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DedPeter [7]
3 years ago
9

Identify the electrical properties of materials that are highly conductive and those that are used in insulation. Use one or two

sentences to complete your answer
Physics
1 answer:
Katena32 [7]3 years ago
6 0

Answer:

This question is incomplete

Explanation:

However, materials that are highly conductive have the capacity to allow heat and electricity to pass through them. This occurs because they have electrons that flow easily in them, these electrons serve as carriers of heat and electric current in these materials. It should be noted here that not only solid metals (like silver, aluminium and copper) are good conductors of electricity, liquid solution such as aqueous NaCl are also good conductors of electricity because of the same movement of electrons (which is being transferred during bonding as charges) in the liquid.

Invariably, materials that are poor conductors of heat and electricity do not have electrons moving freely in them. Examples are sand, rubber and plastic.

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A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from
Alla [95]

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

6 0
3 years ago
How many electrons does it take to make up 4.33 C of charge?
Dovator [93]

Answer:

Number of electrons, n=2.7\times 10^{19}

Explanation:

It is given that,

Charge, q = 4.33 C

We need to find the number of electrons that make 4.33 C of charge. According to quantization of charge as :

q=ne

n = number of electrons

e = electron's charge

n=\dfrac{q}{e}

n=\dfrac{4.33\ C}{1.6\times 10^{-19}\ C}

n=2.7\times 10^{19}

So, the number of electrons are 2.7\times 10^{19} Hence, this is the required solution.  

7 0
3 years ago
Winds are the main cause of ___ currents in the ocean
raketka [301]

Winds are the main cause of ocean currents in the ocean.

5 0
3 years ago
Two spherical objects have masses of 6.2 x 105 kg and 13 x 103 kg. The gravitational attraction between them is 130 N. How far a
Neporo4naja [7]

Answer:

r = 6.4 cm

Explanation:

F = GMm/r²

r = √(GMm/F)

r = √((6.674e-11)(6.2e5)(13e3)/130)

r = 0.06432... m

Those are some high density materials!

8 0
3 years ago
A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

5 0
3 years ago
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