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jolli1 [7]
3 years ago
11

Which terms describe the purpose of antennas on devices that use radio waves

Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

The terms are

1. Transmit

2. Receive

Explanation:

What is an antenna

According to NASA

An antenna is a metallic structure that captures and/or transmits radio electromagnetic waves. Antennas come in all shapes and sizes from little ones that can be found on the roof to watch TV to really big ones that capture signals from satellites millions of miles away.

How does an antenna work?

Antennas are much more than simple devices connected to every radio. They're the transducers that convert the voltage from a transmitter into a radio signal. And they pick radio signals out of the air and convert them into a voltage for recovery in a receiver

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What is a Cell?<br><br> Don't look it up PLEASE... I WILL GIVE BRAINLIEST
Travka [436]

Explanation:

Cell is a structural and fundamental unit mass of the body

6 0
2 years ago
What mass of silver (in grams) is solidified when 749 joules of heat are released by a sample of molten silver at its freezing p
Anastasy [175]

heat released Q = 749 joules

heat of fusion of silver L = 109 J/g

Here phase of silver is changing from liquid to solid

so temperature will remain same

all heat will be released due to its phase change

and in this case we use Q=mL

where m is the mass of silver in gram

Q= mL

749 = m * 109

m = 749/109

m = 6.87 gram

4 0
3 years ago
Read 2 more answers
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
2 years ago
A water pump pumps 3 kg of water every second from the bottom of a well to the top of the well 50 m above if the pump used 2000
vesna_86 [32]

watts = work per second.

work is mgh = 3x10x50=1500

watts out = 1500

Watts used = 2000

eff=1500/2000=75%


8 0
3 years ago
Help please I have to turn this in tonight!!
inna [77]

Answer:

True

Explanation:

i searched it up and well this thing is making me do it up till 20 characters long so yea

3 0
2 years ago
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