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2 years ago

Which type of electromagnetic radiation is commonly used for long-distance communications?

Physics

1 answer:

Citrus2011 [14]2 years ago

5 0

Answer:

B. Radio waves

Explanation:

Radio waves are transmitted easily through air. They do not cause damage if absorbed by the human body, and they can be reflected to change their direction. These properties make them ideal for communications.

Hope this helps!!

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How many times did john glenn orbit the earth

galben [10]

Answer:

2 times

Explanation:

1st time on Feb. 20, 1962 at age 41

2nd time on Oct. 29, 1998 at age 77

8 0

3 years ago

A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela

larisa86 [58]

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

Now we will write the equation for a Change in Potential energy ΔU,

ΔU = Uf - Ui

= 0-m*g*h

ΔU = -m*g*h --Equation 2

Now compare the both equation

Wint = -ΔU

we can rewrite the above equation

ΔU = -W.

So our Answer is ΔU = -W. .

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3 years ago

What is sir Richard Branson's personal dilemma ?

kakasveta [241]

Answer:

Sir Richard Branson's personal dilemma is that he is concerned about the environment and climate change, but he has made his fortune with an airline industry that contributes to the greenhouse gases.

brainliest plz

4 0

3 years ago

gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the

igomit [66]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

$y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\$

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

$d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\$

d = 68.5 x 10⁻⁶ m = 68.5 μm

7 0

3 years ago

State the law of conversation of energy in your own words. Give an example of a situation that you have either encountered or kn

RSB [31]

That Energy Cannot Be Created Nor Destroyed

4 0

2 years ago