Physics College

Which of the following statements correctly explains the relationship between FM and AM radio waves?


A.
FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a shorter wavelength compared to AM radio waves.


B.
FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength compared to AM radio waves.


C.
FM radio waves carry more energy than AM radio waves. This is supported by the fact that FM radio waves have a shorter wavelength compared to AM radio waves.


D.
FM radio waves carry more energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength

Answers

Answer 1

The relationship between the AM and the FM waves is that the wavelength of the FM waves is longer than that of the AM waves.

What are radio waves?

The radio waves are one of the waves that are part of the electromagnetic spectrum and they are used in the process of communication . The reason why the radio waves could be used for the process of communication is that the radio waves do have a long wavelength this they can travel for very long miles without a significant decrease in intensity of the waves.

Thus, FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength compared to AM radio waves.

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Related Questions

Hello! So I need some help with this question. I don’t quite understand it and I did have another tutor figure it out and the answer was wrong. It’s not his answer of 58.04. Can you help me?

Answers

ANSWER:

58 m/s

STEP-BY-STEP EXPLANATION:

Given:

Initial velocity (u) = 13 m/s

The correct answer is 58 m/sDistance (d) = 400 m

Acceleration (a) = 4 m/s²

We apply the following formula to determine the final velocity:

[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ We replacing:} \\ \\ v^2=13^2+2(4)(400) \\ \\ v^2=169+3200 \\ \\ v=\sqrt{3369} \\ \\ v=58.04\cong58\text{ m/s} \end{gathered}[/tex]

The correct answer is 58 m/s

A block of mass M is hung by ropes as shown. The system is in
equilibrium. The point O represents the knot, the junction of the
three ropes. The angle theta is given to be 30o
. Which of the
following statements is true concerning the magnitudes of the three
forces in equilibrium?

Answers

Their magnitude of force must be the same for the tensions to equal out.

An object at equilibrium has zero acceleration so both the magnitude and direction of the object's velocity must be constant. When the body is in equilibrium the forces are balanced. Balanced is a keyword used to describe a balanced state. Therefore the net force is zero and the acceleration is 0 m/s/s. The acceleration of a body in equilibrium must be 0 m/s/s

For a body to be in equilibrium it must not be accelerating. This means that both the net force and net torque on the object must be zero. Clearly when in equilibrium, the net force on the object is zero. According to Newton's second law of motion if the net force is zero the acceleration is also zero. When the acceleration is zero the velocity and thus the velocity is constant by definition. Option d does not apply to equilibrium in this case.

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A string 0.50 m long is stretched under a tension of 2.0 x 102 N and its fundamental frequency is 400 Hz. If the length if the string is shortened to 0.35 m and the tension is increased to 4.0 x 102 N, what is the new fundamental frequency?

Answers

Given,

The initial length of the string, L₁=0.50 m

The tension on the string, T₁=2.0×10² N

The initial fundamental frequency of the string, f₁=400 Hz

The length of the string after it was shortened, L₂=0.35 m

The increased tension on the string, T₂=4.0×10² N

The fundamental frequency of the string before it was shortened is given by,

[tex]f_1=\frac{\sqrt[]{\frac{T_1}{\mu}}}{2L_1}[/tex]

Where μ is the mass per unit length of the string.

On rearranging the above equation,

[tex]\begin{gathered} 4L^2_1f^2_1=\frac{T_1}{\mu} \\ \Rightarrow\mu=\frac{T_1}{4L^2_1f^2_1} \end{gathered}[/tex]

On substituting the known values,

[tex]\begin{gathered} \mu=\frac{2.0\times10^2}{4\times0.50^2\times400^2} \\ =1.25\times10^{-3}\text{ kg/m} \end{gathered}[/tex]

The fundamental frequency after the string was shortened is given by,

[tex]f_2=\frac{\sqrt[]{\frac{T_2}{\mu}}}{2L_2}[/tex]

On substituting the known values,

[tex]\begin{gathered} f_2=\frac{\sqrt[]{\frac{4\times10^2}{1.25\times10^{-3}}}}{2\times0.35} \\ =808.1\text{ Hz} \end{gathered}[/tex]

Thus the fundamental frequency after the string is shortened and the tension is increased is 808.1 Hz

What voltage was needed for 2 amps to flow through a 600 lamp?​

Answers

Answer:

I don't kno..................................................w

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