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61

What is important for tidal forces is not the absolute gravity, but the differential gravity across the planet, that is, how different the gravity is at a point on the surface near the sun relative to a point on the other side. Because the Sun is far away, gravity doesn't change much between the two extremes on earth. However, if you compare it with the ...


45

To deal with this type of problem, you must be careful to define exactly what system you are dealing with, and then not change that system part way through the problem. This definition allows you to be very clear about whether the "system" has any external forces acting, and thus whether the momentum of the system is constant or not. In this case, you seem ...


35

Tides are caused by the gradient of the gravitational field - so the tidal "force" experienced drops with the third power of the distance. This means that the relative strength of the tides should go as $$ratio = \frac{M_{moon} \cdot D_{sun}^3}{M_{sun} \cdot D_{moon}^3}\\ =\frac{7 \cdot 10^{22}\cdot (1.5 \cdot 10^{11})^3}{2\cdot 10^{30}\cdot (3.7\cdot ...


34

The buoyant force on a body immersed in a fluid is equal to the weight of the fluid it displaces. In other words, $$ F_B = \rho_{\text{fluid}} V_{\text{body}} ~g $$ The force of gravity on the body is equal to $$ F_g = m_{\rm body} ~g $$ The apparent weight of this body will therefore be equal to the sum of these two forces. $$ W_{\rm app} = \rho_{\rm ...


32

When the raindrops hit the wagon's surface, they aren't moving relative to the tracks. Friction is required to accelerate the raindrops to the wagon's speed. By Newton's third law, there must therefore be a reaction force on the wagon surface by the raindrops.


29

Look at it this way: Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball ...


28

Just my two cents to complement the other answers. The mistake in your reasoning is that: the state above (where the top touches the bottom) is equivalent to having a box like A (just a box holding a vacuum). is incorrect, it rather equivalent to a box full of air (there is air in between the walls, regardeless of the vertical position of the top). In ...


16

Work is calculated as force times distance. $$W = Fd$$ The purpose of a simple machine like a screw jack is to lessen the force required. However, the work needed is still the same, so the distance over which you exert the force has to increase. Halving the force requires doubling the distance. In this problem, you want to lift 2000 lbs a distance of 1 ...


13

1) Technically while you push it down it will cause an increase in pressure (so the weight will change) but assuming the box has a hole and the air can equalise then the weight of your initial and end state will be the same. This is because the air is equalised within the box at all stages and at the end state the air that was in the box is now above it. ...


13

Yes, unfortunately. Because of the equivalence of inertial reference frames, the the physical laws are the same in both reference frames. However, another possibility, which is non abelian, is that instead of feeling the same amount of pain, you could be feeling the opposite amount of pleasure. It depends if pain (X) are fermions or bosons, that is, if ...


11

We start by noting that force is the rate of change of momentum. Let's suppose you and I are floating in space (so we are the only two interacting bodies) and you're pushing me so I feel a force $F_{me}$, then: $$ F_{me} = \frac{dp_{me}}{dt} $$ where $p_{me}$ is my momentum. But we know that momentum is conserved, so since you are the only thing ...


11

As stated in other answers it is how much the gravitational force is different by on opposite sides of the earth that creates the tides. You can still show this using $a=\frac{GM}{d^2}$ but you need to consider the difference, not the absolute force on the earth. The sun while much more massive is just far enough away that it is getting to a much flatter ...


9

You will be punching the feather with a really small force. That doesn't mean your arm is punching lightly. Your arm can have a lot of momentum and internal tension due to internal forces that makes you punch hard, but the actual force is defined on the interaction with another object, in this case: the feather, so the force you apply to the feather is ...


9

Suppose you and the table are floating in space. If you push the table it will go in one direction and you will go in the other direction. Your and the table's acceleration will be different so you will end up travelling at different speeds. This is obvious from conservation of momentum. The momentum in the centre of mass frame is initially zero, so after ...


9

The mass of the cart is changing! This is the variable-mass system, which says, $$ F_{ext}+v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ where $v_{rel}$ is the relative velocity of the escaping/entering mass. In your case, there are no external forces so, $$ v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ So the change of velocity comes from the change in the mass.


8

Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation) Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa.


8

On the quantum level, force is not acceleration. The concept of "fictitious force" makes no sense on a QFT level, because forces are interactions between quantum states, not the classical forces you might imagine. Quantum forces are not vector fields in space. The notion of "fictitious force" would mean that, e.g., the strong force is something influencing ...


8

The classical theory of electrodynamics can indeed be written as a geometrical theory in a similar way to general relativity. As it happens there is a question and answer addressing just this, but it's in the Maths SE: Electrodynamics in general spacetime. Classical electrodynamics is an example of a class of theories called classical Yang-Mills gauge ...


8

The highly upvoted answer is right but to make things much simpler: Tides are based on the change in gravity, not the gravity. That means they drop off at the cube of the distance rather than the square of the distance like gravity itself does. Thus the object with the most gravity isn't necessarily the one that causes the most tide.


7

Let me highlight lhree points: 1) The mass of a box is the sum of the mass of the box structure, and the mass of the box contents; 2) The force of gravity downward on the box depends only on the mass; (and the local acceleration of gravity) 3) The buoyant force upward depends only on the density of the surrounding medium and the volume of the box. So, ...


7

The reason behind buoyancy is the pressure difference in fluids. More specifically, the difference in hydrostatic pressure on different levels, since the hydrostatic pressure of water increases with depth ($p=\rho gh$ wher $h$ is the distance below the surface). The part of the body which is subject to higher hydrostatic pressure will be pushed more upwards ...


6

A box filled with helium would weigh less than a box filled with air, because helium is less dense than air. A rigid box containing a vacuum would weigh even less than the same box filled with helium, because it is even less dense.


6

Just consider the same situation in water. A box filled with water, immersed in water will have more weight on a weighing machine than an empty box (with vacuum or air)--which might even float depending on the mass and volume of the box. The only difference in this case is the density of the medium.


6

Suppose a glass could sustain 100N force and that my muscles can exert up to 200N force: if I went all out, I couldn't punch the glass with a 200N force because the glass would break, which means it's not able to apply a 200N force on me. I apply F = 200N and the reaction is only f = 100N. The glass table-top in the picture can support a weight ...


6

How can the work energy theorem be valid in presence of non-conservative forces since conservation of energy is not there? The work-energy principle simply states that work is the net increase of KE The principle of work and kinetic energy (also known as the work-energy principle) states that the work done by all forces acting on a particle ...


6

Solid water (ice) is one of the few known solids whose density is lower than that of its liquid form. Yes water is special! (but very much so in its chemical properties too) Due to the crystal structure of the solid phase of water, the molecules arrange themselves in a rigid, ordered fashion and end up being, on average, farther apart from each other (than ...


6

Actually, your book is correct. Even if the most usual uses of angular momentum involve circular or rotating motion, this is not the general case. An object moving in a straight line has angular momentum in a reference frame in which the origin does not fall on the the line. To see this simply remember the definition of angular momentum ...


5

Basically, it has to do with the density of the material as a function of temperature. The density of iron increases as it cools, that is, solid iron is more 'packed tight' than when it is melted. This is understandable, since the kinetic energy of the iron atoms decreases as the temperature drops (ie: the average velocity of the atoms decreases), allowing ...


5

What would happen if axis of rotation pass through centre of mass of an object? Will the object rotate when we will apply force to the object? For the sake of future readers, I'll reply to the original question: Every body has a Centre of Mass, whatever its form. If an object has a regular shape and uniform, homogeneous distribution of mass its ...


5

When the block started to compress the spring, the restoring force was opposing its movement. And the spring ought to have kinetic energy otherwise the momentum of the spring-block system wouldn't be conserved. Ok, if I say the wall can't be neglected, then ok, if it had moved due to the force by the block, hadn't the KE of the block gone to the KE ...



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