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emmainna [20.7K]
3 years ago
7

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart

h with enough speed to make it to the moon. Some information that will help during this problem: mearth = 5.9742 x 1024 kg rearth = 6.3781 x 106 m mmoon = 7.36 x 1022 kg rmoon = 1.7374 x 106 m dearth to moon = 3.844 x 108 m (center to center) G = 6.67428 x 10-11 N-m2/kg2 1) On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?
Physics
1 answer:
Trava [24]3 years ago
5 0
This is honestly not something I’ve learned. The answer is something I don’t know.
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Ryan runs 50 m north and then 50 m east. What is his displacement?
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Explanation:

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3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
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