3 years ago

I WILL GIVE BRAINLIEST TO WHO ANSWERS CORRECTLY

Physics

2 answers:

Jlenok [28]3 years ago

5 0

Dynamic equilibrium.

REY [17]3 years ago

3 0

Answer:

Static Equilibrium

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A child has two red wagons, with the rear one tied to the front by a (non-stretching) rope. If the child pushes on the rear wago

Sergeeva-Olga [200]

Answer:

The rear wagon gains the kinetic energy, but the front wagon will remain at rest.

The two-wagon system will gain a kinetic energy $\dfrac{1}{16}$ of the kinetic energy gained by the rear wagon.

Explanation:

Let's consider that the masses of the wagons to be 'M'. When the child pushes the rear wagon let's assume that the velocity of the rear wagon be 'v'.

Therefor the kinetic energy gained by the rear wagon be $K_{r} = \frac{1}{2}Mv^{2}$ .

Now let's assume that the velocity of the centre of mass (C), as shown in the figure, be 'V'. So from momentum conservation law we can write,

$&& Mv = (M + M)V\\\\&or,& V = \frac{v}{2}$

Now the centre of mass ( $M_{C}$ ) is given by

$M_{C} = \frac{M \times M}{M + M} = \frac{M}{2}$

So the kinetic energy ( $K_{C}$ ) of the system will be

$K_{C} = \frac{1}{2}M_{C}V^{2} = \frac{1}{2}\frac{M}{2}(\frac{v}{2})^{2} = \frac{1}{16}Mv^{2}$

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3 years ago

A lamp hangs from the ceiling at a height of 2.9 m. If the lamp breaks and falls to the floor, what is its impact speed

ser-zykov [4K]

To solve this there is this website that I found that helps
I am in middle school so I have no idea how to solve this
but
this website may help considering u are in high school and u
(hopefully mind u)
know how to solve this
so to get there u google
"whats impact speed"
and click on the first thing there the website is ehow

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3 years ago

An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False

Umnica [9.8K]

Hi There! :)

An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P

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3 years ago

Read 2 more answers

Determine the mechanical energy of this object a 1-kg ball rolls on the ground at <br> m/s

dedylja [7]

Mechanical energy = potential energy + kinetic energy
The ball is on the ground so it has no potential energy. that's all i know.

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