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Jlenok [28]
3 years ago
14

A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits th

e ground? (Disregard air resistance. a = −g = −9.81 m/s 2 .)
Physics
1 answer:
Reptile [31]3 years ago
6 0

Initial speed of the coin (u)= 0 (As the coin is released from rest)

Acceleration due to gravity (a) = g = 9.81 m/s²

Time of fall (t) = 1.5 s

From equation of motion we have:

\boxed{ \bf{v = u + at}}

By substituting values in the equation, we get:

\longrightarrow v = 0 + 9.81 × 1.5

\longrightarrow v = 14.715 m/s

\therefore Speed of the coin as it hits the ground/Final speed of the coin = 14.715 m/s

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Answer:

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Explanation:

Given that,

Mass of astronaut A, m_A=110\ kg

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We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :

m_Av_A=m_Bv_B\\\\v_A=\dfrac{m_Bv_B}{m_A}\\\\v_A=\dfrac{74\times 0.4}{110}\\\\v_A=0.26\ m/s

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In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge

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