# Tag Info

6

I suspect this may not make much sense to non-GR heads, but the Einstein equation relates the curvature of spacetime to an object called the stress-energy tensor. The stress-energy tensor describes the properties of the matter/energy that is causing the curvature. In most cases we're only interested in the amount of matter/energy present i.e. its density, ...

4

Particle A is subject to the centripetal force $F=m\omega r^2$ just the same as particle B. In both cases the wall provides the reaction force to keep both particles in uniform circular motion. There can be no circular motion without a centripetal force.

4

It isn't clear from your question exactly what you are integrating and how, but this is the way to tackle problems like this. You know that: $$\frac{dv}{dt} = -kv$$ The way to solve equations like this one is to rearrange it by dividing both sides by $v$ and multiplying both sides by $dt$ to get: $$\frac{1}{v}dv = -k\,dt$$ Now we can integrate both ...

3

The universe is described by a scale factor, normally indicated by the symbol $a(t)$, that is a function of time. We take the scale factor to be one right now, so in the past $a$ was less than one and in the future $a$ will be greater than one. Roughly speaking, if $a$ has the value $\tfrac{1}{2}$ it means everything was half as far apart as it is now, and ...

2

Let me try to add a missing intuitive piece in the explanation to the discussion to @Gert's answer of why there must always be radial acceleration for any circular motion. Remember what acceleration is: Change in velocity. $\vec a=d\vec v/dt$. More mathematically it is change in the velocity vector. If velocity is changed - either magnetude or direction - ...

2

In method two,you are writing $F_{net}=ma$ separately for each block.So it should be $F_t=m_1a$ and $m_2g-F_t=m_2a$,that is,you must use the mass of just the one block you are considering,not both the blocks together.This agrees with method 1,which is also correct.

2

How does dark energy allow the universes expansion to accelerate? I hope that it is clear to the questioner and the readers that the horse pulling dark energy is the experimental observation that the expansion of the universe is accelerating. Dark energy is proposed as the reason why the expansion is accelerating. It is called "dark" because it is not ...

2

I think you're mentally confusing "acceleration" and "force. I think your thought process is "If I'm going to apply an acceleration to an object, that acceleration will be fighting against the acceleration of gravity, so the two accelerations will partially cancel each other out". The problem is that acceleration isn't something you can "apply" to an ...

2

The equation you're using is a kinematic equation. Kinematics, you might recall, is how we mathematically describe motion; it doesn't deal with the causes of that motion. If an object is accelerating at $70 \sqrt{2} \text{ m/s}$ and decelerating at $-24.5 \text{ m/s}^2$ due to a force or a combination of forces, then it will come to rest after traveling ...

2

You throw the ball upwards with velocity $v$ and it returns to your hand with velocity $-v$. Let's draw a graph showing the velocity as a function of time: Acceleration is defined as: $$a = \frac{dv}{dt}$$ so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is ...

1

Read the article on this link https://www.scribd.com/doc/279174920/Decreasing-Mass-Cosmology-and-the-Accelerating-Expansion-of-the-Universe and you'll see that 'dark energy'isn't neccessary for the accelerating expansion of the universe

1

So, it turns out that I am out of coffee. Anyway, since we already know that: $w = m \times g$ (Newton's Second Law) We know that $w = \frac{Gm_Em}{R_E^2}$ Dividing the entire equation by $m$, we get: $\frac{w}{m} = \frac{mg}{m}$ Which is equal to: $\frac{Gm_Em}{R_E^2m} = g$ (Substituting the values) And finally: $\frac{Gm_E}{R_E^2} = g$

1

The rate of acceleration is called jerk. The reason you don't see a name for derivatives of acceleration in classical mechanics text is that they are not particularly important concepts for the study of Newtonian mechanics. It is not important if we can or can't apply acceleration instantaneously. However allowed jerk rates are analyzed and studied in ...

1

Given a particle moving uniformly on a circular path $\vec r(t)$, without loss of generality, we can parametrize the motion as: $$r(t) = R \big( \vec e_x \cos(\omega t) + \vec e_y \sin(\omega t) \big),$$ where $\omega$ is the angular velocity of the motion. By using Newton's axiom $\vec F = m \vec a$ we can calculate the force necessary for the particle ...

1

You're confusing cause and effect. The better equation here is $A={F \over M}$ When a force is applied to an object, the object is accelerated. The object gains kinetic energy from the force, but it doesn't have a property of acceleration. The gun accelerates the bullets with some acceleration for some time period. One gun could accelerate the bullet ...

1

An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation: $$v = u + at$$ So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved ...

1

Your equation 1.1 can be used with constant velocity. Here you have to use the $2^{\text{nd}}$ equation. ie $a = 2d/(t^2)$. So, the answer is $118.4 \, \text{cm}/s^2$.

1

When you shoot the ball upwardly, gravity acts on it with a force $mg$ where $m$ is the mass of the ball and $g=9.81 ms^{-2}$ the Earth's gravitational acceleration. If the initial upward velocity was $v_0$ then the instantaneous velocity $v$ is given by: $v=v_0-gt$, so after some time $t=\frac{v_0}{g}$ the balls's velocity becomes $v=0$. However, we know ...

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