Answer:
-3.5 m/s²
Explanation:
- Initial Velocity = 21m /s
- Final velocity = 0m/s
- Distance = 63 m .
- Acclⁿ = ?
<u>We </u><u>know</u><u> </u><u>that</u><u> </u><u>:</u><u>-</u><u> </u>
Stopping distance = u²/2(-a)
63m = (21m/s)² / -2a
a = - 21 * 21 / 63 * 2 m/s²
a = - 3.5 m/s²
<em>*</em><em>*</em><em>Edits</em><em> </em><em>are</em><em> </em><em>welcomed</em><em>*</em><em>*</em>
The net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.
To find the answer, we have to know more about the basic forces acting on a body.
<h3>How to find the net force on the box?</h3>
- Let us draw the free body diagram of the given box with the data's given in the question.
- From the diagram, we get,

where, N is the normal reaction, mg is the weight of the box,
is the net force, f is the kinetic friction.
- We have the expression for kinetic friction as,

- Thus, the net force will be,

Thus, we can conclude that, the net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.
Learn more about the basic forces on a body here:
brainly.com/question/28061293
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Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²