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3 years ago

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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?

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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From the question we are told that

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3 years ago

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Answer:

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$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

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Now we can substitute the provided values into our equation. So we get:

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2 years ago

A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically

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Answer:

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P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

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W =

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W = -67.47 (5.698 - 8.4)

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3 years ago