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Karo-lina-s [1.5K]
4 years ago
10

Deshawn made two diagrams to show the Moon in the same position at different times of the year. In his diagrams, he included vie

ws from above (top view) and views from Earth. He says that some of the time the Moon looks bright from Earth, as shown in Diagram A, but other times the Moon looks completely dark from Earth, as shown in Diagram B.
Is Deshawn correct? If he is correct, explain why light on the Moon changes in this way. If he is incorrect, explain how light on the Moon should look in each of his diagrams.

Geography
1 answer:
Darya [45]4 years ago
6 0

Answer:

Deshawn is correct.

Explanation:

The moon reflects the sunlight and this directly influences the way that light reaches the earth and the way we can see this moon, from here on earth.

With that, we can say that the way we see the moon, from here on Earth, changes (even if it is in the same position). This change occurs in relation to the position of the moon in relation to the sun, because we can only see the side that is being illuminated at the moment, if there is an illuminated side.

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<h2>GIVEN DATA</h2>

Mass of Rocket = m = 215 kg

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(a) Kinetic Energy at Final Height = K.E_f

(b) Maximum Height of Rocket above Earth's Surface = H_{max}

<h2>EXPLANATION</h2>

Part (a):

As drag air is negligible, energy will be conserved.

\therefore ΔE = 0

K.E_i + U_i = K.E_f + U_f

K.E_f = U_f - K.E_i - U_i

where, U is the potential energy of the system.

K.E_f=\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}\;\;\;\;----------\;(1)

Substituting Values and simplifying,

K.E_f = 3.665*10^9 J

Part (b):

The rocket will come to rest after reaching the maximum height. Therefore, its final velocity and consequently final kinetic energy will be zero.

i.e.\;\;v_f = 0\;\;\;\&\;\;\; K.E_f=0

Equation (1) will become,

0\;=\;\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}

or\;\;\frac{1}{2}v_i^2= GM(\frac{1}{R_i}-\frac{1}{R_f})\\\\\frac{1}{2GM}v_i^2= (\frac{1}{R_i}-\frac{1}{R_f})\\\\\therefore\; R_f = \frac{2GMR_i}{2GM-v_i^2R_i}

Substituting values and simplifying,

R_f = 10.20*10^6 m

which is the distance from the Earth's centre. To find the height of rocket from Earth's surface, we simply subtract the Earth's radius from above result.

H_{max} = R_f - R

H_{max} = 10.20*10^6\;-\;6.37*10^6\\\\H_{max} = 3.83*10^6\;\;m

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