Answer:
The speed is the same at 1.5 m/s while
The work done by the force F is 0.4335 J
Explanation:
Here we have angular acceleration α = v²/r
Force = ma = 2.8 × 1.5²/r₁
and ω₁ = v₁/r₁ = ω₂ = v₁/r₂
The distance moved by the force = 600 - 300 = 300 mm = 0.3 m
If the velocity is constant
The speed is 1.5 m/s while the work done is
2.8 × 1.5²1/(effective radius) ×0.3
r₁ = effective radius
2.8*9.81 = 2.8 × 1.5²/r₁
r₁ = 0.229
The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J
Answer:
The growth of crack formation in a corrosive environment.
Explanation:
Answer:
0.75 kg
Explanation:
c = Damping coefficient = 3 kg/s
x = Displacement of spring = 0.5 m
F = Force = 1.5 N
From Hooke's law we get
In the case of critical damping we have the relation
Mass that would produce critical damping is 0.75 kg.