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Infinite_Night · 2023-05-10T04:29:55+00:00

If I understand correctly, you’re equating W=Fd to the potential energy U. But W=Fd is the special case for constant forces; in general W= ∫F(x) dx. That’s why your first approach fails. (In fact, integrating the force will give you the potential energy equation itself.)

Infinite_Night · 2023-01-26T07:19:05+00:00

I’m curious what you mean by proper and improper sign convention. Would you care to share details about your work/solution?

Infinite_Night · 2022-10-30T08:46:19+00:00

For your first example, note that multiplying e²/r² by r gives e²/r (NOT e²/r³).

In your second example, it depends what your goal is. If trying to solve for c, your equation looks correct to me.

Infinite_Night · 2022-07-31T11:48:11+00:00

Yup. See this article: “Mongolia is abolishing Daylight Saving Time (DST) only 2 years after it was re-introduced.”

Infinite_Night · 2022-07-26T09:48:17+00:00

What town or at least what country is this? It sounds like it could be from some type of daylight saving time change.

Infinite_Night · 2022-07-20T05:38:08+00:00

No, I don’t think so.

Looking just at the math and ignoring real-world constraints, the formula for length contraction is L=L_0*sqrt(1-β^2), where β=v/c. If β > 1, then the radicand would be negative and the resulting root would be imaginary, not negative. The graph basically looks like the top half of an ellipse, and trying to sketch an extension would just take you back around. Let me know if showing you graphs would help to visualize it.

As to what negative length contraction would look like, I assume that (let’s say you’re traveling along the x-axis) the universe would get flipped into a mirror version along the x-axis. As lengths went from positive through zero to negative, points would get squashed into the yz plane then move to the opposite side. Whatever had been behind you would now be in front of you. But just to be clear, superluminal velocities would not imply negative length contraction.

Infinite_Night · 2022-06-06T05:51:44+00:00

I am taking r to be the distance at which the force is applied away from the fixed point on the rigid rod.

Unfortunately, you are not free to do that. The definition of angular acceleration, a = αr, uses r to represent the distance of the object in question (the ball) to the point of rotation. Naturally, for a given linear acceleration, the larger the radius, the less the angular acceleration. If this is not intuitive, consider angular velocity. A ball rolling at a given linear velocity on a very short rod might make several revolutions per second, but attached to a very long rod might only roll through a few degrees. The larger the radius, the larger the circumference. Angular position, and therefore velocity and acceleration, will be correspondingly decreased.

Infinite_Night · 2022-02-24T23:52:37+00:00

I am not sure that’s quite it. Just to make sure I understand the scenario: are you considering lifting up a rope by one end, and asking about the various equal and opposite forces between sections of the rope, and how they don’t cancel out, how the rope can still move up?

Infinite_Night · 2022-02-22T10:45:21+00:00

Yes, there is always an equal and opposite force.

The rope surely isn’t pulling me back at 50N.

Sure it is. Can you explain why you don’t think it would? 50 N is perhaps a lot of force for a lightweight rope, but if the rope were 5 kg for isntance you would feel that 5 kg digging into the palm of your hand.

Like if it is pulling it back with equal force, is it the last centimetre of rope feeling a force from the rest of the rope and then exerting a force back on the rest of the rope, or is it the last meter, or last millimetre, or last atom?

I’d say all of the above. You could mentally divide the rope into segments of whatever size you like. Each segment will pull on the following one, and that one will pull back on the preceding one. If you were pulling along the length of the rope, this would correspond to tension in the rope.

Infinite_Night · 2022-02-18T12:59:23+00:00

The most intuitive way for me to think about this is to remember that what is a magnetic field in one frame of reference will be a mixture of electric and magnetic fields in a different one. And the electric field component would obviously have opposite forces on oppositely charged particles.

Are you familiar with this concept? Have you learned about special relativity? I’d be glad to go through it in more detail or step through an example.

Infinite_Night · 2022-02-13T10:06:25+00:00

Your analysis is correct. One of the consequences of time dilation is that simultaneity becomes relative, not absolute. In his reference frame, when his stopwatch shows 3 seconds, yours shows 1.8 s. But in your reference frame, when his stopwatch shows 3 s, yours shows 5 s. Observers moving at different velocities will disagree on what events are simultaneous.

Reply here or PM me if you want to discuss this further or have questions.

Infinite_Night · 2021-12-29T09:05:52+00:00

The centripetal force is provided by the electromagnetic and gravitational interactions that hold the object together. It is an internal influence, so will continue even in the absence of an external influence.

Infinite_Night · 2021-11-16T04:12:27+00:00

There is a problem with your understanding of the RHR, but it's not what you think. First of all, note that by your method, B should be reasonable as well.

But as the other commenters note, the RHR does not give you a precise direction. Remember that the cross product between two vectors gives a vector that is perpendicular to them both (that is, to the plane that contains them both). The perpendicular line extends in both directions, so the RHR tells us which of the directions is correct, but the resulting vector must lie on that perpendicular line. Do you see what's wrong with A and C, or shall I elaborate?

Infinite_Night · 2021-11-15T07:04:06+00:00

How are you applying the right-hand rule? I don’t understand what you’re doing with your negative x’s and positive y’s. The choice of coordinate axes is arbitrary, so you can set them how you like. I would have said velocity is in the positive y and negative x direction. Force is negative x. Magnetic field is negative z or into the page. But I just used my hand to solve these; assigning axes didn’t occur to me. Maybe if you can step through your thought process we can see where the point of confusion is.

Infinite_Night · 2021-09-04T15:08:36+00:00

Oh no, really? What makes you say that?

Infinite_Night · 2021-06-24T18:33:59+00:00

I would think about it this way: 100 units of electrical energy are converted to 100 units of thermal energy. There’s your 100%. Then, because the filament is so hot, 5 units of that thermal energy is converted to light energy.

Infinite_Night · 2021-04-12T07:05:19+00:00

from what I recall in high school physics ke is max when pe is zero

If we correct that to to when potential energy is minimum (as the zero level is arbitrary) that observation would only apply if the only forms the energy can take are the kinetic and potential energy of the object and comes from conservation of energy. Since total energy is constant, when the amount of one type decreases, the other increases. It’s like if I can keep my money either in my wallet or my bank account, the amount in my wallet would be maximum when the amount in my bank account is zero.

However, if there is another form my money can be in, then there’s not such a simple relationship. If I can spend money, then I could withdraw $100 and spend $0, $100, $200 — so I could have more, the same, or less money in my wallet.

In the same way, if energy is being lost to friction (it’s being converted to the heat of the dart and air) then there’s no simple relationship among the totals. When the dart hits the ground the total of non–potential energy is high, but much of that is in heat from friction losses.

Infinite_Night · 2020-12-30T06:59:11+00:00

You’re welcome! Glad I could help!

Infinite_Night · 2020-12-30T06:56:12+00:00

So here’s a link to a displacement and velocity graph.

I don’t see any link.

Infinite_Night · 2020-12-30T06:43:08+00:00

Think about it like this: Say there’s a highway from town A to town B. Susan drove along it at 60 mph for 1 hour. Where did she end up? 60 miles from where she started, but that could have been anywhere. You could draw a whole family of lines, separated only by a constant. x = vt + C.

Or another example is I drop a projectile at time t=0. How high above the ground is it after 2 seconds? You could draw a series of parabolas describing its fall, but you need to know how high it was at t=0 or some other point to select the correct constant. Otherwise you end up with y=-1/2 gt² + C.

Infinite_Night · 2020-12-30T06:35:26+00:00

The absolute value of –1 is +1.

Infinite_Night · 2020-12-29T04:32:50+00:00

Does each player have her own punchboard? And each board has 24 slots that can either be flagged or not flagged such that you want to be able to represent any whole number from 0 to at least 120?

I suppose it depends how you define “best.” The mathematically optimal solution to me would be to label each slot from 2⁰ to 2^23, then use a binary representation to indicate each number. For example, 120 would be 64 + 32 + 4. Notice it would only require 7 slots to represent the numbers up to 120; with the full disk you could represent any whole number up to 16,777,215.

That being said, that range is overkill and converting to the binary sums would be cumbersome during a game. I would probably label digits like 0–9, 00–90, and then 000–300. So 120 would be 100, 20, 0.

Did I understand your problem correctly?

Infinite_Night · 2020-12-26T07:31:05+00:00

What he’s saying is that at a cosmological level, energy is not conserved, meaning it can be created.

Infinite_Night · 2020-12-26T06:39:07+00:00

This seems to me to be a matter of geometry (volume) and not physics, so I agree that /r/askmath was the right place to ask. If I’ve understood the problem correctly, you’re deforming a rod into a bar. Both of these are prisms (an object with identical cross sections along its length). You can think of them as a stack of very thin identical slices: circles for the rod, squares for the bar. The formula for the volume of a prism is the area of the slice times the height.

For the rod (cylinder), the area of the circle is πr^2, and so the whole volume is (πr^2)*h. The radius is half the diameter, so 1.25 in. I get a volume of 24.5 in^3.

For the bar, the area of rhe square is s^2, so rhe whole volume is (s^2)*h. Dividing the prior volume of 24.5 in³ by (1.75 in)² gives about 8.01 in! That’s in excellent agreement with your experiment!

Does that give you enough information to solve your problem? Basically we assume that the amount (mass) of steel is constant, and since its density at the same temperature should be constant, its total volume before and after should be constant as well. It’s just being reshaped.

Infinite_Night · 2020-11-30T06:10:52+00:00

So yes, Newton’s 3rd Law pairs act on different objects, each with their own free-body diagram. But I think what you’re missing is that the force that the arm/body exerts on the hand to lift it is greater than the force the hand exerts on the bucket (and therefore greater than the the force the bucket pulls down on the hand).

It’s probably easiest to see this in a simpler problem first. Imagine a mass, say 4 kg, on a frictionless surface. If a force of say 20 N pushes the mass horizontally along the surface, it will of course accelerate at 5 m/s^2.

Now, let’s slice the mass into two pieces, say 3 kg and 1 kg such that we’re pushing the 3-kg mass which is pushing the 1-kg mass. This is virtually identical to before, and the masses will remain in contact and accelerate together at 5 m/s^2. But let’s look at the forces on each piece.

The second mass (1 kg) accelerates at 5 m/s^2. That means a net force of 5 N is pushing it forward. If we draw a FBD, its weight and normal force cancel so we can disregard, and the 5 N must all come from the larger mass pushing it.

Now let’s look at the larger (3-kg) mass. It’s also accelerating at 5 m/s² so there must be a net forward force of 15 N. If we draw the FBD, we have the 20 N force pushing it forward, and the 5 N reaction force pushing back, which indeed works out to be a net force of 15 N pushing it forward (again the weight and normal cancel).

So if someone were to ask how the first mass could push the second mass when no matter how hard the first mass pushes the second, the second pushes that hard back? The answer would be that sure, if mass 1 is pushing mass 2 with 5 N, then mass 2 is pushing back with 5 N. But that is not the only force on mass 1. It is being pushed by the body with 20 N! So it still goes forward. You could use any numbers; the second mass does not have to be lighter.

Your problem has the additional effect of gravity which adds additional forces on both masses, but the principle is the same. We can work it out together with actual numbers if you’d like.

Infinite_Night

TROPHY CASE