If you have a positive electrical force, and a negative electrical force, what do you expect will happen?

1 answer:

6 0

I expect that they will <em>add</em>, and their effect at every location will be the <em>sum</em> of their individual effects at that location.

For example:

If they're acting at the same point and in opposite directions, the effect will be the same as a single force at that point, with strength equal to their difference, and in the direction corresponding to whichever one is stronger.

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Which of the following statements are true concerning electromagnetic radiation fields?Select all that apply.a.) The electric an

ale4655 [162]

Answer:

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

Explanation:

Electromagnetic wave is a transverse wave in which magnetic field and electric field both induces each other as both changes with time

Here magnetic field induces electric field and similarly magnetic field induces electric field.

As we know that this is a transverse wave so here magnetic field and electric field lies in perpendicular planes. but they both propagate in same direction in such a wave that both fields reaches their maximum position and minimum positions simultaneously

So the correct answer is

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

6 0

3 years ago

How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

Natalka [10]

By using Ohm's law, we can find what should be the resistance of the wire, R:
$R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega$

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
$r=35 mm=0.35 \cdot 10^{-3} m$
So the area is
$A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2$

And by using the resistivity of the Aluminum, $\rho=2.65 \cdot 10^{-8} \Omega m$ , we can use the relationship between resistance R and resistivity:
$R= \frac{\rho L}{A}$
to find L, the length of the wire:
$L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m$

4 0

3 years ago

6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if it

Salsk061 [2.6K]

Answer:

Flux is 21 Nm^2/C.

Explanation:

Electric field, E = 6 N/C along X axis

Electric filed vector, E = 6 i N/C

Area, A = 4 square meter

Area vector

$\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\$