(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
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Answer:
Sonia's grandfather was not able to see clearly after returning from his walk on a bright sunny day. He was about to hit a chair when Sonia held him and guided him to the nearby sofa, in sunny environment grandfather's eye is set according to high brightness, his pupils become small to lower the number of light rays entering his eye, when he returned from sunny environment to his house which was having low brightness so the pupils should enlarge to absorb more light to see clearly but due to old age his Ciliary muscle of the eye would have worn out and due to this poor coordination the image was not clear.
A becuz its at da it dont got no wa
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Hope this helps
Answer:
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