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stellarik [79]
3 years ago
14

A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla

ced at the location P, which is at a distance d from the origin.
If L = 2.00 ? 104 m, d = 18.0 ? 104 m,M = 14.0 ? 106 kg,and m1 = 8.00 ? 106 kg,what is the value of this potential energy?
Physics
1 answer:
Yuri [45]3 years ago
5 0

Answer:

= - 5.65\times 10^{-2} J

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18  *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

U =- G \frac{m_1 m_2}{r}

mass per unit length is given as

\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m

calculating potential energy

dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}

=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}

=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}

=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}

=-G*m_1*\sigma ln(1.125)

=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)

= - 5.65\times 10^{-2} J

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Wich statment belongs to Daltons atomic theroy?
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Answer:

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A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
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The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

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<u>Solution</u>:

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vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

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