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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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An object is 15 cm in front of a diverging lens with a
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A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

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where

q is the distance of the image from the lens

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\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

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Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

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4 0
3 years ago
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x = 7.14 meters

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Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :

B=\dfrac{\mu_oI}{2\pi r}

Let the distance is x from wire 1. So,

\dfrac{I_1}{r}=\dfrac{I_2}{(0.25-r)}

\dfrac{2}{r}=\dfrac{5}{(25-r)}

x=\dfrac{50}{7}\ m

x = 7.14 meters

So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.

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