A motor spins upward the flywheel with a persistent torque
of 50N⋅m.
What time does it take the flywheel to get to the top speed?
From the equation:
Tj = J*dω/dt
you can get the two equations:
Δt1= J1*Δω/Tj = 240*125.66/50 = 603.17 sec
Δt2= J2*Δω/Tj = 120*125.66/50 = 301.58 sec
Answer:
Efficiency = 10.2 %
Explanation:
Given the following data;
Mass = 70 kg
Height = 50 m
Velocity = 10 m/s
We know that acceleration due to gravity is equal to 9.8 m/s².
To find the efficiency of energy conversion from potential to kinetic;
First of all, we would determine the potential energy;
P.E = mgh
P.E = 70 * 9.8 * 50
P.E = 34300 J
For the kinetic energy;
K.E = ½mv²
K.E = ½ * 70 * 10²
K.E = 35 * 100
K.E = 3500
Therefore, Input energy, I = 34300 J
Output energy, O = 3500 J
Next, we find the efficiency;
Efficiency = O/I * 100
Substituting into the formula, we have;
Efficiency = 3500/34300 * 100
Efficiency = 0.1020 * 100
Efficiency = 10.2 %
Momentum equation is
change in momentum = mass•initial velocity•final velocity
so....
p=700(15) because your initial is 30, and your final in 15, so you subtract! hope that helped!
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