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2 years ago

what is the magnitude of the vector described below? 13 m/s to the east a. east b. meters per second c. 13 m/s d. meters

Physics

1 answer:

Katen [24]2 years ago

3 0

Answer:

Explanation:

Magnitude of any quantity is the measurable value of the quantity. While the direction of the given quantity is the specific pointing direction of position or the angle at which it move.

The magnitude of the vector described below? 13 m/s to the east will be 13 m/s

While the direction will be eastward.

Therefore, the magnitude is 13 m/s

The correct answer is option C

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is $\epsilon_{max}= 26.8 V$

The emf induced at t = 1.00 s is $\epsilon = 24.1V$

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Explanation:

From the question we are told that

The number of turns is N = 44 turns

The length of the coil is $l = 15.0 cm = \frac{15}{100} = 0.15m$

The width of the coil is $w = 8.50 cm =\frac{8.50}{100} =0.085 m$

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$\epsilon = 26.8 sin (64 * 1)$

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The maximum induced emf can also be represented mathematically as

$\epsilon_{max} = \frac{d \o}{dt}|_{max}$

Where $\o$ is the magnetic flux and $\frac{d \o}{dt}|_{max}$ is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

$\frac{d \o}{dt}|_{max} =26.8V$

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