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3 years ago

What is the gravitational potential energy of a 3 kg ball that is 1 meter above the floor?

Physics

1 answer:

Sveta_85 [38]3 years ago

7 0

Answer:

PE=mgh

M= Mass (kg)

G= Gravitational field strength (N/kg)

H= Hight (m)

PE= Gravitational Potential Energy (J)

Explanation:

Gravitational Potential Energy is the energy stored in a object due to its position above the Earth's surface.

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A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

3 years ago

Lee pushes horizontally with a force of 75 n on a 36 kg mass for 10 m across a floor. calculate the amount of work lee did. answ

svlad2 [7]

To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.

So you would do 75 N x 10m x Cos (0 degrees)= 750 J

8 0

3 years ago

Read 2 more answers

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

3 years ago

Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through th

Anit [1.1K]

Answer:

The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.

Explanation:

The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components. While in a series connection, the components are connected to each other end-to-end.

As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.

4 0

4 years ago

Describe a ball's motion as it rolls up a slanted

emmasim [6.3K]

The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

initial velocity of the ball, u = 1.25 m/s
time of motion of the ball, t = 4.22 s

As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.

Thus, the ball decelerate as it moves upwards.

The acceleration of the ball is calculate as;

$a = \frac{v_f -v_0}{t} \\\\$

<em>at the highest point on the incline plane, the final velocity </em> $v_f$ <em> is zero</em>

$a = \frac{0-1.25}{4.22} \\\\a = -0.3 \ m/s^2$

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

Learn more here:brainly.com/question/23860763

4 0

3 years ago