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3 years ago

All of the following involve waves of electromagnetic energy except _______.

Physics

2 answers:

r-ruslan [8.4K]3 years ago

6 0

All of the following involve waves of electromagnetic energy except the rumble of thunder during a storm.

Electromagnetic waves<span> <span>are used to transmit long/short/FM wavelength radio </span>waves, and TV/telephone/wireless signals or energies. They are also responsible for transmiting energy in the form of microwaves, infrared radiation<span> (IR), visible light (VIS), ultraviolet light (UV), X-rays, and gamma rays.</span></span>

The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

Sunny_sXe [5.5K]3 years ago

4 0

Answer:

B. The rumble of thunder during a storm

Explanation:

A. Sunlight shining on a plant leaf

All visible light coming from different sources is a part of electromagnetic waves.

C. Radios receiving signals through air

Radio waves are part of electromagnetic waves which is of least energy and largest wavelength

D. Watching videos on a computer screen

While watching videos we are receiving the light waves from the television and other ultraviolet waves coming from the source. These both are part of electromagnetic waves

So correct answer will be

B. The rumble of thunder during a storm

because it is a part of sound waves which travel through air

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3 years ago

Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150

alexgriva [62]

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation $F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

$\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

$\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

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Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$

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3 years ago

Why are yet planes faster tharn. cargo planes?

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2 years ago

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

4 years ago