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39

You've caught a non-intuitive part of Newton's 3rd law. It's actually applying in the case you mention, but because the objects involved are of dissimilar hardness it's easy to perceive the impact as a violation of the law. Impacts are actually really complicated. Consider this slow motion video of a punch to the gut. We won't be able to cover all of the ...


10

What makes you think that the maximum force you applied to the dry wall was anything like the maximum force you applied to the brick? It certainly wasn't. The dry wall gave way much before you were able to attain the same force as applied to the brick. Try punching the air and see how much force you are able to apply. The experimental evidence that the ...


8

There is no doubt the Newton's third law holds in this case. The source of confusion is the fact that you are neglecting the time interval of the collision as well as the momentum change the colliding body. As we shall see it is incorrect to assume you applied the same force in both cases just because you started with the same initial conditions, i.e. the ...


7

The gravitational potential field can be found by a full volumetric integration from the overall volume, or planet, or whatever: $$ \Phi(\mathbf r) = -G\int_V\frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|}dV $$ This just comes from having a distribution for $M$ in the potential formula: $$ \Phi = \frac{GM}{r} $$ Also, its easy to see that, if you are ...


7

TL;DR: The physics of hitting things are not as easy as exerting a constant force on something. What I am trying to say with that is that Newton's law of course applies, but it would be more obvious to see it if you were just pushing/leaning against the wall with your weight. Then I'd say the two walls probably feel roughly the same. So what is different ...


6

You are correct. The car has no net force on its environment, and the environment has no net force on the car. This is true of any object traveling with a constant velocity. This is even true in the vertical direction. There is a force of gravity pulling down on the car, and there is a force caused by the road pushing up on the car. If the car is not ...


5

Recall the difference between the weak and strong Newton's third law, cf. e.g. this Phys.SE post. If the internal forces satisfy the weak Newton's third law (but not the strong Newton's third law, i.e. without the collinarity assumption), then it is not guaranteed that the internal forces do no work, cf. e.g. Fig. 1. ^ F | | 2 x-----------...


5

Car collision "damage" usually goes with the energy in the zero momentum frame. In both cases that is (since in the zero momentum frame, the two cases are equivalent, assuming the masses of the cars are equal): $$E_1 = 2 \times \frac{1}{2} m v_{rel}^2 = m \left( 30 \frac{km}{h} \right)^2$$ Therefore a priori there is no difference between the two situations....


4

It can't fall slower as the first cosmical speed (7.8 km/s), which is still very high. Although it would cause much smaller destruction as it would hit directly with the mean speed of the meteors (10-70km/s). The lower angle of the hit doesn't play a significant role, because considering its mass, the interaction with the atmosphere will be probably ...


4

If a body is moving, this doesn't mean that a net force certainly must be exerted on it. It can move without any net force (First Law of Newton). You might say "How that body has started its motion without any net fore?" The answer is: "Equations of motion are moment equations, i.e. they are stated for moments not for a time interval ($F(t_0)=ma(t_0)$). So ...


4

The car can be moved provided it isn't on a frictionless surface, or more precisely the centre of mass of the car/occupant system can be moved provided the car isn't on a frictionless surface. As you say in the question, assuming no external forces are acting then because the momentum of the car + occupant is conserved the occupant cannot move the centre of ...


3

The force the fluid does on the bottom piece does not depend on the height of the water column of the reservoir. It does depend on the height $h$ of the water column in the plate. This can be easily seen by the fact the water is static so the pressure at any horizontal plane is the same. The pressures in $a$ and $b$ are the same. The force of the water on ...


3

You are correct in your definition of force. A car, not accelerating, has zero net force associated with it. However, if the car were to hit something--let's say it's me standing in the middle of the street--it would exert a net force on me, and by Newton's Third Law experience a net force equal in magnitude and opposite in direction. So how can an object ...


3

All you need to do is calculate the perigee distance $r_p$ that is the distance of closest approach. Then if $r_p < R_A$ your satellite will crash and burn. Once again we start from the vis-viva equation: $$ v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) \tag{1} $$ The parameter $a$ is the semi-major axis of the ellipse, and it is related to the ...


2

The motion (perpendicular to the wind) is called Ekman Transport. It results from a combination of the wind and the Coriolis Force (due to the rotation of the Earth), which together produce an underwater motion called the Ekman Spiral. Icebergs, which extend large distances below the surface, experience forces from the whole depth of the spiral. The ...


2

When you move the pen at constant velocity, you are correct that you apply $F=\mu_kmg$. This is true regardless of what constant velocity you choose. The difference is you have to supply more force initially to accelerate the pen to a faster speed, but once it is up to speed you only need $\mu_kmg$ to keep it at that speed. So it really isn't the force ...


2

The more fundamental thing to understand are the conservation laws, particularly the conservation of momentum and of energy. The availability of energy gives rise to force. When you move your fist towards an object at a particular velocity you contain within it a kinetic energy and momentum. Materials are held together with binding forces that, at the ...


2

This is a two-body problem with the center of mass at rest, much like a stationary nucleus emitting an alpha particle. Therefore if the person were to move in one direction the car would go in the opposite direction.


1

Newtons laws are a good approximation for how the world works when the velocity is less than 1% the speed of light, the gravity isn't too strong, when the number of elementary particles an object is composed of isn't too small as an object needs to be large enough for quantum uncertainty to be insignificant, and the amount of space is small compared to the ...


1

Regardless of relativistic effects: Newton's First Law is invalid because friction exists in real life. False, the first law talks about the case when no forces are present, if forces are present go to the second law. Newton's second law is invalid due to the same reasons. False, you add friction to the total force. Newton's ...


1

Newton's laws are valid for all situations where velocities are small (compared to the speed of light, ie relativity is not important) and where quantum effects are negligible (mostly where objects are much bigger than elementary particles). The problem with your argument is that you and your friend are using idealized expressions for Newton's laws, not ...


1

In SI units: $1~\mathrm{N} ~\mathrm{m}=\mathrm{1~kg~ m^2 ~s^{-2}} $ and $1~\mathrm{J}=\mathrm{1~kg~ m^2 ~s^{-2}} $. So the units Newtonmeter and Joule are the same in SI-Units and there dimensions are actually equal to. This is a fact and just a matter of definition. Work along a curve $C$ is $W=\int_C \vec{F} \cdot d\vec{s}$. So in the simplest case where ...


1

seeing the person and the car as one system you can say that the center of mass always stays at the same point if no external force is applied. However, if the person jumps in one direction inside the car, the car will move in the other direction. The center of mass stays at the same position. Of course the floor applies an external force to the system and ...


1

Assuming no air resistance, the only force acting on the satellite is the force of gravity, $\bf{F_g}$. The stated circular orbit also means that centripetal force, $\bf{F_c}$ is involved, which causes the satellite to follow the circular path. Due to this, the centripetal acceleration can be directly equated to g, which allows you to immediately solve for ...


1

If we are dealing with the gravitational field of the earth, then because of the inverse square law, you should, in theory, need a tiny bit less force to raise an object from 50 m to 100 m, as you do lifting it from ground level to 50 m. But this difference in force is minute, I don't know offhand if we have instruments to measure it. But you can work it ...


1

I thought about this in the molecule level, but then if molecules are at constant velocity then there is no "pushing" force that they apply on each other. To clarify my misunderstanding - I imagined the system at constant velocity at molecule level as if it was accelerating so molecules of A push molecules of B and also the opposite. If we are speaking about ...


1

If the blocks initial velocities are zero (i.e. the blocks start to move from rest), then it is impossible for block B to move with constant velocity. Because the only horizontal force acting on it (if there was) is friction force due to block A. We have: $$\Sigma F_B=m_Ba_B$$ If block B moves with constant velocity, then $a_B=0$ Thus, friction force acting ...


1

There is extensive (100s of articles) lit. on the web regarding atmospheric drag and drag in general. A spherical bob and a very thin rod makes the problem very simple. One may use this graph: http://eis.bris.ac.uk/~memag/Teaching/Multi/dragcurve.pdf , and then calculated the Reynolds number to find the drag force using the formula given by the first ...


1

You have two questions. Here is the answer to your first question, "So why stone does not come back/reach to the centre of circle?" The answer is that the stone is accelerating toward the exact center of the circle. Velocity is composed of speed AND direction. As you noted, the velocity is always changing, however the speed is not. This is related to your ...


1

The above isn't drawn correctly. And it is far too complicated in my opinion. I think I was able to solve this using high school physics: This bugged me a long time. I am surprised no one ever posted an answer on the internet. The length of his rope was 60.8 meters, assuming a drop above the target building of about 12 meters. The difference in height ...



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