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AlexFokin [52]
3 years ago
8

A mini rail gun made of a copper rod of mass m and radius r rests on two parallel copper rails that are a distance L apart and o

f length d. A magnetic field of magnitude B is directed perpendicular to the rod and the rails. When current is applied to the parallel copper rails, the rod rolls along the rails without slipping. If the rod starts from rest, what is the speed of the rod as it leaves the rails?
Physics
1 answer:
ira [324]3 years ago
7 0

Answer:

Explanation:

Magnetic Force on the rod  F = Bi L

Work done by this force = F X d

=  Bi Ld

This energy is converted into both rotational and linear kinetic energy

= 1/2 I ω² + 1/2 mv²

= 1/2 x 1/2 m r²ω²+ 1/2 mv²

= 1/4 m v² + 1/2 mv²

= 3/4 mv²

So according to conservation of energy

Bi Ld =  3/4 mv²

v = \sqrt{\frac{4BiLd}{3m} }

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A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
Advocard [28]

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

  • t=0.476v
  • t=1.967v
  • V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

at

  • t = 1.0 seconds
  • 5.0 seconds
  • 20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

  • t=0.476v
  • t=1.967v
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Read more about  Voltage

brainly.com/question/14883923

Complete Question

A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.

Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.

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