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0

Yes indeed, all what i know is that the Poisson's equation is a consequence of Gauss's law But i'm having a trouble to show that the flux over a closed surface is independent of the surface itself. Thanks for responding


-1

The answer is no. The net force at the maximum elongation points has the same magnitude. This is because the rest point of the spring is modified by the gravitational weight of the mass. The mass oscillates around this new rest point, and at the points of maximum amplitude the net force is the same. One can tell that the gravitational force can be ...


2

In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at $-A$ and $+A$.


5

The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a $\Delta x$ amount with $F_{spring}(0) = k\Delta x$. So the spring force on point A is a little smaller than in point -A, since $ F_{spring}(A) = -k(A-\Delta x)$ and $ F_{spring}(-A) = k(A+\Delta x)$ so it compensates the "extra" force. You have ...


0

yes it would, because of the moons atmosphere, even the craters have not faded away so the footsteps would not fade away. Best type of legacy one can leave


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 ...


0

Couple of things to point out here, and I'm not sure whether either of those answer your question, so here are some ideas to help you think about the problem and guide you towards the full answer: The lowest possible value of Earth's gravitational potential is met: a)if the object is at Earth's exact center (r=0, namely there is no mass "below" him (m1=0), ...


1

The equation you cited makes use of "big G", the universal gravitational constant. Generally, this equation is used if you want to calculate the attractive force between two bodies, such the moon and the Earth, or a satellite and the Earth, or the Earth and the Sun, or if you want to calculate escape velocity from he Earth's gravitational field. The center ...


0

We treat a mass producing gravity as if all the mass is concentrated at the centre. Does that make sense to you? Also potential energy values are relative, it is only the difference in two values of P.E. that counts, not any absolute value. But anything placed on the surface of earth will never fall any deeper into the earth. We can assign a value of ...


1

If the slope is truly frictionless, then the ball will never stop moving. You are assuming that after passing through the V at the bottom, it will roll back up the other slope. But It is not rolling but sliding (no friction - no torque to make the ball roll) When it hits the V, it will bounce - so it will lose contact with the surface Exactly how it ...


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 ...


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

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

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 ...


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 ...


0

The local gravitational field increases (slightly) as you descend under the earth because much of the earths mass is in the core, and you're getting closer to the core. As the moon has a smaller core, this effect would be reduced. Even on earth, the effect is not really noticeable at depths we can dig to. So the effect may not be present on the moon, ...


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 ...


0

When you breath in helium, you gain mass, volume, and buoyancy, but lose density and weight.


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.


1

$dx$ is measured along the $OP$ axis and $ds$ is measured along the surface. The relationship between them follows from simple geometry - $ds$ is at an angle $\pi/2-\theta$ to $dx$. From this diagram, $\frac{dx}{ds}=\sin\theta$. The result follows from rearranging. $ds$ is the width of a strip of the shell of constant surface density: $dx$ is the ...


1

You're right that the kinetic energy of the spacecraft is the same both before and after the planetary encounter—in the reference frame of the planet (or, technically, the frame of the planet-spacecraft CM.) But the fact that the kinetic energy is the same before & after in one frame does not mean that the kinetic energy will be the same before & ...


1

Feynman's trajectory The trajectory discussed by Feynman is shown below in red for the blue path, which is a hyperbolic deflection of a small particle around a large star centered at $(0, 0)$. Discussion Feynman's trajectory here trying to answer the question: how much has the speed increased between A and B. He is answering that by saying that there 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 ...


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 ...


0

If a mass M is at a point travelling at a given velocity (= speed and direction vector) then it has a given amount of energy (potential and kinetic energy summed) relative to being stationary* at the same point with respect to a given frame of reference. ie solely by knowing position and velocity vector the PE & KE are defined. (* or at some zero ...


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 ...


0

Here is what I think he means: first we have a planet going around the sun in some orbit, then we change the direction of the velocity to go radially outwards, for example by letting the planet go inside some pipe we put in it's path (notice that a normal planet would never do this, because there are no big pipes in space and also there would be quite a lot ...


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 ...


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 ...


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 ...


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 ...


-1

Your calculations show that just separately computed the components of the net force on 2. Nevertheless, force is a vector, and your final answer should read so. Generally speaking, the gravitational force on $m_1$ due to $m_2$, located at positions $\vec r_1$ and $\vec r_2$ respectively, is \begin{equation} \vec F_{1,2} = \frac{Gm_1m_2(\vec r_2- \vec ...


0

It seems you pretty well understand how to find the pairwise forces and you are just having trouble adding them to get the net force. Since the forces are vectors, you need to add them like vectors. I assume this is difficult to imagine because you are just learning what a force is and you are just learning what vector addition is. So think about something ...


0

It means that he might as well be in outer space. He will float weightless in the air. If he jumps toward the center of the earth he will gradually slow down and stop due to air friction, and will then be trapped, unable to move in any direction. Unless he drank a lot of fluids before he jumped.


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 ...


-1

It just oblique projectile motion U= speed of truck h= height of hill If you want to calculate how far from hill the truck land after flight use U multiply by the whole root of 2H/g g is gravitational acceleration =10m/s2


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 ...


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 ...


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

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 ...


0

When a body is kept in ground, and is at rest, the downward gravitational pull is balanced exactly by the Normal Reaction if we assume the earth to be an inertial frame. This is not correct. The normal force is a constraint force that acts in one direction (upward). Suppose you exert an upward force of five newtons on an object that weighs eight newtons ...


0

What you've proven is that the gravitational field along the axis of a uniform rod is in fact not proportional to $1/r^2$, and diverges as you approach the end of the rod. Both of these are true statements, so well done! But you seem confused about these answers, so I should probably elucidate a bit more. The shell theorem (for spherically symmetric ...


0

I think the apparent paradox is driven by a misunderstanding of the very nature of reaction forces. A reaction force is a force $F_N$ perpendicular to a surface which counterbalance the gravitational pull, or any other force perpendicular to the surface. This is because reaction force are basically what prevent two bodies to occupy the same volume, and their ...


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 ...


0

You need to draw yourself a diagram to get the correct mass of the "slice" in your expression, and get the correct distance. There are two problems with the way you defined your slice: although it is of constant width (in the x direction) $dx$, the distance is not simply $(x-r)$ because most of the material is off axis and thus further away (so you need a ...


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 ...


0

Even without gravity, momentum conservation will still hold. If you elastically scatter an unknown mass $m$ with an initial (known) velocity $v$ against a known mass at rest, for instance we can take the SI standard of $1$ Kg, then from the resulting measured velocities you should be able to find $m$. Thus Yes, there will still be a mass.


0

This scene was clearly shot with a stationary bus oriented (nearly) nose down and a green screen. A backpack and book are clearly dropped on the rider and accelerate at the normal rates. The rider hits seats and can't hold them as he falls. There is no way that air resistance on Earth could provide enough force to change the movement of the bus (and ...


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 ...



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