Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
Answer:
the propagation velocity of the wave is 274.2 m/s
Explanation:
Given;
length of the string, L = 1.5 m
mass of the string, m = 0.002 kg
Tension of the string, T = 100 N
wavelength, λ = 1.5 m
The propagation velocity of the wave is calculated as;

Therefore, the propagation velocity of the wave is 274.2 m/s
Answer:
Explanation:
Given
mass of spring 
extension in spring 
downward velocity 
Position in undamped free vibration is given by

where 
also 



it is given


substituting values we get







I think that in order for work to be done, the object must move in the direction of the force and move over a distance.