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LuckyWell [14K]
1 year ago
6

In the Roman soldier model for refraction, what happens to the distance between the rows of each soldier once they enter the str

eam?
A. Not enough info
B. They are closer together
C. They stay the same distance
D. They are further apart
Physics
1 answer:
Anon25 [30]1 year ago
4 0

They pick up speed and move away from the usual.

Using meter sticks, a group of pupils makes a straight line (shoulder to shoulder) and connects to their closest neighbor. Two media are separated in the room by a strip of masking tape. Students can be seen moving normally in one of the media (on one side of the tape). The kids in the other medium (or on the opposite side of the tape) move very slowly and in baby stages. The class moves in the union in a straight line in the direction of the diagonal piece of masking tape. As they get close to the masking tape, the pupils stay in line. Each pupil immediately changes the speed of her or his walk as she or he reaches the tape.

learn more about distance: brainly.com/question/17273444

#SPJ1

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Luda [366]

Answer:

the relative speed should be 40mph

Explanation:

8 0
2 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
Stan does 178 J of work lifting himself 0.5 m. What is Stan’s mass? The acceleration of gravity is 9.8 m/s 2 . Answer in units o
Rom4ik [11]
Work= (force)(distance)
178= m(9.81)x0.5
178=m(4.905)
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His mass is 36.3 kg
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athe answer to this question is Synex BA-315

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