Around how many miles is a light year
2 answers:
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Here we can use Newton's II law to find the acceleration of the car
here we know that
as we know that two forces are equal in magnitude but opposite in direction so in order to find net force we need to subtract it
now from above equation
so here option A is correct
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!!!
