Answer:
Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W
Explanation:
As we know that the current in the circuit at given instant of time is
i = 2.0 mA
R = 10 k ohm
now we know by ohm's law
![V = iR](https://tex.z-dn.net/?f=V%20%3D%20iR)
![V = (2 mA)(10 kohm)](https://tex.z-dn.net/?f=V%20%3D%20%282%20mA%29%2810%20kohm%29)
![V = 20 volts](https://tex.z-dn.net/?f=V%20%3D%2020%20volts)
so voltage across the capacitor + voltage across resistor = V
![V_c + 20 = 50](https://tex.z-dn.net/?f=V_c%20%2B%2020%20%3D%2050)
![V_c = 30 V](https://tex.z-dn.net/?f=V_c%20%3D%2030%20V)
Now we know that
![U = \frac{q^2}{2C}](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7Bq%5E2%7D%7B2C%7D)
here rate of change in energy of the capacitor is given as
![\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdU%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bq%7D%7BC%7D%20%5Cfrac%7Bdq%7D%7Bdt%7D)
![\frac{dU}{dt} = (30)(2 mA)](https://tex.z-dn.net/?f=%5Cfrac%7BdU%7D%7Bdt%7D%20%3D%20%2830%29%282%20mA%29)
![\frac{dU}{dt} = 0.06 W](https://tex.z-dn.net/?f=%5Cfrac%7BdU%7D%7Bdt%7D%20%3D%200.06%20W)
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π
Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
= (15π rad/s)*(1 rev/2π rad)
= 7.5 rev/s
Answer: 7.5 revolutions per second.
Answer:
a) ![V_{2/1}=0.8m/s](https://tex.z-dn.net/?f=V_%7B2%2F1%7D%3D0.8m%2Fs)
b) The second runner will win
c) d = 10.54m
Explanation:
For part (a):
![V_{2/1} = V_{2} - V_{1} = 0.8m/s](https://tex.z-dn.net/?f=V_%7B2%2F1%7D%20%3D%20V_%7B2%7D%20-%20V_%7B1%7D%20%3D%200.8m%2Fs)
For part (b) we will calculate the amount of time that takes both runners to cross the finish line:
![t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s](https://tex.z-dn.net/?f=t_%7B1%7D%20%3D%20%5Cfrac%7BX_%7B1%7D%7D%7BV_%7B1%7D%7D%3D%5Cfrac%7B250%7D%7B3.45%7D%3D72.46s)
![t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s](https://tex.z-dn.net/?f=t_%7B2%7D%20%3D%20%5Cfrac%7BX_%7B2%7D%7D%7BV_%7B2%7D%7D%3D%5Cfrac%7B250%2B45%7D%7B4.25%7D%3D69.41s)
Since it takes less time to the second runner to cross the finish line, we can say the she won the race.
For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:
X1 = V1*t2 = 239.46m Since the finish line was 250m away:
d = 250m - 239.46m = 10.54m
Answer:
(a) Current flowing through truck battery is 180 A
(b) Time taken in calculator is 333.33 s
Explanation:
(a) Given:
The charge on the truck battery,q = 720 C
Time, t = 4.00 s
Consider I be the current flowing through truck battery.
The relation between current, charge and time is:
I = q/t
Substitute the suitable values in the above equation.
![I=\frac{720}{4}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B720%7D%7B4%7D)
I = 180 A
(b) Given:
The charge on the calculator,q = 7.00 C
The current flowing through calculator, I = 0.3 mA = 0.3 x 10⁻³ A
Consider t be the time.
The relation between current, charge and time is:
t = q/I
Substitute the suitable values in the above equation.
![t=\frac{1}{0.3\times10^{-3} }](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B1%7D%7B0.3%5Ctimes10%5E%7B-3%7D%20%7D)
I = 333.33 s