(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: 7.41 m/s
Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.
Kinetic energy = mv²/2, potential energy = mgh
mv²/2 = mgh
v²/2 = gh
v² = 2gh
v = √2gh
Where g = 9.8 m/s², h = 2.80m
v = √2×9.8×2.8 = 7.41 m/s
Answer:
t = 0.1111 s
Explanation:
Let's reduce the magnitudes to the SI system
d = 120 mm (1m / 1000 mm)
d= 0.120 m
w = 540 rpm (2pi rad / 1 rev) (1 min / 60s)
w= 56.55 rad / s
When at maximum speed we can use angular kinematic relationships to find the time for a sperm revolution with zero angular acceleration
W = θ / t
t = θ / w
t = 2π / 56.55
t = 0.1111 s
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I hope I helped! -Wajiha</span>
