Answer:
The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.
Explanation:
As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.
By definition, the kinetic energy is:
k = 0.5*mV^2
Expressing the initial and final kinetic energy for cars A and B:


Since the masses are equals:

For the known velocities, the kinetics energies result:




The lost energy in the collision is the difference between the initial and final kinectic energies:


Finally the relation between the lost and the initial kinetic energy:


Answer:
150J
Explanation:
work output/work input=100%
so just make work output the subject
C) When both objects have the same temperature.
<em>Hope this helps!</em>
Answer:
a)Amplitude ,A = 2 mm
b)f=95.49 Hz
c)V= 30 m/s ( + x direction )
d) λ = 0.31 m
e)Umax= 1.2 m/s
Explanation:
Given that
![y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]](https://tex.z-dn.net/?f=y%3D2%5C%20mm%5C%20sin%5B%2820m%5E%7B-1%7D%29x-%28600s%5E%7B-1%7D%29t%5D)
As we know that standard form of wave equation given as

A= Amplitude
ω=Frequency (rad /s)
t=Time
Φ = Phase difference
![y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]](https://tex.z-dn.net/?f=y%3D2%5C%20mm%5C%20sin%5B%2820m%5E%7B-1%7D%29x-%28600s%5E%7B-1%7D%29t%5D)
So from above equation we can say that
Amplitude ,A = 2 mm
Frequency ,ω= 600 rad/s (2πf=ω)
ω= 2πf
f= ω /2π
f= 300/π = 95.49 Hz
K= 20 rad/m
So velocity,V
V= ω /K
V= 600 /20 = 30 m/s ( + x direction )
V = f λ
30 = 95.49 x λ
λ = 0.31 m
We know that speed is the rate of displacement

![U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]](https://tex.z-dn.net/?f=U%3D2%5C%20mm%5C%20sin%5B%2820m%5E%7B-1%7D%29x-%28600s%5E%7B-1%7D%29t%5D)
![U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s](https://tex.z-dn.net/?f=U%3D1200%5C%20cos%5B%2820m%5E%7B-1%7D%29x-%28600s%5E%7B-1%7D%29t%5D%5C%20mm%2Fs)
The maximum velocity
Umax = 1200 mm/s
Umax= 1.2 m/s