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3 years ago

14

What describes how the spring constant affects the potential energy of an object for a given displacement from an equilibrium po

sition?

1 answer:

Anit [1.1K]3 years ago

8 0

Answer:

The higher the spring constant, the greater the elastic potential energy.

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Read the following questions and answer them using complete sentences. Be sure to fully explain your answers.

astra-53 [7]

Wave power can be regarded as a reliable source of energy because the ocean currents are always moving.

<h3>What can be the challenges of wave power?</h3>

Wave power is a device that can be used to convert the mechanical energy of the ocean waves into electrical energy based on the principle of conservation of energy.

The major challenges that face the use of wave power in electricity generation is the unreliability of the waves which leads to uncertainty in the quantity of power generated Also, the wave direction and direction of ocean currents all limit the amount of power generated by this method. However, in spite of challenges, it can be regarded as a reliable source of energy because the ocean currents are always moving.

Learn more about wave power:brainly.com/question/1362067

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6 0

1 year ago

according to this passage when making healthcare decisions, how should a person approach the decision making process

Aleksandr-060686 [28]

Answer:

What is your name?

Explanation:

no bro I don't know I am New of this app so I don't know

8 0

2 years ago

What is the physical state of water at 250°C?

mars1129 [50]

It is probably gas as pure water starts boiling at 100d degrees C

7 0

3 years ago

Waves arriving at a fixed boundary are a. neither reflected nor inverted. c. reflected and inverted. b. reflected but not invert

nirvana33 [79]

My freind the correct anwer is C

4 0

3 years ago

Moon effect. Some people believe that the Moon controls their activities. The Moon moves from being directly on the opposite sid

Hatshy [7]

Answer:

6.9%

Explanation:

We are given that

Distance between Earth and Moon=d= $3.82\times 10^8$ m

Radius of Earth=r= $6.37\times 10^6 m$

a.Initially gravitational pull

$F=\frac{GmM}{(r+d)^2}$ .....(1)

After changing

Gravitational pull

$F'=\frac{GmM}{(d-r)^2}$ .....(2)

Equation (2) divided by equation (1)

$\frac{F'}{F}=\frac{(r+d)^2}{(d-r)^2}$

$\frac{F'}{F}=\frac{(3.82\times 10^8+6.37\times 10^6)^2}{(3.82\times 10^8-6.37\times 10^6)^2}=1.069$

Increases in gravitational pull= $\frac{F_2}{F_1}-1=1.069-1=0.069\times 100=6.9$ %

4 0

3 years ago