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larisa86 [58]
3 years ago
13

The main difference between a radio wave and a light wave is its:.

Physics
2 answers:
TEA [102]3 years ago
8 0

Answer: Radio waves and light waves are both electromagnetic raidiations.

Explanation:

Does this help?

vesna_86 [32]3 years ago
6 0

Answer:

Speed.

Their speeds are different; Have you ever heard the metaphor <em>the speed of light?</em> For this case, we must turn that into a reality. So, people refer to light speed as the fastest travelling thing in the world, and that is right, unless there is a thing faster than light that was never classified. Unlikely the case. The speed of sound through air travels a lot slower than light. Look at the diagrams below to see what I mean by that.

                                                             • • •

As shown in the diagrams, sound speed is WAY slower than light speed!

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A jogger accelerates at a constant rate as she travels 5.0
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To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


7 0
4 years ago
How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00
klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

            C = \frac{\epsilon_{o}A}{d}

It is known that the dielectric strength of air is as follows.

               E = 3 \times 10^{6} V/m

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = \frac{\epsilon_{o} A}{d} \times E \times d

                   = \epsilon_{o}AE

                   = 8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}

                   = 1.062 \times 10^{-8} C

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0
4 years ago
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