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3 years ago

What is the practical applications of radio waves

1 answer:

5 0

Radio waves have many uses—the category is divided into many subcategories, including microwaves and electromagnetic waves used for AM and FM radio, cellular telephones and TV.

The lowest commonly encountered radio frequencies are produced by high-voltage AC power transmission lines at frequencies of 50 or 60 Hz. These extremely long wavelength electromagnetic waves (about 6000 km) are one means of energy loss in long-distance power transmission.

Extremely low frequency (ELF) radio waves of about 1 kHz are used to communicate with submerged submarines. The ability of radio waves to penetrate salt water is related to their wavelength (much like ultrasound penetrating tissue)—the longer the wavelength, the farther they penetrate. Since salt water is a good conductor, radio waves are strongly absorbed by it; very long wavelengths are needed to reach a submarine under the surface.

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What are the folds called inside the mitochondria?

max2010maxim [7]

The inner membrane has many overlapping folds called cristae. Inside the inner membrane there is the mitochondrial matrix, it contains enzymes that are used in creating ATP.
You're welcome. :)

7 0

3 years ago

Read 2 more answers

A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti

Korvikt [17]

Answer:

The strength of the electric field is $1.35\times10^{4}\ N/C$ .

Explanation:

Given that,

Speed $v= 5.05\times10^{5}\ m/s$

Time $t= 3.90\times10^{-7}\ s$

Angle = 45°

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$5.05\times10^{5}=0+a\times3.90\times10^{-7}$

$a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}$

$a=1.29\times10^{12}\ m/s^2$

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

$F= ma=qE$

$ma = qE$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}$

$E=1.35\times10^{4}\ N/C$

Hence, The strength of the electric field is $1.35\times10^{4}\ N/C$ .

3 0

3 years ago

Your GPS shows that your friend’s house is 10.0 km away. But there is a big hill between your houses and you don’t want to bike

kati45 [8]

Answer:

The value is $c = 8 \ km$

Explanation:

From the question we are told that

The distance of friends house from your point is $a = 10 \ km$

The distance of your friends street from your street is $b = 6 \ km \ in the \ direction \ towards \ the \ north$

The diagram illustrating this question is shown on the first uploaded image

From the diagram we can apply by Pythagoras theorem as follows

$a^2 = b^2 + c^2$

=> $c = \sqrt{^2 - b^2}$

=> $c = \sqrt{ 10^2 - 6^2}$

=> $c = 8 \ km$

6 0

4 years ago

Find the mean of set values 12g 9g 13g 12g 20g 17g 15g

nalin [4]

${\huge{\boxed{\mathcal{\green{Answer}}}}} \\ \frac{12 + 9 + 13 + 12 + 20 + 17 + 15}{7} \\ = \frac{98}{7} \\ = 14 \\ {\huge{\boxed{\mathcal{\green{Hope \: it \: Helps}}}}}$