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denpristay [2]
3 years ago
6

Give a physical description of radio waves and compare and contrast it with the frequency, wavelength, energy and wave speed wit

h those of other regions of the EMS.
Physics
1 answer:
Ad libitum [116K]3 years ago
8 0
There is nothing physical about radio waves. They are pulses of energy caused by periodic variations in electric and magnetic fields. These are caused by electrons “jumping” between energy levels. Radio waves have the lowest energy, longest wavelengths and lowest frequencies. These values change gradually as we go through microwaves, infra red, visible light, ultra violet, x-rats abd gamma rays which have the highest energy, highest frequency and shortest wavelength.
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The picture shows an object resting on a balance.
Maslowich

Answer:

4.90kgm^-2

Explanation:

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2 years ago
A car with a momentum (impulse) of 20,000 kg m/s collides with a wall and comes to a rest in 0.1 seconds. How much
Pani-rosa [81]

Answer:

» Force is 200,000 Newtons

Explanation:

{ \tt{force =  \frac{impulse}{time} }} \\  \\ { \tt{force =  \frac{20000}{0.1} }} \\  \\ { \tt{force = 200000 \: newtons}}

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Read 2 more answers
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0
3 years ago
Question 8
FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

8 0
2 years ago
For problems that involve an object accelerating along an inclined plane, how can the weight be used to determine the force comp
irakobra [83]

Answer:

C

Explanation:

Ed2020

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3 years ago
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