New answers tagged acceleration
-4
The facts:
No antigravity has been observed ... and no DE has been observed also.
It was discovered (measured) in 1998 an acceleration in the expansion of space.
The mass (M) is the source of gravity (G).
Every force has a source of it, charges, masses,... .
The source of antigravity (AG) is ??????????? (SoAG).
In the absence of apparent source of AG we ...
0
It explodes with a force of 500N.
This sentence is nonsensical. It can explode and release some energy. It can also explode and impart a very high force onto the fragments (which sill be different for each fragment) for a very short time interval.
Once the explosion takes place, the fragments will not accelerate (they may decelerate due to air drag, ...
0
The impact of a bunker buster nuke on rock is going to be very very hard. For the penetrator to have reliable terminal effects, electronics have to be designed with some extra margin of performance.
4
Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough
$$
E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}
$$
For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order ...
4
I've always loved the Orders of magnitude articles on wikipedia. They list examples for the whole range of magnitudes for many physical quantities. There is one for acceleration too.
Among them: Rating of electronics built into military artillery shells: 15 500 G
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I thought gravity is uniform acceleration, not increasing acceleration..
Position: $y(t) = \frac{1}{2} g t^2$
Velocity: $y'(t) = gt$;
Acceleration: $y''(t) = g$;
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There is no such thing as a non-inertial reference frame in SR at all, actually. Attempts to build such a frame (Rindler coordinates) meet some essentially GRstic issues, such as imcompleteness, event horizon, particle horizon, coordinate singularity, some peculiar (gravitationally-affected) kinematics and physics as seen by the observer.
However, ...
0
Suppose two bicycle chains on sprockets and a frame in a J-shape were pulled inwards in a mirror image of each other? They would cancel out both linear and rotational forces and mass would be transfered linearly "round the U bend" without a linear opposite reaction(?)
The structure could then be rotated to a new position, and the process repeated.
2
The only force which works is gravity$^1$. So, change in gravitational potential energy equals final Kinetic energy(assume initial is zero).
$$mgh=mv^2/2$$
$$v=\sqrt{2gh}$$
here $h$ is vertical height traversed.See the velocity does not depend on angle of string, mass of body too..
Let's see the kinematics of body.
The length of string is $h ...
2
If $h$ is the height about the earth then
$$ \ddot{h} = -\frac{G M}{(R+h)^2} $$
$$ \ddot{h} = \frac{{\rm d} \dot{h}}{{\rm d}t}= \frac{{\rm d} \dot{h}}{{\rm d}h} \frac{{\rm d} h}{{\rm d}t} = \frac{{\rm d} \dot{h}}{{\rm d}h} \dot{h} $$
$$ \int \ddot{h}\; {\rm d} h = \int \dot{h}\; {\rm d} \dot{h} = \frac{1}{2} \dot{h}^2 + K$$
$$ \int -\frac{G M}{(R+h)^2}\; ...
4
So you were on the right track with integrating over r and over t. Here's how you could do it:
The acceleration at any radius, r (if we assume Earth is a point mass) is:
$$a=-{GM\over r^2}$$
The minus sign is because the acceleration is anti-radial. Then you can do the following:
$$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$
$$thus$$
...
0
You will calculate the velocity for the x, y and z components separately.
So if the object has $v_x$ velocity initially, and it spontaneously gains acceleration in the negative y direction, this problem becomes very simple. Over time this object will gain velocity in the y ($v_y$) direction. It will also always have its' velocity in the x direction, and ...
0
You must have to think about the mass of the object. Same force will be applied on two different objects of different mass. Velocity (movement) will depend on the mass. Whose mass is more it will be at greater velocity than other higher mass object.
Let's say, i jump on the earth. Same Force is applied both on earth and me. Since my mass is very less ...
1
Newton's third law doesn't imply that things can't move but it does imply conservation of momentum and energy.
Imagine a scenario where an astronaut is in orbit so they don't feel the affects of gravity. If there is an object floating and they push on it (apply a force to it) we know intuitively that the object will start to accelerate in the direction of ...
1
Your premise violates Newton's first law of motion:
If there is no net force on an object, then its velocity is constant. The object is either at rest (if its velocity is equal to zero), or it moves with constant speed in a single direction.
For an object (a body) to be accelerating there must be an external force applied. One of the reasons for ...
0
Let's see with help of an example.
Let the particle is at $(0,0)$ moving with speed of $2m/s$ at $t=0$ and is subjected to acceleration $-2\hat i\ \text{m/s}^{-2}$.Now see after $2 seconds$.
We see displacement is zero,and distance travelled =$2m$ .Also acceleration is constant but still $\langle speed\rangle\not=0$ whereas $\langle velocity \rangle=0$
...
2
If an object is moving in a circular motion, its velocity $\vec{v}$ changes. The centripetal acceleration is just a formula that gives you the length of the derivative $\frac{d\vec{v}}{dt}$ which is the acceleration. It must be caused by some force, according to Newton's second law. If you are holding the object with a rope, then it is the tension of the ...
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When the acceleration is a function of position use the following
$$ a(x) = \frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} u $$
$$ \int a(x)\,{\rm d} x = \int u\,{\rm d} u = \frac{1}{2} u^2 + K_1 $$
which is solved for $u(x)$.
The the position is found from
$$ t = \int \frac{1}{u(x)}\,{\rm d} ...
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You certainly want to numerically integrate your motion equation givent your expression of the force.
I will assume that the mass $m$ is attached at one side of the rope, while the other side is attached to a wall or something that won't move during the integration. Something like this, where the spring is in fact your rope, with $x$ the extension of the ...
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Displacement of elevator with Constant velocity:
$s=ut$
displacement of elevator with uniform acceleration:
$s=ut+0.5at^2$
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