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4 years ago

When a generator rotates a coil of wire in a magnetic field, which of the following is produced?

Physics

1 answer:

Sloan [31]4 years ago

4 0

The correct answer is an ''electric current''.

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The basic unit of time is second - is it true or false ??

Lorico [155]

It is true. Time is always measured in seconds.

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3 years ago

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A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the

Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

$a = \frac{v^{2} }{r}$

Where $a$ is the centripetal acceleration

$v$ is the speed

and $r$ is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be $v_{2}$ , that is

$v_{2} = 2v$

and without changing the length of the string means radius $r$ is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be $a_{2}$ .

Then we can write that

$a_{2} = \frac{v_{2}^{2} }{r}$

From

$a = \frac{v^{2} }{r}$

$v = \sqrt{ar}$

Recall that

$v_{2} = 2v$

∴ $v_{2} = 2\sqrt{ar}$

Now, we will put the value of $v_{2}$ into

$a_{2} = \frac{v_{2}^{2} }{r}$

Then,

$a_{2} = \frac{(2\sqrt{ar}) ^{2} }{r}$

$a_{2} = \frac{4ar }{r}$

$a_{2} = 4a$

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0

3 years ago

What is machine<br>and type of machine

vitfil [10]

Printer could be a machine

3 0

3 years ago

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A bicycle slows down when the rider applies the brakes.

aniked [119]

This is kinetic energy to heat energy

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3 years ago

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A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended

Alik [6]

Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
$k= \frac{F}{\Delta x}$
where F is the force applied, and $\Delta x$ is the stretch of the spring with respect to its equilibrium position. Using the data, we find
$k= \frac{15 N}{3.0 cm}=5.0 N/cm$

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
$\Delta x = \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm$

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3 years ago

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