Answer:
(a) A = m/s^3, B = m/s.
(b) dx/dt = m/s.
Explanation:
(a)

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.
(b) 
This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.
Answer:
993.52 Hz
Explanation:
The frequency of sound emitted by the stationery train is 1057 Hz.
The car travels away from the train at 20.6 m/s.
The frequency the observer hears is given by the formula:

where v = velocity of sound = 343 m/s
vo = velocity of observer
f = frequency from source
This phenomenon is known as Doppler's effect.
Therefore:

The frequency heard by the observer is 993.52 Hz.
To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

Where,
P = Axial Load
l = Gage length
A = Cross-sectional Area
E = Modulus of Elasticity
Our values are given as,
l = 3.5m
D = 0.028m

E = 200GPa

Replacing we have,




Therefore the change in length is 1.93mm
The μs between the clock and floor is 650(M*g) and the μk between the clock and the floor is 560(M*g)
C I believe that would be it