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3 years ago

How does the work needed to lift an object and the gravitational potential energy of an object compare

1 answer:

8 0

When you do work to lift the object, the amount of work you do BECOMES the object's gravitational potential energy. It GETS its potential energy from the work you do to lift it. They're equal. You lose it, and the object gains it. Energy is not created or destroyed. It's just transferred from you to the object.

Later, when you DROP the object, GRAVITY does the same amount of work on it, to pull it to the ground. Again, no energy is created or destroyed. Every time a force acts to move anything, the energy to do it comes from somewhere, and the energy goes somewhere.

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What three sprites are needed to create the Space Invaders game that was discussed in the unit? spaceship, laser, enemy gnome, a

Lilit [14]

Answer: anlien, enemy gnome, spaceship

Explanation:

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2 years ago

What is free nitrogen

Alenkasestr [34]

Answer:

It is simply molecular nitrogen (N2). Nitrogen, in its molecular form, consists of two nitrogen atoms bound together with a tripple bond

Explanation:

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3 years ago

Read 2 more answers

A stretched string has a mass per unit length of 5.12 g/cm and a tension of 19.3 N. A sinusoidal wave on this string has an ampl

Sati [7]

Answer:

a. 0.143 mm b. 77.6 rad/m c. 483.18 rad/s d. +1

Explanation:

a. ym

Since the amplitude is 0.143 mm, ym = amplitude = 0.143 mm

b. k

We know k = wave number = 2π/λ where λ = wavelength.

Also, λ = v/f where v = speed of wave in string = √(T/μ) where T = tension in string = 19.3 N and μ = mass per unit length = 5.12 g/cm = 5.12 ÷ 1000 kg/(1 ÷ 100 m) = 0.512 kg/m and f = frequency = 76.9 Hz.

So, λ = v/f = √(T/μ)/f

substituting the values of the variables into the equation, we have

λ = √(T/μ)/f

= √(19.6 N/0.512 kg/m)/76.9 Hz

= √(38.28 Nkg/m)/76.9 Hz

= 6.187 m/s ÷ 76.9 Hz

= 0.081 m

= 81 mm

So, k = 2π/λ

= 2π/0.081 m

= 77.6 rad/m

c. ω

ω = angular frequency = 2πf where f = frequency of wave = 76.9 Hz

So, ω = 2πf

= 2π × 76.9 Hz

= 483.18 rad/s

d. The correct choice of sign in front of ω?

Since the wave is travelling in the negative x - direction, the sign in front of ω is positive. That is +1.

4 0

3 years ago

A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball,

Strike441 [17]

Momentum = (mass) x (velocity)
Original momentum before the hit =
(0.16 kg) x (38 m/s) this way <==
= 6.08 kg-m/s this way <==
Momentum after the hit =
(0.16) x (44 m/s) that way ==>
= 7.04 kg-m/s that way ==>
Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .-----------------------------------------------
Change in momentum = impulse.
Impulse = (force) x (time the force lasted)
13.12 kg-m/s = (force) x (0.002 sec)
(13.12 kg-m/s) / (0.002 sec) = Force
6,560 kg-m/s² = 6,560 Newtons = Force
( about 1,475 pounds ! ! ! )
Hoped this helped!! ☺

4 0

4 years ago

A particle is located on the x axis 4.9 m from the origin. A force of 38 N, directed 30° above the x axis in the x-y plane, acts

Juli2301 [7.4K]

Answer:

Torque is 93 Nm anticlockwise.

Explanation:

We have value of torque is cross product of position vector and force vector.

A force of 38 N, directed 30° above the x axis in the x-y plane.

Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j

A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.

Position vector, r = 4.9 i

Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm

So Torque is 93 Nm anticlockwise.

7 0

3 years ago