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3 years ago

Why are radio waves considered harmless while ultraviolet waves and x-rats considered harmful

Physics

1 answer:

Anvisha [2.4K]3 years ago

7 0

They are usually thought as harmful.

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499.523. meter

I hope it's helps you

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3 years ago

H o m o zygous what does it mean

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Explanation:

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3 years ago

What is the unit for force?

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3 years ago

Read 2 more answers

1600 AM broadcasts radio waves with wavelengths of about 187.37 m. Convert this to miles.

nikklg [1K]

Answer:

The wavelength in miles is 0.1165 miles.

Explanation:

Given:

Wavelength of the radio wave is 187.37 m.

Now, the wavelength is given in meters.

We need to convert the wavelength from meters to miles.

In order to convert meters to miles, we have to use their conversion factor.

We know that,

1 meter = $\frac{1}{1609}\ miles$

Therefore, the conversion factor is given as:

$CF=\frac{1}{1609}\ miles\ per\ meter$

So, the wavelength in miles is given as:

$Wavelength=\textrm{Wavelength in meters}\times CF\\\\Wavelength=187.37\ m\times \frac{\frac{1}{1609}\ miles}{1\ m}\\\\Wavelength=\frac{187.37}{1609}\ miles\\\\Wavelength=0.1165\ miles$

Hence, the wavelength in miles is 0.1165 miles.

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3 years ago

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

3 years ago