Answer:
(A) Torque required is 21.205 N-m
(b) Wok done will be equal to 1199.1286 j
Explanation:
We have given moment of inertia
Wheel deaccelerate from 135 rpm to 0 rpm
135 rpm =
Time t = 8 sec
So angular speed and
Angular acceleration is given by
Torque is given by torque
Work done to accelerate the vehicle is
Answer:
Work done = 838470 Joules.
Explanation:
Given the following data;
Power = 19 hp
Time = 1 minute to seconds = 60 seconds.
Next, we would convert the unit of power in "hp" to "Watt."
1 Watt = 0.00135962 horsepower
x Watt = 19 horsepower
Cross-multiplying, we have;
19 = 0.00135962x
x = 19/0.00135962
x = 13974.5 Watts.
Now, to find the work done in moving the mower;
Work done = power * time
Substituting into the formula, we have;
Work done = 13974.5 * 60
Work done = 838470 Joules.
Answer:
a) Q1= Q2= 11.75×10^-6Coulombs
b) Q1 =15×10^-6coulombs
Q2 = 38.75×10^-6coulombs
Explanation:
a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as
1/Ct = 1/C1 + 1/C2
Given C1 = 3.00 μF C2 = 7.75μF
1/Ct = 1/3+1/7.73
1/Ct = 0.333+ 0.129
1/Ct = 0.462
Ct = 1/0.462
Ct = 2.35μF
V = 5.00Volts
To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance
Q = 2.35×10^-6× 5
Q = 11.75×10^-6Coulombs
Since same charge flows through a series connected capacitors, therefore Q1= Q2=
11.75×10^-6Coulombs
b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2
C = 3.00 μF + 7.75 μF
C = 10.75 μF
For 3.00 μF capacitance, the charge on it will be Q1 = C1V
Q1 = 3×10^-6 × 5
Q1 =15×10^-6coulombs
For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5
Q2 = 38.75×10^-6coulombs
Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.
PE = mgh = (60 kg)(9.81 m/s²)(10 m) = 5,886 J
which is pretty close to 5900