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Arada [10]
3 years ago
15

A mass on a spring will oscillate vertically when it is lifted and released how will the values of elastic potential energy and

gravitational potential energy vary during the oscillation
Physics
1 answer:
Rama09 [41]3 years ago
6 0
The answer to this is

<span>A mass on a spring will oscillate vertically when it is lifted and released. The gravitational potential energy increases from a minimum at the lowest point to a maximum at the highest point. ... The total mechanical energy at those points is the sum of the elastic and gravitational potential energies.</span>

Hope This Helps!
:D
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The electric force between electric charges is much larger than the gravitational force between the charges. Why, then, is the g
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Assuming you are in the Northern Hemisphere, how would you expect the location of the Sun in your local sky at noon to differ fr
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Which sound waves in the electromagnetic spectrum are considered low energy waves
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7 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
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