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tatiyna
1 year ago
14

Describe how sound signals can be transmitted from a radio station and received on a radio at your home. Create a labeled diagra

m that shows this process
Physics
2 answers:
vlada-n [284]1 year ago
6 0
A radio wave is generated by a transmitter and then detected by a receiver. An antenna allows a radio transmitter to send energy into space and a receiver to pick up energy from space. Transmitters and receivers are typically designed to operate over a limited range of frequencies
MrRissso [65]1 year ago
4 0

Answer:

The person above is right.

Explanation:

You just need to type in sound signals transmitted from a radio station and received on a radio at your home to find a diagram https://flypaper.soundfly.com/wp-content/uploads/2019/03/am.gif

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Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a
Furkat [3]

Answer:

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Explanation:

6 0
2 years ago
A food department is kept at â12°c by a refrigerator in an environment at 30°c. the total heat gain to the food department is
slava [35]

As per energy conservation in the reversible engine we can say

Q_2 + W = Q_1

here we know that

Q_2 = 3300 kJ/h

Q_1 = 4800 kJ/h

now from above equation

3300 + W = 4800

W = 1500 kJ/h

now we can convert it into kW

W = 1500\times \frac{kJ}{3600s}

W = 0.42 kW

so above is the power input to the refrigerator

now to find COP we know that

COP = \frac{Q_2}{W}

COP = \frac{3300}{1500} = 2.2

so COP of refrigerator is 2.2

3 0
1 year ago
Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s
PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0
2 years ago
A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If
sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)=50 \Omega

Voltage(V)=15 V

Time constant(\tau)=\frac{L}{R}

\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}

current i_0=\frac{V}{R}

i_0=\frac{15}{50}=0.3 A

Current as a function of time is given by

i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )

i=0.3\times 0.9998

i= 299.95 mA

6 0
2 years ago
A common small-molecular weight (and therefore fast diffusing for an organic molecule) ingredient in perfumes is vanillin, the p
natima [27]
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7 0
2 years ago
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