Answer:
Explanation:
First of all we shall find the velocity at equilibrium point of mass 1.2 kg .
It will be ω A , where ω is angular frequency and A is amplitude .
ω = √ ( k / m )
= √ (170 / 1.2 )
= 11.90 rad /s
amplitude A = .045 m
velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s
= .5355 m /s
At middle point , no force acts so we can apply law of conservation of momentum
m₁ v₁ = ( m₁ + m₂ ) v
1.2 x .5355 = ( 1.2 + .48 ) x v
v = .3825 m /s
= 38.25 cm /s
Let new amplitude be A₁ .
1/2 m v² = 1/2 k A₁²
( 1.2 + .48 ) x v² = 170 x A₁²
( 1.2 + .48 ) x .3825² = 170 x A₁²
A₁ = .0379 m
New amplitude is .0379 m
The correct answer should be bounced off since the light ray hit the surface and reflected towards a new location. It therefore bounced off.
Answer:
1) It expresses the rate (top speed) at which it can move with time.
2) P = 20 W
3) h = 18 km
Explanation:
1) Power is the rate of transfer of energy.
⇒ Power = 
i.e P = 
Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.
2) m = 400g = 0.4 kg
t = 20 s
h = 100m
g = 10 m/
P = 
= 
= 
P = 20 W
3) u = 600 m/s
g = 10 m/
From the third equation of free fall,
=
- 2gh
V is the final velocity, U is the initial velocity, h is the height.
0 =
- 2 x 10 x h
0 = 360000 - 20h
20h = 360000
h = 
= 18000
h = 18 km
The maximum height of the bullet would be 18 km.