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:

V=IR however in this case we will rearrange it into R= V/I. Which leads to R=15.0/0.30. And your answer becomes 50.
Answer:
A. T=15.54 °C
B. Q/A= 0.119 W/m2
Explanation:
To solve this problem we need to use the Fourier's law for thermal conduction:

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

Then, to find the rate of heat flow per square meter, we have:


The complete question was calculate the period T assuming the smallest amplitude.
Using the equation;
T = 2 π√(L/g)
Where T is the period in seconds, L is the length of the rod or wire in meters and g is the acceleration due to gravity.
Hence; T = 2×3.14 × √(2/9.81)
= 6.28 × 0.4515
= 2.836 seconds