Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks.
W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ
Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)
Answer:
Explanation:
In Michelson interferometer , two light waves from different directions are made to overlap so that fringes are formed on the screen due to interference . In it, two monochromatic and coherent light are made to overlap which have some path difference or phase difference. They form dark and bright fringes .
Now when a match stick is lit in the path of a wave , the fringes will disappear and an general illumination will be observed on the screen as the light from the lit match stick will not be coherent . Incoherent light can not form stable fringes.
There are 10⁹ picoseconds in 1 Ms
1 picosecond= 10¹² s
1 Ms = 10⁻³ s
so the number of picoseconds in one Ms=(10⁻³ s/1 Ms) * (10¹² Ps/ 1 s)=10⁹
Thus there are 10⁹ picoseconds in 1 Ms
Answer:
d. 6.0 m
Explanation:
Given;
initial velocity of the car, u = 7.0 m/s
distance traveled by the car, d = 1.5 m
Assuming the car to be decelerating at a constant rate when the brakes were applied;
v² = u² + 2(-a)s
v² = u² - 2as
where;
v is the final velocity of the car when it stops
0 = u² - 2as
2as = u²
a = u² / 2s
a = (7)² / (2 x 1.5)
a = 16.333 m/s
When the velocity is 14 m/s
v² = u² - 2as
0 = u² - 2as
2as = u²
s = u² / 2a
s = (14)² / (2 x 16.333)
s = 6.0 m
Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.
The correct option is d
