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2 years ago

Acar start from rest and go with velocity of 4m/s for 4second what is acceleration of car

Physics

2 answers:

lesya [120]2 years ago

7 0

Answer:

1m/s

Explanation:

Given

initial velocity(u)=0

final velocity(v)=4m/s

timetaken (t)=4

Aceleration(a)=v-u÷t

now

=4-0÷4

=1m/s

aleksklad [387]2 years ago

6 0

Answer:

1 m/s^2

Explanation:

The formula for accleration is a=Δv/Δt

where, Δv = final velocity - initial velocity = 4 - 0 = 4

initial velocity = 0 since the car starts form rest and final velovity is 4 as the car goes from rest till 4 m/s

Δt = 4 since the car takes 4 seconds to reach a velocity of 4m/s

Hence, a = 4 m/s / 4 s = 1 m/s^2

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A stone is thrown upward at an angle. what happens to the horizontal component of its velocity as it rises? as it falls?

Ne4ueva [31]

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5 0

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, $\theta=28^{\circ}$

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

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

$97\times v_c=1.35\times 4.61\cos28$

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

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.

$p_e=m_b.v_b\sin\theta$