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Inessa05 [86]
3 years ago
15

Which point has the most Potential energy? [Select]

Physics
1 answer:
Margaret [11]3 years ago
3 0

Answer:

<em>Most potential energy: A</em>

<em>Most kinetic energy: D</em>

Explanation:

Kinetic Energy is the type of energy an object has due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

\displaystyle K=\frac{1}{2}mv^2

Where:

m = mass of the object

v = speed at which the object moves

The gravitational potential energy is the energy stored in an object because of its height h in a gravitational field.

It can be calculated with the equation:

U=m.g.h

The point where the object has the most potential energy is that where it has more height. It corresponds to point A.

When the object is at zero height, all of its potential energy was transformed to kinetic, thus the point where the kinetic energy is D.

Most potential energy: A

Most kinetic energy: D

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1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

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(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
3 years ago
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