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kicyunya [14]
3 years ago
6

1 example of a conductor and 1 example of a insulator in your EVERYDAY world.

Physics
1 answer:
ratelena [41]3 years ago
4 0

Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.

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NASA developed a reflective foil barrier to help shield spacecraft from heat transfer in space. These reflective barriers are no
morpeh [17]

Answer: Radiation

Explanation: Radiation is the energy that comes from a source in form of electromagnetic waves, subatomic particles, light, or heat which travels through space.

Examples of radiation include the light, heat, and particles emitted from the Sun.

Using a foil barrier to prevent heat transfer is possible because foil has a silver color, and silver reflects light and heat instead of absorbing them. This is the opposite of black surfaces that absorb heat.

So in homes where these foil reflective barriers are used, the transfer of heat through Radiation is highly reduced.

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I wanna say withdrawal? Ron needs to go to rehab
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3 years ago
A sledgehammer is an example of what kind of simple machine?
alisha [4.7K]

I am 95 percent sure the answer in C

3 0
3 years ago
An LED is useful because when a current passes through it, it gives out... what?
Mkey [24]

An LED is useful because when a current passes through it, it gives out light.

4 0
2 years ago
A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe
lana [24]

Answer: V = 15 m/s

Explanation:

As  stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately

3 0
3 years ago
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