Ask the Physicist!

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QUESTION:
Something that confuses me is that scientists say it takes approx. 8min for light to travel from the sun to earth. So if the sun were to explode we wouldn’t know till 8min after. However Einstein gathered that the faster an object travels the more time slows down so as you approach the speed of light time stops. But isn’t this just relativity? If you are moving that fast everything seems like it’s not moving cause your moving so fast. If time were to actually stop how could it take 8min for us to realize the sun exploded wouldn’t we realize it instantly? So why can’t you move faster than the speed of light? Wouldn’t that just mean things relative to you would seem to be stopped for longer, meaning time doesn’t stop when you reach the speed of light?

ANSWER:
See the faq page for an explanation why you can't move faster than the speed of light. Suppose that someone was moving with speed 99% of the speed of light from the sun to the earth, a distance of about 8 light minutes. Because of length contraction the distance would, for her, be shortened 8x√(1-0.99²)=1.13 light minutes and the time for her to get to earth would be 1.1/0.99=1.14 minutes; so she would observe the sun's demise to hit earth in a time of 1.13 minutes, and she would arrive 0.01 minutes later. But, just because she observed the time to be 1.13 minutes does not mean someone on earth does; the earth-bound do not see the distance to the sun to be shorter but see it as 8 light minutes.

QUESTION:
Suppose you have two persons watching universe expanding, one of them is on earth, the other is moving fast towards the end of universe. Both of them are watching it all the time. Does this mean, that the "end" of universe exists twice and each is in a different "position"?

ANSWER:
You miss the whole point of special relativity, that the time and position of some event do not have absolute values. That the traveler will observe the end of the universe to be in a different location and at an earlier time than you do does not imply that the universe ends twice at two different locations; rather, it implies that time and position are different for him than for you.

QUESTION:
I was telling my kids that time is space and explaining the block universe and they asked me, if time is space, then how many minutes are in a mile? It sounds stupid, like it does not have an answer, but the more I think about it, the more it seems like if space is time, it should have a specific answer, and if not, why not?

ANSWER:
I will first give the simple answer without much detail since you are dealing with kids. But your kids must be fairly sophisticated to be thinking about space-time at all, so I will then give a little more background on how space-time (four dimensions) arises. When special relativity is mathematically represented as suggesting a four-dimensional space, the fourth dimension is not represented by time (t) but by x₀=ct where c is the speed of light. So the fourth dimension has units of length just like the original three, x₁=x, x₂=y, x₃=z. So, if x₀=1 mi=t·186,000 mi/s, then t=1/186,000 s=5.38x10^-6 s=9x10^-8 minutes, approximately one nanominute.

Before Albert Einstein developed special relativity, it was generally assumed that time and length are universal and independent. One second for me is the same for anyone else in the universe; similarly, one meter is the same regardless of the position or motion of the meter stick I am measuring. If someone is moving with a speed v along my x axis and I measure the length of his meter stick, I find it is shorter; similarly, if I measure the rate his clock runs, I find that it is slower than mine. We are talking about meter sticks and clocks which, if located in my reference frame, are identical to mine. Mathematically this is expressed in what we call a Lorenz transformation,

x'₀=γ(x₀-βx₁)

x'₁=γ(x₁-βx₀)

where the primed coordinates are in the moving frame as measured by the stationary frame, β=v/c, and γ=1/√(1-β²). So, two of the coordinates in our 4-space get mixed up together if the other frame is moving. Does anything analogous happen in everyday 3-space? The answer is yes; if you take a coordinate system with three axes, (x,y,z) and rotate it through an angle θ about the z-axis, the new x'- and y'-axes both depend on what θ is and they depend on both x and y are:

x'=xcosθ+ysinθ

y'=-xsinθ+ycosθ.

Hence the Lorenz transformation may be interpreted as a rotation in the x-t plane. This is the impetus of mathematically working in a 4-dimensional space (space-time).

QUESTION:
I teach high school physics, and I recently performed a lab to investigate Lenz's law. Given the current state of things, this was done in my own home, with students on video chat timing my experiment with their phones. My experiment involved dropping a cylinder of neodymium magnets (about 3 cm long and 1 cm in diameter) down a 1.5 m long copper pipe about 2 cm in diameter. I marked out different lengths on the outside of the cylinder, and used a steel nail to hold the magnet in place at these various “starting lines.” So I could directly drop the magnet 153 cm from the top, and 130, 110, 90, 70 and 50 cm using the nail. For each distance, we did three trials. For each of those trials, seven students recorded through the Google Meet. I expected a bit of a hectic disaster, but the precision of the results was amazing. My idea was to use the graph of distance vs. time to discuss what was happening (they haven’t learned about Lenz’s law yet) in terms of forces. I talked about different types of graph shapes for different situations and how they relate to what is happening with forces: uniform acceleration, uniform velocity and free-fall with air resistance (terminal velocity). My fingers were crossed that our data would approach linearity like the latter, and it did! A valuable learning experience for all. The one hitch is that the z-intercept of our linear trendline is positive, not negative. When an object goes from rest to terminal velocity, I had shown them that the z-intercept of the linear portion of the graph is negative, because the speed only increases to this section of the graph. With our results, I am forced to conclude that the magnet slows down before achieving linearity, because the average slope of the graph is steeper prior to the domain of our results.

ANSWER:
The questioner said in his initial message that the graph of the data would be included but it was not. I have not been able to get him to reply to any messages, but I spent some time working this all out, so I thought I would go ahead and post it. In doing my calculations I approximated some of the variables with reasonable guesses. I had three neodymium approximately 1 cm magnets which when stacked were about 1 cm; the mass of three was 10 grams so three cm would be about m=30 grams=0.03 kg. I took the acceleration to be about g=10 m/s². I made a wild guess at the terminal velocity by watching videos on the internet and chose v=20 cm/s=0.2 m/s. I also note that many videos seem to indicate that terminal velocity is achieved very quickly. The resistive force is given by F=-bv (not proportional to v² as most air drag problems are) where b is a constant to be determined. I choose the coordinate z to increase in the positive vertical direction and the initial position to be at z=0 and the initial velocity to be v=0. This is a classic problem and I will just state the results which result from solving the Newton's second law equation, -mg-bv=m(dv/dt):

v=(mg/b)[e^-bt/m-1]

=0.2(e^-50t-1)

z=(mg/b)[(m/b)(e^-bt/m-1)-t]

=0.2[0.02(e^-50t-1)-t]

Note that, for large bt/m, v≈-mg/b=-0.2 which is the terminal velocity. This can be solved for b, b=1.5 N/(m/s).

The plot above for v shows that it takes only about 0.1 s to approximately reach terminal velocity. The plot for z shows z(t) in black and z(t) if the magnet moved with the terminal velocity the whole time; the linear portion of the curve is shifted up by 4 mm, possibly responsible for the "positive z-intercept" observed by the questioner. Of course these plots are not what is being plotted by the questioner because each new datum starts at rest; but every datum plotted has the same behavior as the one I have shown, so all the individual runs are shifted by the same 4 mm. The shift is so small compared with the total fall distances, the non linear time is so small compared with the total fall times that the experiment should be able to extract a reasonably good measurement of the constant b. Unfortunately, I do not have the times of the various data points, so I cannot be more specific; the time of fall from the top if the terminal velocity is 20 cm/s would be 7.5 s.

QUESTION:
Is following the line of steepest topographic descent (as stream would do) an example of conservation of energy?

ANSWER:
Let's take a stream as the specific example to discuss the question. The reason the water flows downhill is that there is a force acting on it, the downhill component of the force of its own weight. Generally a stream of water will tend to follow the path where it finds the greatest force impelling it down, that is the steepest terrain. Now, if that were the only force on the water, it would accelerate and the water would be going faster at the bottom. But, in order to do this, the stream would have to get narrower as you went down because water is basically incompressible and if you have, say, 100 gallons per minute flowing at the top of the hill you need the same flow rate at the bottom where it is flowing faster. (Think of a hose nozzle where the water flows much faster out the nozzle than in the hose.) But this is not what happens because there are drag forces which impede the flow. What you generally find is that, if the stream is about the same depth and width as it moves down the hill, the speed is pretty constant. Therefore the kinetic energy at the bottom is the pretty much the same as at the top but the potential energy has decreased; mechanical energy has decreased, not been conserved. However, if you consider the whole system of the stream, the stream bed, the banks, etc., the energy is conserved because the lost potential energy of the of the water will show up as heat in the environment and the water. However, this would be true regardless of whether or not the water flowed down the steepest path. So, I would not use the choice of steepest path to be an example of energy conservation.

QUESTION:
I tried posting this question on a different website and didn't get much of a response. For some reason there doesn't seem to be a lot of information available about this subject, or maybe I just can't find it. Anyway, I am hoping you can help shed some light on the subject. Please forgive the formatting - I've posted a couple of different questions, but if you need a single, concise question to answer, the main one I have is this: What's going on here?

I have a bottle of water that I put in a freezer for a period of time, until the water becomes super-cooled and is still liquid. Now we know that when we shake or disturb the bottle, some of it crystallizes. But it doesn't turn into solid ice. Instead it turns into slush - a mixture of tiny ice crystals and liquid water. Here are my questions:
What determines what parts of the water will become ice, and which will stay liquid?
What is the temperature of the slush mixture?
Is the water part slightly warmer than the ice part?
or is it all pretty much the same exact temperature?
If I poured out the liquid water into a different bottle, and left the slush / ice in the original bottle, would there be any chemical differences between the two?
Is the explanation that the liquid water has impurities such as minerals, and the frozen water is pure H20?
If that is the case, then if we performed this experiment with pure water, then it should turn directly into solid ice, and there'd be no slush, right?
I have read about Fractional Freezing which seems relevant. But this doesn't seem to fully describe what is happening. I am hoping for more of a general explanation rather than a list of individual answers.

ANSWER:
I recommend that you google supercooled water videos and watch a few. It seems pretty clear to me that your problem with slush is that your water is not pure enough. Possibly not cold enough either so that generated heat warms the water just above zero before it freezes. This is a very touchy sort of experiment, one of the videos has the teacher admitting it took him a number of tries to get a good enough take to post. By the way, I thought the answer on Reddit was pretty good.

QUESTION:
Hi, I have a bee in my bonnet about a real world home improvement matter concerning wall tiles and the surfaces they are stuck to. There is much ado about the supposed weight of tiles per square metre that the sub surface can support and arguements are made for using concrete backer boards which can take greater weights than plasterboard (or drywall if you prefer). So plasterboard can take 30kg/m2 and backer boards 50kg/m2 or more. The base of wall tiling always finishes at a horizontal surface, such as a floor etc. So given there are no gaps between tiles and that the wall is truly vertical, what is the relevance of the supporting weight figures for the subsurface board? I can't see that there would be any force acting that would cause any board to fail assuming the tiles end on a horizontal surface that can take the weight. I follow that there will be compressive force down through the tiles and thus through the backing board by virtue of the adhesive to some degree. But surely this compressive force can't affect the backing board as the weight is carried down through the tiles to the horizontal surface that they connect with.

ANSWER:
Well, this isn't really physics, but having had a recent experience (a backsplash behind our range), maybe I can make a few comments. If the wall is sheetrock and sealed or painted, it is best; this is because the surface is just paper and if it is necessary to provide vertical force, it could be too weak. But, as you state, if the tiles extend all the way to the floor (or other support like a cabinet as in my backsplash), there should be no reason to apply the concrete boards. On the other hand, I would not put tiles on a bathroom sheetrock wall because if moisture gets behind the tiles you would have a mess. The same thing goes for bare wood. I once had laid large terra cotta tiles on my kitchen floor on top of an old pine floor; it turns out I had a slow leak in the ice maker which I did not discover until the tiles started coming up from the saturated spongy wood floor. I had to rip up half the floor and leave it to dry for several weeks before reinstalling (too late for a subfloor because it would have raised the floor too far). When I tiled our laundry room I did use the backing board and it is still in pristine condition after 40 years.

QUESTION:
This is really not a question about astrophysics but something simple I am missing. I know that as two massive objects orbit each other they lose energy by emitting gravitational waves and that eventually the orbit decays and they will collide. However as the Earth and Moon rotate around each other and the system loses energy by tidal forces the Moon moves AWAY from the Earth. Why the difference?

ANSWER:
It is understandable that you would think of this because it would appear that the moon is gaining energy rather than losing energy. In fact, it doesn't just appear to gain energy, it is gaining energy. The reason is rather subtle. The moon causes a bulge in the oceans which is responsible for tides. The bulge is actually two bulges, one toward and one away from the moon as shown in the figure (greatly exaggerated); this is because of the tidal force. Because there is friction between the oceans and the ocean floors, the rotation of the earth drags the bulge forward, so it is slightly ahead of the moon as shown in the figure (again greatly exaggerated). Now the bulge exerts a force on the moon which causes it to accelerate a tiny bit; that's where it gets the added energy to increase its orbital radius. Similarly, the moon pulls on the bulge which causes the earth to slow down. If this were the whole story, the energy would be conserved, the earth losing energy and the moon gaining the same amount. But, as noted above, there is friction between the earth and the oceans which takes energy away from the system, giving a net loss of energy but the moon ends up with more than it started with.

QUESTION:
If subject A is traveling through space at a high rate of speed and subject B is traveling at a much slower rate of speed would you expect to see a difference in the rate at which either subject can process information. Example: Would it be easier for subject A to process information than subject B or vice versa?

ANSWER:
This is too vague. What does "process information" mean? What do you mean by "easier". Now, whatever the answers to these questions are, if each subject is presented with the same information stream in his frame of reference and has the same equipment to process it, they will be equally easy to process. But if you mean that A sends an information stream which he processes to B to process (or vice versa), they will not process the information at the same rate. For simplicity's sake, I will choose the "slow" frame B to be at rest; in relativity, only the relative velocity is important. Let's choose a simple concrete example of information stream: the information is a stream of ten pulses separated by 1 second as seen in the reference frame B; B will process them in 10 seconds. Now, how does A process that same information which B processed in 10 seconds? Suppose A is moving away from B; then the time between pulses will be longer than 1 second because he moves some distance away while waiting for the next pulse to catch up with him—the information is processed more slowly. You should be able to see that if you interchanged A and B everything else would be the same so B would process the information more slowly. Suppose A is moving toward B; then the time between pulses will be shorter than 1 second because he moves some distance toward B while waiting for the next pulse to arrive—the information is processed more quickly. You should be able to see that if you interchanged A and B everything else would be the same so B would process the information more quickly. An earlier answer would be of interest to you.

QUESTION:
Matter wave of a particle depends on its momentum,doesn't that mean different observers with different speed will measure different wavelength?

ANSWER:
Yes. This is called the Doppler shift.

QUESTION:
If I am driving in a vehicle in a straight line at a constant speed and activate a drone in the vehicle so the drone is hovering inside the vehicle, open the back of the vehicle and increase my speed so the drone drifts out the back of the vehicle, what then happens to the drone once it is not longer inside the vehicle?

ANSWER:
There will be a period when the drone is leaving the car when there will be lots of turbulence and other currents of air; I am answering neglecting this transition period and also assuming the air outside the car is still. You have adjusted the drone so that, even though it is moving forward with the car at speed v, it hovers in still air. When it suddenly finds itself outside it sees a headwind of v, not still air. This headwind will exert a force which will tend to slow the drone down so that eventually the drone will be hovering at rest relative to the ground.

QUESTION:
Once a spacecraft breaks free from earth's gravity, how much propulsion does it actually need to reach a destination, say Mars? Would the craft have to continually accelerate through the vacuum of space, or once it reaches its maximum velocity, would its propulsion be turned off?

ANSWER:
Technically, you never "break free from earth's gravity". If you acquire a speed greater than the escape velocity, then you will never fall back to earth, but you will slow down continually forever (assuming you have no other objects exerting forces on you). Nevertheless, the force gets very small pretty quickly. If you are 100 earth radii away from earth, the weight (force of earth gravity) is 1/100²=10^-4 times smaller than on earth. And 100 earth radii is not all that far in the whole scheme of things; the distance to the moon is 60. Anyhow, a spacecraft always coasts nearly all the time on a mission somewhere in the solar system. Propulsion is used mainly for getting up to speed and then slowing down to orbit or land on your destination; a much smaller use is to steer, changing direction.

QUESTION:
I have had this question in my mind for over 10y, and I would be very grateful if you could help me to get an answer. The theoretical experiment goes as follows:
1) Bring to outer space 1 round glass chamber, 1 pure iron ball, and a source of heat. 2) Place the iron ball in the center chamber, and heat it until it gets bright red.
3)Go out of the chamber, close it, and generate a total vacuum.
As far as I have been able to go:
Q1: Can we see the bright red iron bar?
A1: Yes, and considering that the ball is in an absolute vacuum the light must be behavior as a particle. *Quantum mechanics dictate that given the duality of light a media is not necessary for light to travel, thus radiation.
Q2: If the light behavior as a particle it must have a mass. Where does this mass come from?
A2. Given that the iron bar is in an absolute vacuum, my logic dictates that the only source of mass for the photons must be the mass of the iron bar itself... Is this correct?
Q3. After an X amount of time when the iron bar has finally cooled down. Will the iron bar be lighter than at the beginning of the experiment? ** Consider that the energy in the iron bar can only be released in the form of light.
A3: I have been stuck in here for several years.....
Thank you very much for your time, please notice I’m not a physicist but a Biology’s with a deep interest in physics.

ANSWER:
I do not see why all the stuff about a glass chamber, heat source, etc. is necessary to ask your question which is, simply: if an object is hot and cools soley by radiation, does the object have less mass after cooling?

First, your A2 is wrong. Photons do not have mass even though they do have momentum and energy. Therefore, your conclusion that the object will be lighter because the photons carry away mass is wrong. However, because mass is just a form of energy, we can say that the photons carry away energy (even though they have no mass). But, where did that energy come from? The object had to provide it. But there are just as many atoms after the cooling as before, so the object evidently changed mass into energy. This change in mass is very difficult to observe in the case of a red hot iron object cooling down. Inasmuch as E=mc², the change in mass would be Δm=ΔE/c² where ΔE is the energy carried off by the radiation. Suppose that ΔE is a real big number, say a 10,000 Joules; then Δm=10⁴/(3x10⁸)²=10^-13 kg=0.1 nanograms! This is approximately the mass of a single yeast cell.

QUESTION:
For a space traveler that leaves Earth headed for a distant star @ 0.9c, wouldn't his onboard clock (that's @ rest relative to him) read the time passes normally......i.e. as if he were still on Earth? He would see no time dilation nor length contraction!

ANSWER:
Wrong. the distance to the star is measured from the earth; think of it as a long stick. The stick is moving with speed 0.9c as seen by the traveler, so he sees the distance shorter by a factor √(1-0.9²)=0.44. Since, as you correctly stated, her clock runs normally, it only takes her 4.4 years to get to a star 10 light years away. It would be worth reading my explanation of the twin paradox which is linked to from the faq page.

QUESTION:
Which is the correct term to refer to the speed of light (scalar) or the velocity (vector) of light. They are both used in books & literature. I think its should be the "speed of light", but just asking.

ANSWER:
It depends on the context. One context is "the speed of light is a constant in all frames of reference and independent of the velocity of the source or observer". Another would be "when a ray of light passes close to a massive galaxy, it is bent and so its velocity changes although its speed does not."

QUESTION:
How do I calculate the kinetic energy of an object moving submerged under water? Specifically if water pressure is involved. Like if a cannonball was fired at 110 m/s under 9,000 feet underwater. Or If a submarine is moving 10 m/s at a depth of 800 feet. I know kinetic energy is factored in and pressure at that depth and possibly water drag. I'm just not sure if there is a specific equation for this. Can you please help me with this?

ANSWER:
The kinetic energy of something depends only on its mass m and its speed v, K=½mv². It has nothing to do with its pressure or its drag force. If there are any forces on the object, those forces do work which might change the kinetic energy. Because of the near incompressibility of water, the depth will have almost no effect in how something moves. The effect of the pressure is simply the buoyant force which is a force equal to the weight of the displaced water which is nearly the same 100 ft down as 10,000 ft down. Similarly, the drag depends only on the density, speed, and geometry, not pressure, the drag will be about the same at any depth.

QUESTION:
What's the physics behind "walking ladders" that move down slopes by themselves? I'm thinking maybe it's the same for slinkys that 'walk' down steps. Gravity and the weight of the ladder being moved around the different legs. I tried Googling for an answer but can't find any scientific answers.

ANSWER:
This was a fun problem to figure out. First, let's think about all the forces on the ladder when it is slightly tipped so that two legs are off the ground. There is the weight W, normal forces (not shown) N_front on the front leg and N_rear on the rear leg, and frictional forces (not shown) f_front on the front leg and f_rear on the rear leg. The weight acts at the center of mass of the ladder which is centered relative to the two sides, but is closer to the front than the rear because the front is heavier. That means that the normal force on the front leg is greater than on the rear leg; this is important to remember as we go further. Now, I want to examine the torque which the weight causes. I have resolved W into components down the incline (W_x) and normal to the incline (W_y). W_y causes a torque about the axis I have labelled DROP AXIS causing the currently-off legs to drop to the ground. The inertia causes the ladder to keep going lifting the other legs off the incline; if none of the legs ever slipped (lots of friction), the ladder would rock back and forth without going forward. Now, the other component W_x exerts a torque about the axis I have labelled TWIST AXIS; if no legs slip, this torque has no effect on the motion. However, for a couple of reasons, the rear leg may slip:

Because the normal force is smaller for the rear leg, the maximum static frictional force will be smaller than for the front leg.

The material contacting the ground may be different for the rear legs than for the front, e.g. the rear legs may have a smaller coefficient of static friction and could slip even if the normal forces were the same.

So, for the video, the rear legs must slip and the ladder, with each "step" twists about a normal axis. There are frictional losses which would cause the ladder to lose energy and therefore eventually stop, but since it is going down an incline, this energy is replenished by the work done by W_x as it moves down the incline. It seems to me that it would be pretty tricky to get this set up since everything must be just right for it to work.ADDED THOUGHT:
A different possibility occurred to me. I have treated the ladder as if it is rigid. However, if it is kind of rickety the lifted legs will tend, if they can, to move forward because of gravity, even if the two down legs do not slip. So when they come back down they will have moved forward a bit.

QUESTION:
My Dad loves physics and always referred to the speed of light as 186,000 miles per second, per second or 186,000 miles per second sq. that is how he always spoke of it. Is that speed of light an excelleration? (even though nothing goes faster than light.)

ANSWER:
The dimensions for speed are length/time, not length/time/time. Therefore, 186,000 mi/s/s is wrong.

QUESTION:
E=m(c)2
What is the reason for twice the speed of light? Einstein tried everything and found it?

ANSWER:
Oh dear, it is not twice c it is c squared. E=m·c·c=mc². And Einstein did not find it by trial and error; how could he, since it was not really appreciated at the time that mass was, in fact, just another kind of energy? It was a natural result of his theory of special relativity.

QUESTION:
According to Enistine space time is like a fabric and all masses round around the sun so why Moon revolve around Earth not sun?

ANSWER:
Not just the sun, but every object with mass, warps spacetime. It is much easier to think about this in terms of Newtonian gravity for which the equivalent statement is that all objects with mass have a gravitational field. The force which a mass M exerts on a mass m a distance r away is F=MmG/r² where G is the universal gravitational constant. You can calculate the ratio of the force the sun exerts on the moon to the force the earth exerts on the moon:

F_sun/F_earth=(M_sun/M_earth)(R_earth/R_sun)=(2x10³⁰/6x10²⁴)(3.8x10⁸/1.5x10¹¹)²=2.1

So the sun exerts a force more than twice the force the earth exerts on the moon. But if you think about it, the moon does revolve around the sun since it moves right along with the earth; but because of the force the earth exerts, it also revolves around the earth.

QUESTION:
according to law of gravitation less mass object attract toward heavy mass object so why we don't attract toward wall because wall have greater mass then a human mass ?

ANSWER:
Actually, the law of gravitation states that two masses attract each other with equal and opposite forces. So the wall does exert a force on you and you exert an equal and opposite force on the wall. However, since gravity is such a weak force, you do not feel it at all. For example, if you and have a mass of 100 kg and the wall has a mass of 10,000 kg, the force you experience, if you are 10 m away, is about 10x10,000x6.67x10^-11/10²=6.67x10^-8 N=1.5x10^-8 lb.

QUESTION:
I'm trying to calculate what effect hitting a 1,500 lb water barrel at 30 mph in a 15,000 vehicle would have upon the truck's momentum? Also, if I increased the truck's weight to 65,000 and drove it at 50 mph, what difference would this make on the momentum too? I'm working on a road safety prototype in my shed and to be honest this kind of math is way beyond me, so any help would be much appreciated.

ANSWER:
I will assume that the barrel and vehicle move off together after the collision; this is called a perfectly inelastic collision. In a collision the total linear momentum, the product of the velocity times the mass, remains constant. Although lb is not a unit of mass, it is a measure of mass. So, before the first situation you state, the momentum before the collision is 15,000x30=450,000 lb·mph and after it is 16,500v where v is the speed after the collision; equating the two and solving for v, I find v=27.3 mph. For the second case, 65,000x50/66,500=v=48.9 mph.

QUESTION:
I have watched Human Universe with Brian Cox. If you haven't seen it, in part one he goes to Russian to meet with the Soyuz spacecraft. He tells his viewers that only knowing two of Newton's equations (F=ma and the law of gravitation F=GmM/r²) he can calculate that the spacecraft only needs to reduce velocity by 128 m/s to safely reenter orbit. I have looked and tried to figure out how the calculations work but can't. Do you have any clue how he would have done the math?

ANSWER:
I will assume that the Soyuz is returning from the International Space Station. First I will give some information needed later:

R=6.4x10⁶ m (radius of the earth)
h=4x10⁵ m (altitude of the space station)
G is the universal gravitational constant (will not need it)
M is the mass of the earth (will not need it)
m is the mass of Soyuz (will not need it)
g=9.8 m/s²=MG/R² (acceleration due to gravity at r=R)
√(MG/R)=√(gR) (this will be useful later and is why I do not need to know G or M)
MmG/r² is, as you note, the gravitational force on m a distance r from the center of the earth.

If orbits are circular, a=v²/r and you can show that v=√(MG/r). I am thinking that we need to find the difference of speeds between orbits with r=R+h and r=R:

v_R-v_{R_+h}=Δv

=√(MG)[√(1/R)-√(1/(R+h))]

=√(MG/R)[1-(1+(h/R))^-1/2]

≈√(MG/R)[½(h/R)]

=½h√(g/R)

=247 m/s

Note that, because h/R=0.0625 is small compared to 1, I have used the binomial expansion (1+x)ⁿ≈1+nx. My answer is about twice what Green got. Of course, in the real world, air drag when entering the atmosphere would significantly reduce the required speed change due to the additional braking it would cause; but you cannot estimate that using only those two equations. So, I am at a loss to understand the difference.

ADDED THOUGHT:
Maybe the Soyuz dropped to a lower orbit before beginning its reentry, h~200 km, after leaving the ISS, but I could find no reference to its altitude in the video.

QUESTION:
i am currently designing a fully adjustable boom arm for a microphone my prototype is fine and it is functional but i am interested on getting more weight suspension from my joints the joints im using are friction based and use clamping force to lock the joints in the desired position what material should i to use as washers to create more friction in my joints

ANSWER:
it is hard for me to be specific since i do not know what the joints are made of what you need to do is find a material with a larger coefficient of static friction with the material currently in the joints i think the very best way would simply apply trial and error try different materials to see how they work i would start with obvious choices like rubber why dont you use punctuation or capitalization it is very annoying reading your question

QUESTION:
In a fire where does the photons come from? Are photons in a superposition of wave and quanta particles?

ANSWER:
Fire is many chemical reactions going on. The reactions are, for the most part, exothermic and most of the energy comes out as heat and light. Photons are particles which have wave properties.

QUESTION:
I am confused in the formula of orbital angular momentum of electron. According to quantum physics the formula of orbital angular momentum is L=l√(l+1)h/2π B.M. while according to Bhor's postulate the formula of angular momentum is L= nh/2π. Both are giving different result. For example for hydrogen putting n=1 in Bohr formula and l=0 in quantum mechanics formula gives h/2π and 0 respectively.

ANSWER:
I believe there is an error in your question, L=√(l(l+1))h/2π. The answer to your question is that the Bohr model is simply incorrect. As luck would have it, it describes the energy levels of the hydrogen atom well, but it is incorrect regarding angular momentum of the levels.

QUESTION:
Why would a heavy object like a bowling ball not float on top of water, but a wooden telephone pole, which is much heavier, would float? I'm sure the shape and the material the object is made from plays a role but I would just like to know the science behind it.

ANSWER:
The science behind this is over 2000 years old. Archimedes was a Greek mathemetician who discovered what we call today Archimedes' Principle. It says that there is an upward force (called the buoyant force) on an object in a fluid which is equal to the weight of the fluid which it displaces. It turns out that, applying this principle, anything with a density smaller than water floats and anything with a density greater than water sinks. Wood is less dense than whatever bowling balls are made of. You might then ask why a steel battleship can float. It is because it is hollow and therefore the total density is much lower than the density of steel; a hollowed out bowling ball would float.

QUESTION:
I read that metals conduct heat well because the electrons of metallic elements have free electrons in its outer shell that move quickly when heated. These quickly moving electrons transfer their kinetic energy to other electrons which vibrate quickly and collide with other electrons and transfer the heat energy and so and so forth. I also read that metals conduct electricity well because the electrons of the outer shell are free to move (as is the definition of electricity - the movement of charged particles). So it would reason to stand that heating a metal should generate electricity or at the least facilitate the production of electricity as both heat conductance and electricity involves the movement of electrons in terms of metals. To my surprise after a google search, I learned that heating a metal does not generate electricity, and furthermore hampers (not facilitate) electric conductance. Please dear physicist help me to understand why hot metals are poor conductors of electrons.

ANSWER:
In a conductor, there are "free electrons" which essentially behave as an electron gas. There is normally no net current because, just as there is no wind in still air, the electrons move in random directions and the net flow is zero. If you apply a voltage across a conductor, these moving electrons, on average, flow from the negative to the positive terminals. Since they are now experiencing a force due to the electric field, they accelerate; but, obviously, they do not really speed up the whole time that they move from one end to the other of the conductor but, on average, move with some constant speed called the drift velocity. The reason is that they are not really free; rather, they accelerate for some short distance and then collide with one of the atoms in the solid which scatters them, then accelerate again, then collide again, etc. Now, all the atoms in the solid are constantly vibrating as if attached to tiny springs, and the hotter the metal gets, the larger the amplitudes of their vibrations becomes. This means that they present bigger targets for the electrons to hit and so collisions happen more often which reduces the drift velocity and therefore the conductivity.

QUESTION:
I was confused about the strict physics definition of a wave. Considering it is a traveling disturbance that carries energy, would wind on a field, an audience created "wave", and dominoes falling be waves? I thought the audience would not be a wave, and I was unsure about the dominoes because there is no wavelength or period.

ANSWER:
The "strict physics definition of a wave" is that it is something which satisfies the wave equation. The solutions of a one-dimensional wave equation are always of the form f(x-vt) where f is any function; for example where x is the position, t is the time, and v is the speed at which the disturbance travels through the medium which supports it. For your examples: dominos falling and people in a stadium sitting and standing are both waves, but wind is not. Wind is an example of the entire medium moving but that would not be described by a wave equation. The stadium wave would not be a wave if the fans all all just ran around together, like a hurricane of people.

QUESTION:
In my understanding of time dilation if one has a clock in constant motion relative to me I will measure the time ticking off on that clock as slower. However from the other clocks view it is just as valid to say that I AM in motion so an observer traveling along with that clock would measure my clock as moving slower. How is that reconciled? Especially if at some time in the future both clocks are brought together at rest with each other to compare clocks? Am I missing the part that acceleration plays here?

ANSWER:
There is no need to reconcile that each observer measures the other's clock to run slower unless you bring the two of them back together into the same reference frame. Acceleration has nothing directly to do with it. So this is the "twin paradox" which we know is not really paradoxical at all. Go to the faq page and find "twin paradox".

QUESTION:
We have been taught that the big bang created all heavenly bodies as we understand them. I'll disagree with that. Were the exo-planets we have discovered thus far created by the same creation event that brought forth us, Mars and Venus or was it something else that lead to the Gliese's and the Kepler's?

ANSWER:
Good grief, nobody claims that the big bang created everything as it is right now. The big bang created a huge amount of energy (from where, nobody knows) and physical laws. From that point the universe evolved over time eventually atoms, stars, galaxies, planets, etc. appeared. To see a timeline of the universe, click here.

QUESTION:
I'm designing a special suction tube for dental practitioners (I'm an orthodontist) and have a basic question I feel I should know. I want to maintain as MUCH air flow through the tip of the suction tube(end of the tube by the patient's mouth) as possible. The lumen of the suction at the dental unit has a cross sectional area of 75mm2. This tube sucks the air from the patient's mouth into the dental unit. If I use a tube that has a much smaller lumen EXCEPT at the patient's mouth where the cross sectional area would be 75mm2, will I lose air flow? Another way of asking this is: am I required to keep the cross sectional area of the ENTIRE tube at 75mm2 in order to maintain the same flow rate at the tip of the suction (patient's mouth) that I have at the dental unit?

ANSWER:
I learned something new: in physics, lumen is a measure of luminous flux. Biolgists, though, use it as a measurement of inner space of a tubular structure; apparently it is just the cross-sectional area of the tube since the questioner specifies the area of the opening. If you were dealing with an ideal fluid, incompressible, no wall drag, no viscosity, you could argue that the flow would be constant, the speed in the narrower part of the tube would be larger, but the volume per unit time flow would be the same. But air is not incompressible and energy loss effects are not negligible. In order for you to maintain the same volume flow rate you would need a more powerful pump which is pulling the air. Think of trying to drink through a pinched straw–you have to suck harder to get the same amount of drink.

QUESTION:
I have a question for you. So from what I've learnt, Einstein's Theory of Relativity says that an object with mass will bend light towards it by way of its gravity and the source and direction of the light will therefore appear differently to an observer. Besides that, Einstein also says that E=mc^2 and thus m=E/C^2 so something like light, with measurable energy, would have a measurable mass. My question is, just like that object with mass pulls light towards it with its gravity, can light also pull that object towards it with its gravity? Similarly to how an object falling to Earth pulls on the Earth itself? Or does light not have gravity in the same sense as objects with actual mass on account that its mass is relativistic and not actual, classical-physics mass?

ANSWER:
For the first part of your question, see an earlier answer. (You could also have found this on my faq page.) The answer to your second question is that the theory of general relativity finds that any energy density warps spacetime, that is causes a gravitational field; a photon has energy and therefore creates gravity; keep in mind how small this is when compared to any macroscopic mass.

QUESTION:
This is not a homework question, I am actually a teacher and saw this question in a textbook and would like to know if this is a valid question to ask a third-grader. I may be wrong but I do not believe that the pound and ounce are units of measurement for mass. I have tried looking it up online and did not find anything useful. Which two units of measurement are used to measure mass?
Gram
Pounds
Ounce
Kilogram

ANSWER:
Kilograms (kg) and grams (g) (1 kg=1000 g) are units for mass. Pounds (lb) and ounces (oz) (1 lb=16 oz) are units for force. This is very confusing because of the ways weight is measured in different countries. In countries where Imperial units are used, weight (the force which the earth pulls down on something) is measured in pounds (which is correct) and we think that this must also be a measure of mass (the amount of stuff there is). In countries where SI units are used weight is measured in kilograms (which is incorrect) and we think that this must also be a measure of force. I will leave it up to you to decide whether third-graders can understand this subtle difference!

QUESTION:
I know that in a solid the particles are still in motion, but I was wandering if their is any point in the state of matter where the particles are completely motionless? (something like before solid)

ANSWER:
The Heisenberg uncertainty principle stipulates that you cannot know to absolute precision both the position x and momentum p (mass times velocity). Stated mathematically, ΔxΔp≤ℏ where ℏ is the rationalized Planck constant (a very tiny number and Δ denotes the uncertainy of the quantity. So if the velocity (momentum) uncertainty is zero (particle perfectly at rest), the uncertainty of the position of the particle is infinite—the particle could be anywhere in the universe.

QUESTION:
how can the calculations for the curvature of space-time under general relativity be correct without the mass-energy associated with dark matter taken into account? I am just a retired attorney trying to educate my self about our physical world.

ANSWER:
Keep in mind that I specify on the site that I do not do astronomy/astrophysics/cosmology, so you may wish to take my answer with a grain of salt. When you refer to the "curvature of space-time" you are actually talking about the gravitational field. And, you are certainly right, dark matter, if it actually exists, would affect the field. I believe that it is actually the other way around: by studying the gravitational field you can infer something about the distribution of dark matter. The most well-known such inference has to do with rotating galaxies and how the angular velocities of the stars varies with distance from the center. If the only gravity in our Milky Way galaxy were from stars and gas which we know are there, the galaxy would not be able to hold together, stars at larger radii would have speeds far too large. The Wikepedia article on dark matter halos would be of interest to you.

QUESTION:
If a toy car hits a wooden block and pushes it along the track, do the wooden block and the car have the same level of energy or does the energy of the car decrease and the energy of the block increase?

ANSWER:
You have described a perfectly inelastic collision, one in which the two colliding bodies stick together after the collision. I am assuming that the block is at rest before the collision. So linear momentum is conserved and energy is not conserved. I will denote the velocity of the car before the collision to be V, the velocity of the car/block after the collision to be U, the mass of the car to be M, and the mass of the block to be m. Conservation of linear momentum gives MV=(M+m)U or U=V[M/(M+m)]. Note that U<V, so the car has lost energy and the block has gained energy.

If you are interested you can calculate the change in energy (ΔE=E_after-E_before) for each:

ΔE_car=½M(U²-V²)=½MV²[ (M²/(M+m)²)-1]

ΔE_block=½MU²=½MV²[M²/(M+m)²]

For example, if M=m, ΔE_car=-(3/4)[½MV²] and ΔE_block=(1/4)[½MV²]. The car lost three fourths of its initial energy and the blocked gained one fourth of the initial energy. In this case, half of the total energy was lost. In this case, half the speed was also lost.

QUESTION:
The question is about Maximum gradient a vehicle can climb. The maximum friction force that can be transferred to the road cannot be more than μmgcosθ. As far as I know, μ cannot be greater than 1 (irrespective of contact patch size) for a Rubber tyre on concrete road. At 45 degrees, the drag due to gradient (mgsinθ) is equal to the Normal force (mgcosθ). Any more than 45 degrees, and the drag force increases and Normal force decreases, thus making the vehicle's tyres slip due to loss of traction. Now, I see videos of RC cars on youtube climbing gradients well over 54 degrees. Could you explain this? Is the μ value greater than 1?

ANSWER:
Your analysis is fine if the vehicle is a point mass. But, the video you refer to has the failure to climb a steep incline due to the vehicle tipping over backwards rather than slipping. I will assume the truck shown above is at rest because it is easier to visualize and exactly the same as if it were moving up with constant speed. The whole weight mg may be considered to act at the center of mass of the truck (white cross) which is located a distance h above the ground, a distance d₁ behind the front axle, and a distance d₂ in front of the rear axel, as shown. There are three equations, sum of the forces along the ramp equals zero, sum of the forces perpendicular to the ramp equals zero, and sum of the torques about the point where the rear wheel touches the road equals zero:

f₁+f₂-mgsinθ=0

N₁+N₂-mgcosθ=0

(d₁+d₂)N₂+hmgsinθ-d₂mgcosθ=0

Now, we are interested in when the truck will start rotating backward about the rear axle; when it is just about to do that, the normal force N2=0. So, the third equation tells us that tanθ=d₂/h; therefore the maximum angle before tipping is θ_tip=arctan(d₂/h). For example, if d₂=1.5 m and h=1 m, θ_tip=56.3⁰. What determines tipping is just where the center of mass is; you want it as far forward and as close to the ground as you can.

But, if you lose traction, slip, before you rotate, you do not need to worry about rotating. As you correctly surmised above, the truck will start slipping when θ_slip=arctan(μ) where μ is the coefficient of static. I agree with you that μ=1 is a reasonable estimate for rubber on dry concrete. However, the video you sent is not done on a concrete ramp, rather the surface appeared to be carpet. Since the tire tread can press into the carpet pile, it is not surprising that μ>1 for those videos.

QUESTION:
How does a single speaker unit play an audio clip consisting of sound frequency of different magnitude? It's the same speaker and has only one 'vibrating membrane' to induce the sound in the air to propagate. How does this single membrane generate sound waves of different frequency and different amplitude simultaneously?

ANSWER:
Sound waves are linear, i.e. if several are acting on a point in space, the net action is simply the sum of all the soundwaves. As an example, I have plotted three waves representing the individual sounds, each with different frequencies and different amplitudes, at some point in space as a function of time. The net sound at this point in space is simply the sum of the three, shown by the black curve. Your speaker need only be driven to vibrate like this; this is pretty easy to do, usually with a magnet attached to the speaker cone and inside a coil, the current in which varies like the black curve.

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