(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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Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds
Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3 meters per second
Answer:
No
Explanation:
In such situations we cannot determine which one is more valid as both serves the purpose well.
Two theories are carried out in different environment and circumstance keeping different parameters and one can opt for any number of ways to carry out that experiment but what matter at the end is the accuracy they bring.
Each of the theory is a new discovery and follows all the possible logical rules hence it is not possible to decide which one is more valid.
Answer:
0.79 s
Explanation:
We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:
The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that 
Now, we find the acceleration:
Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:
