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3 years ago

Link between mass and force of gravity?

2 answers:

8 0

Mass x Force of gravity = Weight

Weight is a force.

On Earth, the force due to gravity is ~9.81 N/Kg. So an object of 100 Kg will have a weight of ~981 N

eduard3 years ago

3 0

Force of gravity is what we call "weight".

It's the product of (mass) x (acceleration due to gravity).

On Earth, acceleration due to gravity is about 9.81 meters per second-squared.
It's different in other places. Example: It's about 1.62 on the Moon.

So the weight of any mass on Earth is (9.81) x (the mass), in Newtons.
The weight of the <u>SAME mass</u> on the Moon is (1.62) x (the mass), in Newtons.

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Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

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3 years ago

An object of known mass M with speed v0 travels toward a wall. The object collides with it and bounces away from the wall in the

Bingel [31]

Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision

Explanation:

The impulse theorem states that the change in momentum of an object is equal to the impulse, which is the product between the average force applied and the duration of the collision:

$\Delta p = F \Delta t$