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24

No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...


5

The tension is not the same for all parts of the rope. If the tension is $T_1$ between the hands and $T_2$ between the top hand and the ceiling then. $$ T_2 = T_1 + F_2 \\ T_1 = F_1 $$ where $F_1+F_2=W$ are the forces acted upon the arms. You arrive at this if you make two free body diagrams, one at each hand. The result is that $T_2 = W$ and ...


3

I have Marion-Thornton 4th ed. around here somewhere. It is an older book and presents some material differently than we are used to in more modern books (for instance they even use the old imaginary time method when discussing some things in special relativity, which I personally dislike). However I agree with DanielSank, different pedagogy does not equal ...


3

What @lemon said is right. While $\bf F = -\nabla \phi\left({\bf r}\right)$ suggests that $\bf F$ depends on $\bf r$, it ultimately depends on time through ${\bf r} = {\bf r}\left(t\right)$. That is, ${\bf F} = -\nabla \phi\left({\bf r}\left(t\right)\right)$. So, there is no inconsistency in writing Newton's second law as (a dot represents a time derivative) ...


3

The perpendicular force is no longer the weight of the box, you need to consider the force by the two workers in the perpendicular direction as well.


3

No. Not necessarily. A common demonstration done in the physics lab is suspending a heavy object by a string, then below that a string of slightly larger diameter (and strength). If the lower string is pulled down slowly, the thinner string above will break. But if the string is pulled quickly, that string will break (the stronger link) Why? Because the ...


3

It is true, although the details depend on the chain. One simple model is as follows: the chain is composed of links that are solid until the force on them exceeds a threshold, $T_\text{breaking}$, the breaking tension. Once this happens, the link comes apart completely and the chain is broken. This threshold can be different for the different links. When ...


3

Yes you are right. The 10kg piston acts as a force over the area of the piston, increasing the pressure and decreasing the volume of the gas inside. When the set up is tilted, the force no longer acts on the gas, but sideways, so the pressure equalizes.


3

In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ...


3

Newton law of gravitation is given by: $$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$ The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so: $$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822$$ Hence, the equation simplifies to $$F =(9.822) ...


2

There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force. The structure of field which causes ...


2

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles. Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p ...


2

I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it. Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground: The spring clearly has work done on it because its kinetic energy increases and ...


2

You have an incorrect understanding of forces, accelerations and velocities. A force of 1 Newton can accelerate a 1 kg mass at a rate of 1 m/s/s (often written as 1 m/s$^2$). A force of 2 Newtons can accelerate the same 1 kg mass at a rate of 2 m/s/s. We relate the accelerations, velocities and positions via the kinematic equations; for this case we can use ...


2

1 Newton can move 1 kilogram 1m But, can 2 Newtons move 1 kilogram 2 meters? Is this because 1 Newton = acceleration of 1/ms, so, 2 Newtons would result in a speed of 2m/s? Or would it be 1m/s lasting twice as long? So what does a joule do: accelerate or just move? A joule does not accelerate, it is the unit, the measure both of ...


2

The logic is simple: a body loses energy because opposing forces cancels themselves out. if you want to move in a direction +x against an opposing force -x the net balance must be in the direction +x. The logic is the one that rules the composition of forces or vector addition If $F_2 > F_1 = -18 N$ then the body will move, accelerate in the -x ...


2

Option 4, none of the above. Your option 1 is wrong because points don't rotate. Your option 2 is closer to correct, but ultimately still wrong. You're overly hung up on points (the origin). It might help to get a handle on what "rotation" is. Points don't rotate. Better said, a rotated point is indistinguishable from the original. What about one ...


2

Assume that the wind hits the panel normally with speed V and is then stopped (or escapes along the panel). That's not the best of assumptions. The air will flow around the plate. Only in the center of the plate will it hit the panel normally and stop. Your analysis does however capture some of the key dynamics in that the drag force is proportional to ...


2

The weight vector, $W=mg$, can be split up into two components, $mg\sin\theta$ and $mg\cos\theta$, where $\theta = 30$ in your case. The $mg\sin\theta$ is the force vector which would generate a torque, $\tau = \bf{L}\times{}mg\sin\theta$.


2

Any force greater than zero can stop the car. Only it will take longer and the distance moved by it by the time it stops also will be greater. If the force is larger these parameters (time to stop and distance traveled before stopping) will decreasing. Theoretically, infinite force is required to stop it instantaneously.


1

It appears to be rolling friction. Even on the Moon, a wheeled vehicle will eventually come to rest because of friction at the axle and friction between the wheel and the ground. The wheels on that vehicle are deformable. That means that the reaction force by the ground is not purely vertical. It has a horizontal component that resists the rolling motion of ...


1

At 1:02 in that video, you see the "people" at the end of the swing do something like a half revolution with a radius of 15 meter in 0.3 seconds (rough estimate of dimensions and time - feel free to come up with your own estimates by counting frames etc - then substitute those numbers in the equations below). The centrifugal acceleration is given by $$a = ...


1

Yes, alternatively you can use that the potential energy of the spring is transformed into kinetic energy of the object. This is simpler than considering the work $L$ done by the spring. The result is however the same: $$ U=E_{kin}\\ \Rightarrow \frac{1}{2} k x^2=\frac{1}{2}m v^2\\ \Rightarrow v=x\sqrt\frac{k}{m} $$


1

This page has a helpful summary of the history--it seems he initially accepted the Aristotelian idea that objects could only continue to move if some "force" inside them was moving them (keep in mind this is before his technical definition of 'force'), and it took him a while to switch to the idea that bodies naturally tend to keep moving unless acted on by ...


1

See the wiki page . There is no difference between the two . Weight of a body of mass $M$ is $M.g$ which is equal to gravitational force on the body.


1

The localized vector is a vector which we know its magnitude and direction and its point of application , which is the point where the force acts , if the force were free vector , how you will calculate for example the moment ? that won't make any sense ok the second principle i found it in Vector mechanics book called PRINCIPLE OF TRANSMISSIBILITY. check ...


1

The ideal shape for the water to just 'hang' is actually the cube that you've proposed. In this case all of the forces on the water are uniform across the interface. For one region to slip down, another needs to move up (as you've indicated in your diagram). But since every location is experiencing the same forces, there's no reason any spot to start ...


1

If multiple forces act on a body at the same time, you should compute the net force before considering whether it is increasing or decreasing the kinetic energy of the object, and thus whether the work done by the force is positive or negative. So, using your example, the net force is $F = F_1 + F_2$, and since $F_1 > 0$, $F_2 < 0$ and $|F_2|< F_1$ ...


1

When two bodies interact, there is a force between them. Positive work is done on the object for which the dot product of force and velocity is positive. It follows that negative work is done on the object for which the dot product is negative. By Newton's law (for each action there is an equal and opposite reaction), the force on one object is the reverse ...


1

Force represents an energy transfered (and applied in general in a straight line). On the other hand Torque represents a force acting on a point on a straight line but the effects are applied elsewhere (in rotating the body). As such there is (at least) this subtle difference. The force is applied to a point of the body which then (by internal constraints ...



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