a) 120 s
b) v = 0.052R [m/s]
Explanation:
a)
The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).
The graph of the problem is missing, find it in attachment.
To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.
The first point we take is t = 0, when the position of the book is x = 0.
Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.
Therefore, the period is
T = 120 s - 0 s = 120 s
b)
The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.
The perimeter of the wheel is:

where R is the radius of the wheel.
The period of revolution is:

Therefore, the tangential speed of the book is:

Answer:
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Explanation:
Total force required = Mass x Acceleration,
F = ma
Here we need to consider the system as combine, total mass need to be considered.
Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg
We need to accelerate the group of rocks from the road at 0.250 m/s²
That is acceleration, a = 0.250 m/s²
Force required, F = ma = 1744 x 0.25 = 436 N
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Answer: 9130 joules
Explanation:
Workdone by wheelbarrow = ?
Time = 11 seconds
Power = 830 watts
Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.
i.e Power = (workdone/time)
830 watts = Workdone / 11 seconds
Workdone = 830 watts x 11 seconds
Workdone = 9130 joules
Thus, 9130 joules of work is required to get the wheelbarrow across the yard.
The final answer is B hope its helps