactly on the surface is not
sliding but rather adheres to the
surface and is at rest. The motion
of remainder of the fluid has some
velocity just outside the surface
increases to a maximum at the center
but certainly not linearly.
Everything about this velocity
profile is due to the viscosity of
the fluid, sort of the fluid version
of friction. More detail can be seen
in
this link and links which it
contains. The figure shows a figure
from that link and shows the shape
of the velocity profile, v (r ).
Eventually, at high enough speeds
the fluid becomes turbulent and "all
bets are off".
QUESTION: How
should I get to calculating the
force that needs to be applied to a
piston to compress a container of
gas? Suppose the temperature is
constant, and by default, as the
volume decreases the pressure
increases. Let's suppose the volume
gets halved.
ANSWER: The force
depends on the area A of
the piston. And, if the gas cannot
be well approximated as an ideal
gas, you would need detailed
properties about the gas. For an
ideal gas, PV /T
must remain constant and if T
is constant, PV is
constant. Now, it seems easy, like
if you halve the the volume you need
to double the pressure; that would
give you the answer for the force to
hold the piston there once you got
it there, F=P 2 A= 2P 1 A .
Be sure that you understand that
this is what you mean because if you
simply started at the beginning with
this force and continued pushing
with that force, when you got to the
half-volume position you would go
right by it and it would start
slowing down until finally it would
turn around and start coming back.
QUESTION: For
reference I'm 31. Am I closer to the
age 60 than I am 20? I think I'm
closer to the age 60 because time is
moving forward and we cannot move
backwards to my age at 20. So
wouldn't I be closer to 60 because
time can only move in a linear
motion towards 60 and we are
constantly distancing ourselves from
the age of 20?
ANSWER: You are
11 years older than 20 and 29 years
younger than 60. You are closer to
20 by almost a factor of 3. If you
had asked which you would be closer
to being, you would be closer to
being 60 than 20; but you didn't.
QUESTION: Is a
mirror VERY different to how people
perceive me in reality? I am
bothered by the fact that I never
realised before but I guess there's
a reason… How impactful is the
reversal effect? I always thought
that's how I look to others.
ANSWER: The image
you see is not how others see you,
but a "mirror image". If you look at
the image of yourself in the mirror,
the image of your right hand is the
left hand of the mirror you. If you
want to see an image of yourself
which is what others see, you need
to use a corner mirror. However,
since most of us are pretty much
left-right symmetrical, there is
seldom much difference.
QUESTION: My eight
year old has a question. Get ready.
If two particles are in entangled,
and one is on the earth and the
other has passed the event horizon
of a black hole. Would the
entanglement be broken or even if
it's unmeasurable still be
entangled. Does the medium "the
ether" transcend the event horizon?
Silas Massicotte (8) and Malachi
Massicotte (6).
ANSWER:
Seriously? Your 8-year old
understands quantum mechanics and
cosmology?! That would be pretty
amazing. I usually don't do, as
stated on the site,
astronomy/astrophysics/cosmology,
but I did a little research and
think I can give you an answer. When
one of the two entangled particles
is totally destroyed, the other one
will not know at all because there
is no communication between the two.
Since the wave function of the
destroyed one simply disappeared,
the wave function of the other will
remain in the mixed state it was in
originally, but it will no longer be
entangled with anything. If a
measurement is made on the survivor,
there will be no change in whatever
remains of the other particle. There
is a pretty good explanation at this
link . There is no such thing as
the ether.
QUESTION: how can
electromagnetic waves travel in
vacuum? how can something without
any physical body and dimensions
travel in nothing? it should need
something to mediate its propagation
like sound waves require air
molecules. if its pure energy then
what is pure energy?
ANSWER: This is
exactly the way physicists were
thinking at the end of the 19th
century. Serious searches for the
medium required for light to be able
to propogate were being made. The
best known search was the
Michaelson-Morley experiment which
failed to find it. It was called the
luminferous æther. The punch line is
that it doesn't exist at all. Light
can travel through empty space. One
way to understand this is through
electro-magnetic theory, quite well
understood at the begining of the
20th century. James Clerk Maxwell
unified all that was known about
electromagnetism into four
equations, today known as
Maxwell's equations . One of
those equations says that
time-varying magnetic fields cause
electric fields; another says that
time-varying electric fields cause
magnetic fields. Using all four you
can show that one possible solution
is a wave equation for both electric
and magnetic fields. Since waves are
time-varying, these waves will "keep
each other going" which has nothing
to do with any medium. If you go to
the link above, you can follow other
links telling you more.
