All categories

3 years ago

What is an essential characteristic of an object in equilibrium?

Physics

1 answer:

7nadin3 [17]3 years ago

3 0

Answer:

Zero Acceleration.

Explanation:

If an object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to describe equilibrium situations. Thus, the net force is zero and the acceleration is 0 m/s/s. Objects at equilibrium must have an acceleration of 0 m/s/s.

You might be interested in

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

3 years ago

If Sean traveled 7 blocks in 200 seconds and each block is 150 meters. What is his velocity?

valentinak56 [21]

Answer:

5.25

Explanation:

tell me if u need an explanation

3 0

3 years ago

A pool ball is rolling at a velocity of 12 cm/s to the right when it is hit by another ball. It continues moving right at a velo

madam [21]

Answer:

Explanation:

Acceleration = rate of change in velocity

a = (21-12)/(0.45) = 20cm/s^2

6 0

3 years ago

2. What is the gravitational force on a 70.0 kg object that is 3.18 x 10 ^6m on the

Vlada [557]

Can you send a pic of the question

6 0

3 years ago

An object of mass 10 kg, initially at rest, experiences a constant horizontal acceleration of 4 m/s2 due to the action of a resu

finlep [7]

So, the total energy value that has been transferred by work is <u>32 kJ</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

$\boxed{\sf{\bold{W = F \times s}}}$

With the following condition :

W = work (J)
F = force (N)
s = shift or displacement (m)

Now, because in this question, the "s" is not directly known, whereas it is known that the initial velocity is zero, the object has an acceleration, and is moving in certain time intervals. Then, use this formula to find the value of "s" !

$\boxed{\sf{\bold{s = \frac{1}{2} \times a \times t^2}}}$