How can you increase the gravitational potential energy (GPE) of an empty shoe box on the bottom shelf of a bookcase

Physics

2 answers:

Anettt [7]2 years ago

5 0

Answer:

Add items to the box

Explanation: I did the test

ivanzaharov [21]2 years ago

4 0

Answer:

Add some Items to it.

Explanation:

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A driver traveled 270 km in 3 hours. The driver’s destination was still 150 km away. What was the driver’s average speed at this

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The driver was going 90 kilometers every hour.

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2 years ago

A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee

Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

Initial Velocity; $u = 0$

Height from which it has dropped; $h = 15m$

Gravitational field strength; $g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2$

Final speed of brick as it hits the ground; $v = \ ?$

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

$v^2 = u^2 + 2gh$

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

$v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s$

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

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1 year ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

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To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$