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-1

You are correct. By F=ma You know that Force is directly proportional to Acceleration . If force is increased 2 times the acceleration WILL increase 2 times Your answer is correct! It will increase as F increases!


1

An object will not accelerate without an non-zero net force. If you accelerate an object with a constant force, and you keep applying that force, then the object will keep accelerating at the same rate. The answer is twice the force. $$a = \dfrac{F}{m}$$ $$2a = \dfrac{2F}{m}$$


0

Since the length of the rope must remain constant the trees must bend towards the center on both sides . If the mass of the person is M , then by balancing forces in vertical Mg = 2.T.sin(theta) : theta being angle ~20 degrees


0

To compute $f(r)$ you have to use Newton's second law which (here) has the form : $$m \left[\left(\ddot{r} - r\dot\phi^2 \right) \vec{e}_r + \left( r \ddot \phi + 2\dot r \dot\phi\right) \vec{e}_\phi\right]= f(r) \vec{e}_r .\tag{1}$$ and simplify this equation. Your $\ddot{r}$ is wrong. To get $f(u)$ use : $$u = \dfrac{1}{r} \Leftrightarrow r = ...


1

Balance torques around the corner of the step, so r x cos(38.7) x mg= F x 0.3. F = 114 N


0

In doing a force analysis on the system, you can note that the tension in the rope at this point is essentially the force which pulls the far half of the rope. So it may be helpful to consider the force as $T=m_{half \, rope} a$. And we can calculate the mass of part of the rope as: $$ \int_{x_1}^{x_2} \lambda(x) dx = m|_{x_1}^{x_2} $$ The only thing we ...


0

When the pseudoscalar invariant $\vec E \cdot \vec B$ is zero, we have three cases. If $E^2<c^2B^2$ then you can switch to a frame moving with speed $E/B$ in a direction mutually orthogonal to $\vec E$ and $\vec B$ where there is no electric field in the new frame. Solve in the new frame. Then bring it back to the original frame. If $E^2>c^2B^2$ then ...


5

A force is defined as a change in momentum over time. In the Newtonian limit, this means a mass times an acceleration. But when dealing with things like photons, the formal definition is applied. Photons have no mass, only momentum. Therefore, if a force is applied to them, their momentum can be changed. This can happen in two important ways, a force can ...


0

When you have a block with a mass of 10 kg, the normal force is approximately 98 N. If the block starts out stationary, it will continue to be stationary until you apply sufficient force to exceed the static friction. At that point it will start to slide, and will continue sliding until the force drops below the force of dynamic friction. For the ...


0

What about the car when there is no force applied to the block? The block will experience no frictional force, so when dealing with static friction, we have to say $f_s \geq \mu_s n$, since there are instances when the frictional force may not act at all or may at at a reduced capacity. So here, $f_s = 18N$, since that is the only force necessary to keep the ...


0

It is the net force, not force.


1

This type of question keeps reccuring. I suppose either teachers don't teach this correctly, or students do not pay attention in class. $F=m a$ is really $$\sum F = m a_{cm}$$ These two distinctions (the sum, and the acceleration of the center of mass) make all the difference in the world (At least read on Newton's Laws of motion). You push on the rock and ...


1

The main thing is that the total force you are applying on the body is not enough to move the rock.This means that the total force on the box is zero because the force of static friction is grater than that of your applied force,which cancels out the effect your force.As your forces increases the static friction also increases until a point comes when you ...


1

Force = mass * acceleration is the basic simplified version of the equation. There are more complected formulas available for this that take into account more complicated scenarios; like taking in the account of different forces coming from different directions like in your scenario. sum(forces) = mass * acceleration force = the derivative of its linear ...


0

Of course the force changes during the impact - so to get close to an answer, you need both the time of the impact and the magnitude of the momentum transfer. As user77567 pointed out, a fairly simple way to measure momentum transfer is with a ballistic pendulum. This would be a heavy steel ball (much heavier than the hammer) hung from a long wire. When you ...


14

The confusion comes from how you have written the equation. If you write it like this $$F_{net} = ma$$ it will be easier to see your error. You are exerting a force on the boulder, but net force is zero. This means that other forces such as friction are canceling your force out. In this case. Friction equals applied force.


2

(i) Roger Shawyer Shawyer's output seems to be mostly available on emdrive.com. Among the theoretical explanations he provides there are A Note on the Principles of EmDrive force measurement Principle of Operation Theory paper None of these appear to be peer-reviewed. (ii) NWPU group Applying Method of Reference 2 to Effectively Calculating ...


1

Because observations made by physicists have found that this is what nature does.


0

There are many different levels of explanation for this question. Strangely enough most of them will dive into quantum electrodynamics, Feynman diagrams and exchange of virtual photons... I will try a simpler path that still carries some explanation. When you put two charges at a distance, they deform the -- otherwise flat -- electromagnetic (EM) potential ...


1

you can draw feynman digrams and then calculate scattering amplitudes and it is in the non relativistic limit is proportinal to potential.so if the potential is positive it means they repel. this sort of claculation is done in peskin book and A.Zee book.in peskin book page no 125. this is the most rigorous work to prove gravity is always attractive. by ...


1

Firstly, I would like to say that there is no particular terminal separation between negative charges and positive charges. Actually you will understand it better if I would clarify in this way that scientists first saw that having even follow the same statistical distribution i.e. Fermi Dirac distribution some of them actually repel others and some do ...


0

The fastest possible way to do a lane change is to fully steer one way on the traction limit and then steer on the opposite way again on the traction limit. The traction limit is $\mu g = \frac{v^2}{r}$ where $g=9.81\;{\rm m/s^2}$ is gravity, $\mu=0.8\ldots0.9$ is the coefficient of friction (half it in the rain), $v$ is the speed in meters per second and ...


4

No. The easiest way to see this without invoking rotating reference frames is to write out Newton's Law's in polar coordinates, which work out to be: \begin{align*} F_r &= m \ddot{r} - m r \dot{\phi}^2 \\ F_\phi &= m r \ddot{\phi} + 2 m \dot{r} \dot{\phi} \end{align*} From these, it's pretty easy to see that if we have $F_\phi = 0$ and $\dot{r} ...


0

In a rotating frame of reference you usually consider the centrifugal force, one of the three possible fictitious forces. This is because you usually consider it as rotating at constant angular speed. There are other forces however: the Coriolis force that appear when the object moves in the rotating frame of reference. The Euler force when the angular ...


0

Two electrons when they move experience these forces $$ F_{electrostatic repulsion } = \frac{ke^2}{r^2}$$ And, $$ F_{magnetic attraction} = \frac{μ_0 . e^2 v^2}{4 \pi . r^2}$$ As you can see from the formulae for attraction there must be a velocity. For the two forces to be the same the speed of the electrons must as fast as light, practically these two ...


0

you can check the discussion here. There is a certain case in which a phonon mediates attraction between two electrons. Indeed, acoustic phonons correspond to a slowly varying in-space displacement of atoms which produces a charge. This charge, in turn, results in an electric potential for the electrons. This means that the electron distorts the crystal ...


0

In certain scenarios there can be a magnetic attraction, but the electrostatic replusion will greatly overpower it.


0

First of all, unless you have sealed the cube, there is technically air under the cube as well, and using the equations of buoyancy you can determine that the effects of air pressure is in fact dependent on the hieght of the cube ($PA = \rho g h A$ is a common approximation). Now, for most cubes in consideration, air pressure is an insignificant effect ...


-1

It is because there aren't any forces acting on the mud keeping turning with the tire that it flies off. At whatever point the mud comes off, it will travel tangent to the tire at first and the follow a parabola due to earths gravity. It is most likely the more loose mud will come off first and at that point the tangential direction of the tire points ...


3

You were probably expected to note that the path of any point of the tire is a cycloid. At the point of contact with the ground, the tire is not moving. As it rises from the ground it is moving faster, with a speed that is $v(1+\sin \theta)$ where $v$ is the bike speed and $\theta$ is measured counterclockwise from horizontal. The centrifugal force is ...


0

Displacement refers to the object's position relative to the observer. The "place in space" of the orange. Distance is the object's position relative to an earlier position. If you pick up an orange, and run 10 miles holding it straight out, no work gets done on the orange. But if you extended your arm 10 miles, you would have to be doing work on the ...


2

Assuming that you mean $R_3$ is the initial distance from one object to the other, and given that gravitational force goes as the inverse square of the distance $F \propto \frac{1}{R^2}$, it follows that if $R$ increases by $3R$ (making the final distance $(3+1)R = 4R$), then the force changes by $\frac{1}{4^2} = \frac{1}{16}$ - it does indeed become 16x ...


17

It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. ...


0

The dimension of force is $\text{mass} \cdot \text{distance} / \text{time}^2$ and this is independent of the dimension of the space (as $\vec F = m \vec a$ holds in any dimension). The relations $\text{distance}^{-d+1}$ for gravitation in $d$ dimension has a different reason. The fundamental relation holding in all dimensions is $$ \nabla \cdot \vec F_G = ...


3

If you 'carry' an object 10 meters in one direction then return it back 10 meters from where you started the work done on the object is not the force you expended times distance walked. The formula you write is often misunderstood and misused. In your example, when you lift the object in a gravitational field, the work being done on the object is its weight ...


0

I used to be really confused by friction and the fact that frictional force = (mu) * normal force. However, this is wrong. Frictional force is actually less than mu times normal force, and below that limit only opposed applied forces parallel to the surface. If there are two blocks on the floor of the same mass except I'm pushing one down, and my friend ...


1

What is meant by this is the force normal to the surface of contact between an object and what it is resting on. In most problems like this, this object boundary is taken to be flat. The question is slightly mischievous: it's asking you to be precise to test your understanding of friction: to make the statement precise, you need to say that if the normal ...


2

Imagine a box on a horizontal surface. If you just push it down should there be a friction force? If yes to which direction? Now imagine a box on an inclined surface. Static friction balances the component of weight that is along the surface. If you apply more force normal to the surface, does that component change?


1

2 points: first, $F=ma$ describes the acceleration of an object due to the sum of all forces acting on the object. If these forces are in different directions, they may partly or fully cancel each other out. In the case where the object is not accelerating (so it's moving with constant speed in a constant direction, or it's not moving at all), the sum of ...


2

Although an object that moves with constant velocity has no acceleration, it has kinetic energy and it has momentum. Acceleration is not a conserved quantity. It is not passed from one object to another. Momentum and energy, however, are conserved quantities that pass from one object to another. If a moving object hits a target, kinetic energy will be ...


7

$\vec{F} = m\vec{a}$ means that an object with a force $\vec{F}$ exerted upon it accelerates by an amount $\vec{a}$, not that an object accelerating with $\vec{a}$ exerts a force $\vec{F}$ on something else. Typically, the force exerted by an object has nothing to do with Newton's second law, but is given by other laws (like Coulomb's law in electrostatics). ...


1

To answer your question directly, when the direction of the velocity of the front of the car does not match the front wheels' direction, the tire is deformed, acting like a spring which exerts a lateral force on the car. The deviation between these directions is called the slip angle, and to a first order approximation the lateral force on the tire is ...


1

The diagram you need to draw is this: Now we can compute the forces $$F_1 = \frac{q\cdot q_2}{4\pi\epsilon_0 (a^2 + d^2)}$$ The horizontal component is given by $$F_{1h} = F_1 \cdot \frac{d}{h}$$ The vertical component is $$F_{1v} = F_1 \cdot \frac{a}{h}$$ and finally, the force between the two green charges is $$F_2 = \frac{q_1\cdot ...


0

If you only found the leftward force on the particle, technically your answer is wrong (or at least obtained by invalid methods). The vertical component may have been small enough that it didn't really affect the magnitude, but it would appear it certainly affects the direction. So what you'll need to do it calculate the vertical component of the force as ...


2

You've inserted the scale between yourself and the planet Earth. The scale's spring is compressed by the force exerted on you by Earth's gravitation. The compressed spring tries to uncompress itself and pushes back on you with linear restoring force. Linear restoring force is exactly equal to the force exerted on you by the Earth's gravitation. By ...


0

when you accelerate the vehicle and you turn, the center of gravity shifts from different parts of the car body. Maybe I am misinterpreting your question though..


1

Energy is required to push something over a distance. In the case of the elevator, when it's not moving a brake can be engaged and the power removed and the elevator will sit just fine. That's because it doesn't take any energy to keep something still. But wait, if all I want to do is keep the helicopter still, then it doesn't require any energy? Sort of. ...


0

It appears that you are only considering the system consisting of the vehicle, i.e. car+wheels+brakes, as @user31782 has pointed out. Examining only this system, there is no way for the car to stop: it is traveling through space at a certain speed, and that speed is independent of the spinning wheels; if you apply the brakes the wheels stop spinning (and ...


2

Without defining what a thing is, it makes little sense to discuss the ontology of a thing. Does an apple exist? First, one must say what an apple is; once we agree on that, it's straightforward to show (by example) that apples exist. Given a definition of force, force certainly does exist; we can point to time derivatives of momentum that we observe, and ...


2

Force is a concept which describes, and can be used to manipulate, real phenomena that exist regardless of the existence of the human race. Newton's second law (Acceleration = Force / Mass) is a definition of force. Mass certainly exists, as do velocity and its time derivative, acceleration. There is no reason to suppose that force does not exist. Force ...



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