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jeyben [28]
3 years ago
11

How do microwaves use the behavior and characteristics of the electromagnetic wave to function?

Physics
1 answer:
serg [7]3 years ago
6 0

Answer:

Radiation moves out of the microwave into waves causing heat.

Explanation:

Inside your microwave oven, electrical energy is transformed into EM energy in the magnetron. When the microwave photons interact with food, food molecules are physically agitated, transforming the EM energy into kinetic energy, or energy of movement   .

Hope it helped!

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Which of the following correctly arranges the planets by the size of their diameter in increasing order?
valentinak56 [21]
Increasing Order..
So, it's from the little one into the biggest one..

My Answer is
C: Mercury < Earth < Saturn < Jupiter

What Shape of The Earth ?
Is it Round or Flat ?
Subjective Answer..
7 0
3 years ago
Read 2 more answers
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

3 0
2 years ago
Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?
oksano4ka [1.4K]

Answer:

Your project goes well.

Explanation:

Because that's how it works.

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In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
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its supposed to be (a) 1W

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3 years ago
I need help ! if anyone answers this its worth 40 points please help !!
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Double displacement...I think
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