Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
Answer:
a) Emf of battery = 15.6 V
b) Current through the starting motor = 104 A
Explanation:
The image for the question seem not to be attached but, when the starting motor wasn't connected, the lights are in series with the battery connection and when the starting motor is connected, it is in parallel with the lights.
a) For this part, we use the information when the starting motor isn't connected.
Let the emf of the battery be E = ?
The voltage across the terminals be V = 15 V
Let the internal resistance of the battery be R = 0.05 ohms
And the current in the circuit be I = 12 A
Since the internal resistance is modelled to be in series with the battery, the loop equation for this circuit will be
E = V + IR
E = 15 + (12)(0.05) = 15 + 0.6 = 15.6 V
From this information now, we can obtain the resistance of the lights.
From Ohm's law,
Voltage across the lights = 15 V
Current through the lights = 12 A
Resistance of the lights = R₀
V = IR₀
R₀ = (15/12) = 1.25 ohms
b) Now, the starting motor is connected and the resistance in the circuit is if the form
Internal resistance in series with the parallel combination of resistance of starting motor and the resistance of the lights.
Current through the lights = 8.0 A
Voltage across the lights = IR₀ = 8×1.25 = 10 V.
Since the starting motor is in parallel with the lights, they will have the same voltage across them, that is, 10 V
But the voltage across the internal resistance = (Emf of battery) - (Voltage across the parallel combination of lights and starting motor) = 15.6 - 10 = 5.6 V
Current through the internal resistance = (5.6/0.05) = 112 A
Current through the starting motor = 112 - 8 = 104 A
Hope this Helps!!!
