Answer:
(a) ![V=11.86\ V](https://tex.z-dn.net/?f=V%3D11.86%5C%20V)
(b) ![V=9.76\ V](https://tex.z-dn.net/?f=V%3D9.76%5C%20V)
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:
![V=R.I](https://tex.z-dn.net/?f=V%3DR.I)
(a) The electromagnetic force of the battery is
and its internal resistance is
. Knowing the equivalent resistance of the headlights is
, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:
![\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)](https://tex.z-dn.net/?f=%5Cvarepsilon%3Di.R_i%2Bi.R_e%3Di.%28R_i%2BR_e%29)
Solving for i
![\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A](https://tex.z-dn.net/?f=%5Cdisplaystyle%20i%3D%5Cfrac%7B%5Cvarepsilon%7D%7B%20R_i%2BR_e%7D%3D%5Cfrac%7B12%7D%7B0.06%2B5.2%7D%3D2.28%5C%20A)
i=2.28\ A
The potential difference across the headlight bulbs is
![V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V](https://tex.z-dn.net/?f=V%3D%5Cvarepsilon%20%20-i.R_i%3D12%5C%20V-2.28%5C%20A%5Ccdot%200.06%5C%20%5COmega%3D11.86%5C%20V)
![V=11.86\ V](https://tex.z-dn.net/?f=V%3D11.86%5C%20V)
(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is
![V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V](https://tex.z-dn.net/?f=V%3D%5Cvarepsilon%20%20-i.R_i%3D12%5C%20V-37.28%5C%20A%5Ccdot%200.06%5C%20%5COmega%3D9.76%5C%20V)
Answer:
nba young bruuhh
Explanation:
have a great day and plz mark brainliest!
Answer:
Magnitude the net torque about its axis of rotation is 2.41 Nm
Solution:
As per the question:
The radius of the wrapped rope around the drum, r = 1.33 m
Force applied to the right side of the drum, F = 4.35 N
The radius of the rope wrapped around the core, r' = 0.51 m
Force on the cylinder in the downward direction, F' = 6.62 N
Now, the magnitude of the net torque is given by:
![\tau_{net} = \tau + \tau'](https://tex.z-dn.net/?f=%5Ctau_%7Bnet%7D%20%3D%20%5Ctau%20%2B%20%5Ctau%27)
where
= Torque due to Force, F
= Torque due to Force, F'
![tau = F\times r](https://tex.z-dn.net/?f=tau%20%3D%20F%5Ctimes%20r)
![tau' = F'\times r'](https://tex.z-dn.net/?f=tau%27%20%3D%20F%27%5Ctimes%20r%27)
Now,
![\tau_{net} = - F\times r + F'\times r'](https://tex.z-dn.net/?f=%5Ctau_%7Bnet%7D%20%3D%20-%20F%5Ctimes%20r%20%2B%20F%27%5Ctimes%20r%27)
![\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm](https://tex.z-dn.net/?f=%5Ctau_%7Bnet%7D%20%3D%20-%204.35%5Ctimes%201.33%20%2B%206.62%5Ctimes%200.51%20%3D%20-%202.41%5C%20Nm)
The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.
Now, the magnitude of the net torque:
![|\tau_{net}| = 2.41\ Nm](https://tex.z-dn.net/?f=%7C%5Ctau_%7Bnet%7D%7C%20%3D%202.41%5C%20Nm)
Answer:
3,00,000
Explanation:
because 1 m =100m so, 3000x100=300000
Average acceleration = (change in speed) / (time for the change) .
Change in speed = (ending speed) - (beginning speed)
= (9.89 miles/hour) - (2.35 yards/second) = 26,839.2 ft/hr
Acceleration = (26,839.2 ft/hr) / (4.67 days) = 2,873.58 inch/hour²