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erastovalidia [21]
3 years ago
14

A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below

the horizontal, what is the magnetic flux through the desk surface?
Physics
1 answer:
AlexFokin [52]3 years ago
3 0

Answer:

The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

\phi=BA\costheta

Where, B = magnetic field

A = area

Put the value into the formula

\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}

\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927

\phi=6.6\times10^{-4}\ T-m^2

Hence, The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

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A spring has a force constant of 310.0 N/m. (a) Determine the potential energy stored in the spring when the spring is stretched
skad [1K]

Answer:

(a) 0.2618 J

(b)  0.1558 J

(c) 0 J

Explanation:

from Hook's Law,

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Ep = 1/2ke² ......................... Equation 1

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(a) When The spring is stretched to 4.11 cm,

Given: k = 310 N/m, e = 4.11 cm = 0.0411 m

Substituting these values into equation 1

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(b) When the spring is stretched 3.17 cm

e = 3.17 cm = 0.0317 m.

Ep = 1/2(310)(0.0317)²

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(c) When the spring is unstretched,

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Hope this helps!

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