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miv72 [106K]
3 years ago
8

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

he surface of the moon (i.e., x = 6r)?
Physics
1 answer:
Cloud [144]3 years ago
7 0

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

W = U_f - U_i

we know that

U_f = -\frac{GMm}{6r}

U_i = -\frac{GMm}{r}

now from above formula

W = -\frac{GMm}{6r} + \frac{GMm}{r}

W = \frac{5GMm}{6r}

so above is the work done to move the mass from surface to given altitude

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Answer:

p = mv

Explanation:

  • The momentum of a body is defined as the product of its mass and velocity. Its physical symbol is 'p'.
  • The formula for momentum is given by

                               p = mv

         Where,

                                m -  the mass of the body in kg

                                v - velocity of the body in m/s

  • Therefore, the unit of momentum is expressed as the kg m/s
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  • If a body is at rest, the momentum associated with the body is zero.
  • The momentum plays a significant role in the kinematics of the body. As similar to the energy conservation law, the total momentum of the body is conserved.
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Answer:

Solution is given in the attachments

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In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x
Ostrovityanka [42]

Answer:

(a) increase

Explanation:

On a line graph; we have the x-axis to the positive side and the negative side .In a positive x-axis direction, the force is usually positive, vice versa the negative side as well.

The change in the potential energy of a charge field-system can be given as:

\delta U= -q(EdsCos \theta)

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q = positive test charge

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Given that:

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Replacing our values in the above equation, we have:

\delta U = -(-q)(60Cos 0)

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If earth was not tilted on a axis, we wouldn't have any seasons. we wouldn't have seasons because the axis tilts us towards or away from the sun, and if you're tilted torwards the sun the sun you would have summer. if you're tilted away from the sun, you have winter. if there was no axis, the temperature would be the same all around the earth, so therefore we would not have seasons. 

hopefully this helps 
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