(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.
(b) The acceleration of the child-wagon system is 0.33 m/s².
(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.
<h3>
Net force on the third child</h3>
Apply Newton's second law of motion;
∑F = ma
where;
- ∑F is net force
- m is mass of the third child
- a is acceleration of the third child
∑F = 96 N - 75 N - 12 N = 9 N
Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;
- the wagon
- the children outside the wagon
<h3>Free body diagram</h3>
→ → Ф ←
1st child friction wagon 2nd child
<h3>Acceleration of the child and wagon system</h3>
a = ∑F/m
a = 9 N / 27 kg
a = 0.33 m/s²
<h3>When the frictional force is 21 N</h3>
∑F = 96 N - 75 N - 21 N = 0 N
a = ∑F/m
a = 0/27 kg
a = 0 m/s²
Learn more about net force here: brainly.com/question/14361879
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Answer: We don’t really ever need to know how to dissect a frog but hey, one day maybe I’ll need to give a frog a discectomy
Answer:
vₓ = xg/2y
Explanation:
In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.
By newton equation of motion we know that
y = v₀ t - ½ g t²
t = 2y / g
This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance
vₓ = x/t
vₓ = xg/2y
vₓ = xg/2y
Where we assume that x and y are known.
The answer is he weighs 187.39 LBS/Pounds
