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denis23 [38]
3 years ago
11

A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire

and rail is 0.35 Ω. Pulling the wire with a force of 1.0 N causes 4.0 W of power to be dissipated in the circuit.
a. What is the speed of the wire when pulled with a force of 1.0 N?
b. What is the strength of the magnetic field?
Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

a)  v=4.0m/s

b)  B=2.958T

Explanation:

From the question we are told that:

Wire Length l=10cm=>0.10m

Resistance R=0.35

Force F=1.0N

Power P=4.0W

a)

Generally the equation for Power is mathematically given by

P=Fv

Therefore

v=\frac{P}{F}

v=\frac{4.0}{1.0}

v=4.0m/s

b)

Generally the equation for Magnetic Field is mathematically given by

B=\frac{\sqrt{PR}}{vl}

B=\frac{\sqrt{4*0.35}}{4*0.10}

B=2.958T

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1,3

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<em> B.0</em>

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Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?
lubasha [3.4K]
1) The equivalent resistance of two resistors in parallel is given by:
\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}
so in our problem we have
\frac{1}{R_{eq}} =  \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1}
and the equivalent resistance is
R_{eq} =  \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega

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3 years ago
Read 2 more answers
1 point
Yuliya22 [10]
PE= 3kg x 10N/kg x 10m
= 300J
8 0
3 years ago
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