Answer:
Explanation:
Work done by torque is given as
Word one = torque × angular displacement
W = τ × θ
Given that,
τ = 2000Nm
Mass of motor = 300kg
Radius r = 55cm = 0.55m
Work done by wheel in first t= 23second.
Now we need to find the angular displacement
We know that,
τ = I•α
Moment of inertia of wheel
I = MR²
I = 300 × 0.55²
I = 90.75 kgm²
Then, τ = I•α
α = τ / I
α = 2000/90.75
α = 22.04rad/s²
Then, using circular motion,
∆θ = wit + ½αt²
wi = 0rad/s
∆θ = ½αt²
∆θ = ½ × 22.04 × 23²
∆θ = 5829.2 rad.
Then,
Work done?
W = τ × θ
W = 2000 × 5829.2
W = 1.17 × 10 ^7 J
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Answer:9.17 m/s^2
Explanation:
mass=1200kg
Force=11 x 10^3 N
Acceleration=force ➗ mass
Acceleration=11 x 10^3 ➗ 1200
Acceleration=9.17
Acceleration=9.17 m/s^2
Answer:
Explanation:
The amplitude of he combined wave is:
A, is the amplitude from the identical harmonic waves
B, is the amplitude of the resultant wave
θ, is the phase, between the waves
The amplitude of the combined wave must be 0.6A:
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
