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2

Hint: You're far enough from the ground that the formula $mgh$, which is an approximation, doesn't really apply anymore. What is the true formula for gravitational potential energy?


0

In this problem you need to consider the difference between the G.P.E. of the mass on the Earth's surface and the G.P.E. at this intermediate point. You haven't said how you have calculated this, but it seems likely that you have made the total gravitational field equal to zero at point P. This is an approximation. You should understand that if you really ...


2

You calculated the specific potential energy at that distance. You were asked to calculate the potential energy needed to reach that point. You did two things wrong in that calculation. You forgot to multiply by 930 kg and you forgot to use the given condition "that the potential energy of a 930 kg object on the Earth's surface is -58.7GJ". What you need ...


1

When we lift those three weights from Y to X, we can use the reversbile machine. So, we ease the machine B, because machine A takes those weights. Is that a right picture? There are two weights with mass $M$ and $3M$. Initially, both weights are at the same height $h_0$ which we can freely set to zero: $h_0 = 0$. Now, machine B lowers mass $M$ to ...


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It's considerably cheaper to fly from ATL (Atlanta, Georgia) to ORD (Chicago, Illinois) than it is to fly from VPS (Valparaiso, Florida) to BPT (Beaumont, Texas). It's not distance; Atlanta and Chicago are considerably further apart than are Valparaiso and Beaumont. It's not wind speed; Chicago is the windy city, after all. It's not oceans; Atlanta and ...


4

They aren't. Major metropolises tend to be near water, as that was the primary method of transportation when said cities were founded. Airports tend (gosh :-) ) to show up near major population centers. I suspect Chicago, St. Louis, Denver, etc. would be amused at your claim that airports are near the sea!


-1

I think that wind direction may have something to do with it. It will be easier to predict prevailing winds close to the coastline.


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That's essentially what a Time On Target attack is all about. There are a couple of ways to use TOT: "George, you take out this target, John, you take out that target, you over there take out that third target..., and I'll take out this Nth target, with our bombs all within 3 seconds of one another." The idea is to take out the officers, command post, comm ...


4

I think your question is a bit too ambiguous to answer properly, but here is one way to consider the situation. Suppose you've got a nail, and you know it takes 10 Newtons (a made-up value) to make it put a dent in a board. If you hit the nail with a hammer, with an impulse force which peaks at 1 Newton, hitting ten times in succession will do nothing. ...


1

Both are correct. The elastic energy of solid is stored in the chemical bound. On the other hand, it also contributes, although by very very tinny amount, to the total mass of the system due to $E=mc^2$. So combining both facts, we arrived at the conclusion that the chemical bound has mass, even though the mass of the chemical bound is about only $10^{-9}$ ...


1

Part 1. Essentially you're right. You can think of it as subtracting infinity from the energy. A better way to view it is that the convention that the zero of energy should correspond to potential at infinity was always an arbitrary choice. Normally it is a very sensible convention, but if the potential diverges at infinity, as is the case for the harmonic ...


1

The distinction to be made here is that, for the quantum harmonic oscillator system, there are no unbound states, only bound states thus, there is no benefit to insisting the states have negative energy, no reason to 'subtract infinity' in order to zero the potential at infinity. However, in systems that permit both bound and unbound states, it is ...


1

Work done on a system increases the energy of the system. But you have to be clear about what system you are talking about. For the system consisting of the Earth and your other object, there is no external force, and no work done on the system. The total energy is constant; it just changes from potential energy to kinetic energy. If your system is the ...


-1

In a closed system there is no where for the energy to go. There is no way for the gravitational force to change the overall energy. I would simply change "the work done equals the energy change, BUT not when the work is done by a conservative/internal force" to "The work done equals the kinetic energy change". This just means that you can convert ...


2

Energy is conserved so it can't be created or destroyed. All we can do is change energy from one form to another. In your example we are changing the potential energy of the mass $m$ into kinetic energy. The increase in kinetic energy must be equal to the decrease otherwise energy wouldn't have been conserved. By an external force I assume you mean some ...


0

Total work done = change in energy. So when you compute the kinetic energy at 30 mph and 60 mph, you find that you needed to do 3x as much work in the second interval. Note that you do need to convert from mph to m/s if you want to work in SI units… Leave potential energy out of it. It is not relevant in this situation. Work done = gain in kinetic energy ...


4

If the particle moves from the point $x$ to $x+dx$, and assume $dx\gt 0$ for simplicity, then its potential energy increases by $$ dU = \frac{dU}{dx}dx $$ Well, it increases if $dU$ is positive and decreases if $dU$ is negative. So far I have only used the definition of the derivative – pure mathematics. However, the total energy is conserved. The sum of ...


4

Potential energy for a point mass (and also for a sphere) is not $PE = mgh $ (this is special case for a uniform field) but rather: $$ PE = - \frac{GMm}{r}$$ where G is the gravitational constant, M and m are both masses and r is the distance between the masses. (in the case of a sphere, the distance is to the centre of the sphere) Can you see the answer ...


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Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a ...


8

Most of the electromagnetic energy is in x-rays which means it is deposited in the bulk material of the bomb and in the surrounding air over a few meters (or tens of meters at most). All that stuff heats up, needs to expand but piles up against other stuff also trying to expand. You get a massively energetic shock-wave of very hot material which is in ...


2

The first the expression $U(r) = -\frac {GM} r$ is a potential, but not potential energy. The units are velocity2. This is a widely used potential in solar system astronomy, geology, geophysics, and in aerospace engineering. For example, see ...


2

Notice that $h$ and $r$ are related in the following way: \begin{align} r = R + h \end{align} where $R$ is the radius of the Earth (the distance from the center to the surface) and $h$ is the height above the surface. Then notice that \begin{align} U = -\frac{GmM}{r} = -\frac{GMm}{R+h} = -\frac{GMm}{R}\left[1 -\frac{h}{R} + O(\frac{h}{R})^2\right]. ...


4

You make a good point. In a battery a chemical reaction (a redox reaction) creates the potential difference, and the potential difference is calculated assuming the electrons start and finish in well defined energy states. If electrons returning to the cathode have some residual kinetic energy then the could affect the reaction and change the EMF. However we ...


0

Are you wondering why your two equations $W=\frac{1}{2}kx^2$ and $W=Fx$ don't match? That's because the latter equation, $W=Fx$, is only true for a constant force. The more general expression is $W=\int \vec F \cdot d \vec x$, similar to what you wrote in differential form $dW=F\,dx$. The expression with $\frac{1}{2}$ in it is correct. The expression with ...



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