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2 years ago

What property of an electromagnetic wave changes when the wave is encoded with analog information?

Physics

1 answer:

Licemer1 [7]2 years ago

7 0

Answer:

Amplitude and Frequency

Explanation:

Analog signals are composed of continuous waves that can have any values for frequency and amplitude. These waves are smooth and curved.

Radio transmissions are a combination of two kinds of waves: audio frequency waves that represent the sounds being transmitted and radio frequency waves that "carry" the audio information. All waves have a wavelength, an amplitude and a frequency as shown in the figure. These properties of the wave allow it to be modified to carry sound information.

The two most common types of modulation used in radio are amplitude modulation (AM) and frequency modulation (FM). Frequency modulation minimizes noise and provides greater fidelity than amplitude modulation, which is the older method of broadcasting . Both AM and FM are analog transmission systems, that is, they process sounds into continuously varying patterns of electrical signals which resemble sound waves.

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What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af

sladkih [1.3K]

Answer:

$L=0.0045\ kg-m^2/s$

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, $\omega=4300\ rpm=450.29\ rad/s$

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

$L=I\omega$

Where I is moment of inertia

For sphere, $I=\dfrac{2}{5}mr^2$

$L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s$

So, the magnitude of the angular momentum of the sphere is $0.0045\ kg-m^2/s$ .

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2 years ago

An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux

zhenek [66]

Answer:

$\phi=1.56\times 10^{-5}\ Wb$

Explanation:

Given that,

Emf, V = 22 mV

Number of turns in the coil us 519

Rate of change of current is 10 A/s.

We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.

Let's find the inductance first. So,

$L=\dfrac{\epsilon}{(dI/dt)}\\\\L=\dfrac{0.022}{10}\\\\L=0.0022\ H$

We have,

$L=\dfrac{N\phi}{I}$ , $\phi$ is magnetic flux

$\phi=\dfrac{LI}{N}\\\\\phi=\dfrac{0.0022\times3.7}{519}\\\\\phi=1.56\times 10^{-5}\ Wb$

So, the magnetic flux is $1.56\times 10^{-5}\ Wb$ .

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