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pav-90 [236]
3 years ago
11

A truck is carrying a steel beam of length 13.0 m on a freeway. An accident causes the beam to be dumped off the truck and slide

horizontally along the ground at a speed of 20.0 m/s. The velocity of the center of mass of the beam is northward while the length of the beam maintains an east–west orientation. The vertical component of the Earth's magnetic field at this location has a magnitude of 38.0 µT. What is the magnitude of the induced emf between the ends of the beam?
Physics
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

emf= 9.88 \times 10^{-3} T

Explanation:

Given:

  • Length of the beam, l=13\,m
  • speed of the beam, v=20\,m.s^{-1}
  • magnitude of the vertical magnetic field, B=38\times 10^{-6} T

According to the Faraday's law the emf induced in a rod passing transversely through a magnetic field is given as:

emf= l.v.B

emf=13\times 20\times 38\times 10^{-6}

emf= 9.88 \times 10^{-3} T

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7 0
2 years ago
Please help me guys never mind the calculations ​
vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

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\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

I hope I helped you^_^

7 0
3 years ago
Discuss the force that exists between the Earth and the moon by referring to the mass of each.
Fudgin [204]
The word gravity is used to describe the gravitational pull (force) an object experiences on or near the surface of a planet or moon. The gravitational force is a force that attracts objects with mass towards each other. Any object with mass exerts a gravitational force on any other object with mass.

Hope it answers your question!

Brainliest would be nice but of course you don’t gotta :)
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Answer:

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Explanation:

Mass of the sportscar= 1500 kg

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Hence, let the Force acting on it be F

We\ know\ that,\\Force=Mass*Acceleration\\F=ma\\\\Here,\\F=1500*5\\=7500 kg m/s^2\ or\ 7500\ Newtons

4 0
3 years ago
what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el
Evgen [1.6K]

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Because of the speeds of protons! :D

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