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3 years ago

Which force acts as an object to move it from rest or a constant straight line motion

Physics

2 answers:

sleet_krkn [62]3 years ago

8 0

Answer:

The force is Inertia

Explanation:

The force that acts on an object to move it from rest or a constant straight line motion is known as Inertia.

In physics the above statement is governed by Newton's first law of motion which is also known as Law of Inertia.

This law states that, an object that is at rest will remain at rest and an object that is moving will continue to move in a straight line with constant speed, if and only if the net force acting on the object is zero.

This implies that, A stationary object will remain motionless if no force acts on it while a object with constant velocity will continue moving with constant velocity until a force acts on it (neglecting resistance from air and friction).

tankabanditka [31]3 years ago

7 0

Answer:

Inertia.

Explanation:

Inertia is the property of matter by which it remains at rest or in uniform motion in the same straight line unless acted upon by some external force or the resistance to change in velocity by an external force.

Inertia is the tendency for an object at rest to remain at rest, or for a moving object to remain in motion in a straight line with constant speed.

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2 years ago

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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.

$p_e=m_b.v_b\sin\theta$

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

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

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3 years ago

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skad [1K]

Correct answer choice is :

B) Upwarped

Explanation:

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