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STALIN [3.7K]
3 years ago
11

Ellus

Physics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

W'=125.44 N

Explanation:

The weight of a person on the surface of Earth is 784 N

Weight is given by :

W = mg

m is mass of the person and g is acceleration due to gravity on the surface of Earth (10 m/s²)

m=\dfrac{W}{g}\\\\m=\dfrac{784}{10}\\\\m=78.4\ kg

The acceleration due to gravity on the surface of Moon, g' = 1.6 m/s²

Weight of the person on the moon is :

W'=mg'

W'=78.4\ kg\times 1.6\ m/s^2\\\\W'=125.44\ N

Hence, the person would weigh 125.44 N on the Moon.

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Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
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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

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(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}}}

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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.

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