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3 years ago

What best describes the angle between a changing electric field and the electromagnetic wave produced by it?

Physics

1 answer:

mel-nik [20]3 years ago

3 0

The correct answer here is Option A, always equal to a right angle.

Keep in mind that the phase difference that exists between the electric field and the magnetic wave is 90 °, this means that the changing electric field and the electromagnetic wave are always perpendicular and will be forming a right angle.

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How much work does a 120 watt motor perform for 300 seconds?

MrRa [10]

Answer:

36,000j

Explanation:

W= P•t

120•300= 36000

I checked my answer with P= W/t and got the right answer

hope this helps

6 0

3 years ago

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

7 0

3 years ago

Processes such as dissolving, melting, freezing, and evaporationg can affect the appearance of a substance, but the identity of

Vlada [557]

The answer to this question is D

8 0

3 years ago

Before Newton and develop reflecting telescopes what instruments did astronomers use?

never [62]

The first telescopes were refracting telescopes ... a tube
with lenses at both ends. The quality of the image depends
on the quality of both lenses, because the light must pass
through both of them.

The reflecting telescope was an improvement, because the
light doesn't have to go through the mirror. The quality of the
primary mirror only depends on the shape and shinyness of
its surface.

If you could carve an optically perfect parabolic surface out of
a piece of wood and coat it with a thin layer of shiny silver, then
you could use a log to make a good reflecting telescope.
II seem to recall that Newton made one out of brass !

7 0

4 years ago

What effect does an unbalanced force have on an object

Volgvan

It changes the speed and direction of an object

7 0

3 years ago

Read 2 more answers