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3 years ago

13

What does gravitational potential energy typically change into? Question 8 options: A-Electric energy B- Kinetic energy C- Radio

active energy D-Radioactive energy

2 answers:

MariettaO [177]3 years ago

7 0

The answer is “B”, Kinetic Energy

nika2105 [10]3 years ago

6 0

<em>Answer:</em>

<em>kinetic energy </em>

<em>Explanation:</em>

<em>i should know i had a test about it.</em>

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2 years ago

A satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther a

irinina [24]

The speed of the second satellite is less than the speed of the first satellite.

<h3>What is speed?</h3>

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Given is a satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet.

When the distance from the center of the orbit increases, the time to complete the orbit will be greater.

Thus, the speed of the second satellite is less than the speed of the first satellite.

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1 year ago

Explain What determines the amount of<br> chemical energy a substance has?

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The kinds of atoms in the substance and their arrangements

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2 years ago

Mining runoff is _____ water that comes from the mining of rock.

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3 years ago

Read 2 more answers

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

soldi70 [24.7K]

Answer:

$v = 7793150 m/s$

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

$v = \frac{kQ}{r}$

Where $k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }$

and it is satisfy the superposition principle, thus

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}$

$v = 222.73v$

The electrical potential at 10 cm from charge 1 is:

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}$

$v = 395.44 v$

Since the work - energy theorem, we have:

$q\Delta v = \frac{mv^{2} }{2}$

where q is the electron's charge and m is the electron's mass

Therefore:

$v = \sqrt{\frac{2q\Delta v}{m} }$

$v = 7793150 m/s$

6 0

2 years ago