Explanation:
It is given that,
Spring constant of the spring, k = 15 N/m
Amplitude of the oscillation, A = 7.5 cm = 0.075 m
Number of oscillations, N = 31
Time, t = 15 s
(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

Total number of oscillation per unit time is called frequency of oscillation. Here, 


m = 0.0895 kg
or
m = 89 g
(b) The maximum speed of the ball that is given by :





Hence, this is the required solution.
A change in momentum is called impulse.
Answer:
P = 40.7kPa
Explanation:
To find the pressure on a surface 6 meter below you use the following formula, which takes into account the heights in which pressures are measured and also the density of the fluid and the gravitational acceleration:
(1)
P2: pressure for a height of -6 m = ?
P1: pressure for a height of -2 m = 1.5kPa = 1500 Pa
ρ: density of water = 1000kg/m^3
g: gravitational acceleration = 9.8 ms^2
y2: -6m
y1: -2m
(the height is measure from the water level, because of that, the heights are negative)
You solve the equation (1) for P1:
(2)
Next, you replace the values of all variables in equation (2):

hence, the pressure on a surface 6 m below the water level is 40.7kPa
Nuclear, thermal and I think cemical idk