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Tpy6a [65]
3 years ago
5

how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?

Physics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = -PE_g = -2058\ kg.m/s^2

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= \frac{1}{2} m\times v^2 = m\times g \times h

Solving for v, we get

v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2

The person will be going at a speed of 7.67 m/s².

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If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
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0.82 mm

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substituting the parameters in the above equation; we have:

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y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

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\delta y = y_2-y^{'}_2

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6 0
4 years ago
arzan, who weighs 700 N, swings from a cliff at the end of a convenient vine thatis 20 m long. From the top of the cliff to the
vekshin1

Answer:

= 7.07 m

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