Answer:
reading would be 5.413 m.
Explanation:
Given:-
- The actual distance from ruler to an object is d = 24.0 m
- The adiabatic bulk modulus, B = 2.37 *10^9 Pa
- The density of seawater, ρ = 1025 kg/m^3
- The preset value of speed of sound in air, v_th = 343 m/s.
Find:-
Determine the distance reading that the ruler displays.
Solution:-
- We will first determine the actual speed of the sound ( v_a) in sea-water which can be determined from the following formula:
v_a = √ (B / ρ )
- Plug in the values in the relationship above and compute v_a:
v_a = √ ( 2.37 *10^9 / 1025 )
v_a = 1520.59038 m/s
- The time taken (t) for for the sound to travel from source(ruler) to an object which is d distance away.
d = v_a*t
t = d / v_a
t = 24.0 / 1520.59038
t = 0.01578 s
- The distance reading on the ruler would be preset speed (v_th) of sound in air multiplied by the time taken(t).
reading = v_th*t
reading = (343)*(0.01578)
= 5.413 m
Answer:
A. Time, t = 4.35s
B. Height of its fall, S = 92.72m
Explanation:
Vo = 0 m/s
Vi = 0.46h m/s
S = h m
a = 9.81 m/s2
To calculate the time taken, we need to get the value of the distance, h.
Using the equations of motion,
Vi^2 = Vo^2 + 2aS
Where Vi = final velocity
Vo = initial velocity
a = acceleration due to gravity
S = height of its fall
(0.46h)^2 = 0 + 2*9.81*h
0.2116h^2 = 19.62h
h = 19.62/0.1648
= 92.722 m
To calculate the time,
S = Vo*t +(1/2)*a*t^2
92.772 = 0 + (1/2)*9.81*t^2
t^2 = 185.44/9.81
= 18.904
t = sqrt(18.904)
= 4.348 s
Answer: observed frequency (f') = 511.97Hz
Explanation: when there is a relative motion between an observer and a sound source, the frequency of sound wave perceived by the observer is different from the frequency of originally sent out by the sound source.
This is called Doppler effect and given mathematically below as
f' = (v + v') /(v- vs) * f
f' = observed frequency
v = speed of sound in air = 340m/s
v' = velocity of observer= 45m/s
vs = velocity of source relative to observer = - 36m/s ( the negative sign came as a result of the fact that the velocity of the source is in opposite direction to the velocity of the observer)
f = original frequency of sound source = 500Hz
f' = (340 + 45)/{340 -(-36)} * 500
f' = 385/ (340 + 36) * 500
f' = 385/ 376 * 500
f' = 1.0239 * 500
f' = 511.97Hz
At that point it is no longer trying to uncompress nor is it trying to stretch. This is the same thing as a pendulum at the bottom of its swing, no longer falling but not yet rising against gravity. Thus the kinetic energy there is the same as the potential energy when it is compressed. The energy of compression is

This gives E=0.5(37)(0.2)²=
0.74JThis is the same as the kinetic energy when it is at natural length