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Yuri [45]
3 years ago
7

Compare and contrast electric motors and generators.

Physics
2 answers:
Marrrta [24]3 years ago
6 0
Electric motors require electricity to move the motor parts and do work (like an electric fan).

Elecgric generators actually burn diesel fuel to spin a motor around and around and around to GENERATE, or MAKE, electricity that you can then use to power your fans and lights in your house.
nata0808 [166]3 years ago
5 0
-- Both devices have mechanical processes going on OUTside,
and electrical and magnetic processes going on INside.

-- Electric motors are intended to accept electrical energy as their
input, and to deliver useful mechanical energy as their output.

-- Electric generators are intended to accept mechanical energy as
their input, and to deliver useful electrical energy as their output.

Depending on their internal design, some styles of each device
can be used "backwards", to function as the other one.
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harkovskaia [24]

Answer:

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Explanation:

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Thank you.

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4 0
3 years ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
Why does the amount of damage to a car depend on speed?
Zina [86]

if you are going slow, there wont be much of an effect or not any damage.

3 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

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3 years ago
What does the ozone layer protect us from?
djyliett [7]
For the future, Put the right subject please, The answer is to protect us from harmful UV rays, Which include UVA, UVB, And UBC.
6 0
3 years ago
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