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:
1027.2 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32.2 ft/s


The height the tomato would fall is 450+577.2 = 1027.2 m
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
Answer:
560 N/m
Explanation:
F = kx
75 N = k (0.61 m − L)
210 N = k (0.85 m − L)
Divide the equations:
2.8 = (0.85 − L) / (0.61 − L)
2.8 (0.61 − L) = 0.85 − L
1.708 − 2.8L = 0.85 − L
0.858 = 1.8L
L = 0.477
Plug into either equation and find k.
75 = k (0.61 − 0.477)
k = 562.5
Rounded to two significant figures, k = 560 N/m.