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Anon25 [30]
3 years ago
7

A man, facing a high wall, notices that an echo is heard 4.0 seconds after he makes a sharp sound. After walking 200m directly t

owards the cliff, an echo is heard after 3.4 seconds. Calculate (i) the speed of the sound (ii) how far he was from the wall when he made his second observation of the echo
Physics
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:2.4

Explanation:basically 2.4 = 2.500 kg and then we add 3.4 seconds in all of them 4.0 + 3.4 = 2.5 ok

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Answer:

B.

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A 27.0-g object moving to the right at 21.5 cm/s overtakes and collides elastically with a 11.0-g object moving in the same dire
7nadin3 [17]

Answer:

final speed of the first object: 17.73 cm/s

final speed of the second object: 24.23 cm/s

Explanation:

Data:

mass of the first object, m1 = 27.0 g

mass of the second object, m2 = 11.0 g

initial speed of the first object, v1i = 21.5 cm/s

initial speed of the second object, v2i = 15 cm/s

final speed of the first object, v1f = ? cm/s

final speed of the second object, v2f = ? cm/s

Elastic collisions formula:

v1f = (m1 - m2)/(m1 + m2) * v1i + 2*m2/(m1 + m2) * v2i

v1f = (27 - 11)/(27 + 11) * 21.5 + 2*11/(27 + 11) * 15

v1f = 17.73 cm/s

v2f = 2*m1/(m1 + m2) * v1i + (m2 - m1)/(m1 + m2)*v2i

v2f = 2*27/(27 + 11) * 21.5 + (11 - 27)/(27 + 11)*15

v2f = 24.23 cm/s

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