All categories

3 years ago

The primary job of a(n) ______ on a receiver is to capture modified radio waves.

Physics

2 answers:

PilotLPTM [1.2K]3 years ago

8 0

The correct answer is antenna.

Mars2501 [29]3 years ago

7 0

Answer: The correct answer is an antenna.

Explanation:

In radio communication, radio receives radio waves which is an electromagnetic wave. It converts the information carried by them to usable form.

An antenna intercepts radio waves. It converts them to tiny alternating current which are applied to receiver. Then, receiver extracts the desired information.

Therefore, the primary job of an antenna on a receiver is to capture modified radio waves

You might be interested in

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago

PLEASEE I NEED HELP FAST!!! .Study the scenario.A small container of water with a low temperature is poured into a large contain

RSB [31]

Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

Q_c = m ce ( $T_{f}$ -T₀)

Heat released by the largest container is

Q_a = M ce (T_{i}-T_{f})

how

Q_c = Q_a

m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

Of the final statements, the one that best describes the process is C

since it talks about the thermal energy and the heat that is transferred in the process

8 0

3 years ago

30 points!

vlada-n [284]

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

6 0

3 years ago

What is the correct definition of refraction?

scoray [572]

i believe it's C but i'm not completely sure

5 0

3 years ago

Which sentences describe central ideas of the video? Select all that apply.

sertanlavr [38]

Answer:

the correct answer is see.

Explanation:

it is wat it isssssssssss

4 0

3 years ago