Answer:
60 rad/s
Explanation:
∑τ = Iα
Fr = Iα
For a solid disc, I = ½ mr².
Fr = ½ mr² α
α = 2F / (mr)
α = 2 (20 N) / (0.25 kg × 0.30 m)
α = 533.33 rad/s²
The arc length is 1 m, so the angle is:
s = rθ
1 m = 0.30 m θ
θ = 3.33 rad
Use constant acceleration equation to find ω.
ω² = ω₀² + 2αΔθ
ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)
ω = 59.6 rad/s
Rounding to one significant figure, the angular velocity is 60 rad/s.
Answer:
F = -4567.40 N
Explanation:
Given that,
The power developed by the engine, P = 196 hp
1 hp = 746 W
196 hp = 146157 W
Speed of the car, v = 32 m/s
Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :



F = -4567.40 N
So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.
Answer:
0.52 Nm
Explanation:
A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T
Angle between the plane of loop and magnetic field = 30 Degree
Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree
θ = 60°
Torque = N i A B Sinθ
Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60
Torque = 0.52 Nm
Answer:
0.231 N
Explanation:
To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

So the force acting on the other end to generate this torque mush be:

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