The current in the second wire is 1.37 A.
<h3>Current in the second wire</h3>
The current in the second wire is calculated from the magnetic force between the two wires.
F/L = (μI₁I₂)/(2πr)
where;
- μ is permeability of free space
- I₂ is current in the second wire
- F/L is force per unit length
- r is the distance between the two wires
μI₁I₂ = (F/L 2πr)
I₂ = ((F/L 2πr) / (μI₁)
I₂ = (7.7 x ⁻⁵ x 2π x 0.082)/(4π x 10⁻⁷ x 23)
I₂ = 1.37 A
Thus, the current in the second wire is 1.37 A.
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Back emf is 85.9 V.
<u>Explanation:</u>
Given-
Resistance, R = 3.75Ω
Current, I = 9.1 A
Supply Voltage, V = 120 V
Back emf = ?
Assumption - There is no effects of inductance.
A motor will have a back emf that opposes the supply voltage, as the motor speeds up the back emf increases and has the effect that the difference between the supply voltage and the back emf is what causes the current to flow through the armature resistance.
So if 9.1 A flows through the resistance of 3.75Ω then by Ohms law,
The voltage across the resistance would be
v = I x R
= 9.1 x 3.75
= 34.125 volts
We know,
supply voltage = back emf + voltage across the resistance
By plugging in the values,
120 V = back emf + 34.125 V
Back emf = 120 - 34.125
= 85.9 Volts
Therefore, back emf is 85.9 V.
Answer:
95m
Explanation:
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Answer:
I dont the child mass...you should substitute that value to (m)
then you can get your answer
Just put a full picture clearly we can’t really see