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39

The bus experiences considerable drag, and will therefore fall more slowly than a person inside the bus. The scenario is possible in principle - but after carefully viewing the clip and doing some calculations, I believe that the details are inaccurate. Assume the bus has a mass of 5000 kg (pretty light for a bus), and is 3 m wide by 3 m tall - so the ...


22

If the bus was in a vacuum (both inside and outside), then the passenger would float. However, the effects of air resistance on the two objects (passenger and bus) are probably not negligible in such an instance. The bus will be moving relative to the outside air, and so will be accelerating towards the ground at a rate less than $g$. If we then released ...


13

I don't think this sounds unreasonable as an estimate at all. Let's check it. One designs a building as a compromise between two competing factors: One needs all of the load bearing materials to be well mildly loaded - working in their linear region so that there is no danger of their undergoing plastic (irreversible) deformation, creeping then ...


10

SECTION A : The example in Feynman's Lectures Let a body P (Planet or Particle or whatever) moving in orbit around a center of attraction called $\:\rm{SUN}$, as in above Figure. Suppose that the attractive force $\:\mathbf{f}\left(r\right)\:$ depends continuously only on the distance $\:r\:$ of the body P from the center $\:\rm{SUN}$. Here it's not ...


6

At first, the bus and the person would accelerate at the same rate due to gravity. However, the situation is more complicated due to air resistance. The bus experiences air resistance as it falls. The person inside the bus experiences less air resistance because the air inside the bus moves with the bus. This means that the person does not experience as much ...


4

The easiest way to calculate escape velocity, is neglicting Earths rotation and assuming the object takes of in a radial direction. Then, indeed, you start from $$E = K_1 + U_1 = K_2 + U_2$$ where $K_1=\frac{mv^2}{2}$ and $U_1=- \frac{GMm}{r}$. Since the range of gravitional forces is infinity, you say (theoretically, not practically) that an object has ...


3

I am not sure what is the path $C$ you are integrating over? In your definition you evaluate $U(C)$ which in the present case of force is independent on the explicit path you choose but still depends on initial and final point, i.e. $U(p_1,p_2)$. In your final result it seems you are actually 'walking' three times the path $p_1=(-\infty,y,z)$ to ...


3

The car becomes "weightless" when the curvature of the road is sufficient that the car does not stay connected to the road. Angles don't really matter - what matters is a change in the direction of the road. As you know, an object going around in a circle needs a force $F = \frac{mv^2}{r}$ to stay in the orbit at radius $r$. Normally, when you drive over a ...


3

The point is that if $\frac{1}{2} mv^2 - GMm/r$ is constant, then $v$ only depends on $r$! This is surprising and very useful; it means that $v$ will be the same no matter what path a planet takes from some $r_1$ to $r_2$. In this case, the two paths he's using are the planet's usual elliptical orbit, and a path that goes straight toward the sun. You don't ...


3

No, no you guys (Except Floris and those who up-voted him) have missed an important observation... Look Carefully at the video again. At first the bus just tilts as the bridge bends. When the bus starts tilting (due to friction with the bridge it has not yet started falling) it has not yet obtained considerable vertical velocity. However as the man loses ...


3

Lunar soil or lunar regolith, is mostly created by meteorite and micrometeorite impacts which directly pulverize the rock, or from the ejecta from the impact. Some amount (I can't seem to find any figures) is also created from high-energy particles in solar wind causing bits of rock to spall. In theory, the bootprints would last until the soil turns over ...


3

This answer has images only (An image is equivalent to 1000 words) If you have glasses red-left / blue-right see image below in black & white 3D. and moreover a colored 3D.


3

Yes, in the situation you describe the ball will collide with the outer wall of the box. This is an example of a tidal force. Suppose you are floating right in the centre of the box (specifically at its centre of mass), then you and the box will fall at the same speed so you'll appear to be hovering weightless at the centre of the box. Suppose you now ...


2

This link explains it: The Earth experiences two high tides per day because of the difference in the Moon's gravitational field at the Earth's surface and at its center. You could say that there is a high tide on the side nearest the Moon because the Moon pulls the water away from the Earth, and a high tide on the opposite side because the Moon pulls the ...


2

They say that gravity decreases as we dig into the earth. That's an immediate consequence of an overly simplistic model of the Earth, that the Earth is of a uniform density throughout. This is very far from the case. But I also read that gravity increases for the first approx. 2000km of distance underground Actually, it's about 2900 km ...


2

If you have a planet of mass $M$, then its self-gravitational binding energy is roughly $-GM^2/2R$ give or take a small numerical factor. So, for the Earth, this would be $-2\times 10^{32}$ J. Something colliding with the Earth, which has a similar mass and size, would do so at velocities of tens of km/s at least. I think the minimum closing velocity would ...


2

First I will make a few assumptions: the hill is flat at the its top (so its surface is perpendicular to gravity); the side of the hill is flat and makes an angle $\theta$ relative to the top; the end of the hill is sufficiently far away such that it will never be reached during the "jump" sideways of the top of the hill; the tradition from the top to the ...


2

Define the origin of your coordinate system at the top of the hill with the vehicle travelling in the $+x$ direction with initial velocity $v$ and initial position $s=(0,0)$. The road then "falls away" from the vehicle at a rate of $v \tan(\theta)$, so the altitude $h$ of the vehicle above the downward sloping road can then be written as a function of time ...


2

I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ...


1

In a reference frame where the center of mass of the Earth-Sun system is at rest, we will have $$ m_\text{Sun} \vec{v}_\text{Sun} + m_\text{Earth} \vec{v}_\text{Earth} = 0 \quad \Rightarrow \quad \vec{v}_\text{Sun} = - \frac{m_\text{Earth}}{m_\text{Sun}} \vec{v}_\text{Earth}. $$ In particular, if this is true at some initial time, then it's true at all ...


1

As Tom mentioned, the normal force has changed. Draw a free body diagram. You should have three forces displayed, assuming no horizontal influences, such as friction. Sintetico discusses horizontal motion below. Considering vertical motion only, as you said, the body is still at rest (and not accelerating), thus $\vec{a}=0$. Newton's second law then says ...


1

When a rocket is fired from Earth with a sudden impulse, its total energy is given by: $$E_k \text{ (kinetic energy)} + E_p \text{ (potential energy)}= \frac{1}{2}mv^2 - \frac{GMm}{r} = constant$$ The potential energy here is taken to be negative because the reference point chosen for potential energy to be zero is when the rocket is unbound in Earth's ...


1

When a rocket is fired from Earth with a sudden impulse, its total energy is given by: $$E_k \text{ (kinetic energy)} + E_p \text{ (potential energy)}= \frac{1}{2}mv^2 - \frac{GMm}{r} = constant$$ where the variables have their usual meaning. The potential energy here is taken to be negative because the reference point chosen for potential energy to be zero ...


1

First of all let's study an imaginary system where both the bus and the person are not subject to drag forces due to the air: If the person is not bounded to anything he will be subject to free falling and thus to a uniform acceleration $g$. Also the bus will be free falling and thus they fall together with the same velocity. If we take the drag forces into ...


1

Anything over 500 miles in diameter, give or take is almost always sphere-shaped, the primary variation being rotation speed, which can give a flatness to the object, for example, Jupiter is visibly flattened by it's high rotational speed. The problem with building a strange shape by very large collision is that the heat generated in a collision of that ...


1

As stated above, your linear calculation is correct and your assumption on compression is correct too. I can try to give a rough answer to the shrinkage. Lets start by looking at pressure. http://cseligman.com/text/planets/internalpressure.htm In simple terms, the pressure is the weight above you, over the surface area, which would be linear, but ...


1

Let us make an estimate. Let the skyscraper be 400 m tall, each storey 4 m high, 500 people per storey, 20 m2 per person, 10000 m2 per storey, let us assume that the building is a 100x100 m2 square in the plan and that it only has 10 cm thick structural walls in a 25x25m2 grid. So the cross-section area of the structural walls is 2x5x100x0.1 m2=100 m2. Let ...


1

Young's modulus is the ratio of tensile stress to tensile strain for a material: E = (F/A)/(∆L/L) = (F * L) / (A * ∆L) F/A is force per area, and (∆L/L) is change in length per original length For structural steel, Young's modulus is 200 gigapascals. This quantity can be used to predict how much the steel will compress under a given weight per unit area. ...


1

The reason for this is that, in your notation, $W=C_1-C_2$. This means that the work performed in moving about a circuit from point $A$ to point $B$ along curve $\mathcal C_1$ and then back along $\mathcal C_2$ is exactly the difference between the work performed in moving the particle along $\mathcal C_1$ versus moving it along $\mathcal C_2$. This is ...


1

Starting from the work around a close curve $$W = \oint \vec F(r)d\vec s = 0$$ now be stokes throem: $$ \oint \vec F(r)d\vec s = \iint_A \vec \nabla \times\vec F(r)d\vec A = 0$$ where $A$ is the area surrounded by the closed curve. Now if you do the math you will see that $ \vec \nabla \times \vec \nabla U( r) = 0 $ for any field $U( r)$ thus we can ...



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