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2 years ago

Which factors affect the resistance of a material? Select three options. Length current thickness temperature voltage.

1 answer:

8 0

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Length, temperature, thickness are the factors that affect the resistance of a material.

<h3>What is resistance?</h3>

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The relation of resistance with length and thickness is given by ;

$\rm R= \frac{\rho l}{A}$

Hence the value of resistance is directly propotional to length and inversly propotional to the area or thickness of the wire.

As the value of temperature increases, the value of resistance in the material is increasing.

Hence length, temperature, thickness are the factors that affect the resistance of a material.

To learn more about the resistance refer to the link;

brainly.com/question/20708652

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A satellite m-500 kg orbits the earth at a distance d 215 km, above the surface of the planet. The radius of the earth is re 6.3

AnnZ [28]

Answer:

7.78 * 10³ m/s

Explanation:

Orbital velocity is given as:

v = √(GM/R)

G = 6.67 * 10^(-11) Nm/kg²

M = 5.98 * 10^(24) kg

R = radius of earth + distance of the satellite from the surface of the earth

R = 2.15 * 10^(5) + 6.38 * 10^(6)

R = 6.595 * 10^(6) m

v = √([6.67 * 10^(-11) * 5.98 * 10^(24)] / 6.595 * 10^(6))

v = √(6.048 * 10^7)

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3 years ago

Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out ov

Mumz [18]

Answer:

The number of excess electrons are on the negative surface is $4.80\times10^{10}\ electrons$

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}$

$C=0.285\times10^{-10}\ F$

We need to calculate the potential

Using formula of potential

$V=Ed$

Put the value into the formula

$V=18000\times1.5\times10^{-2}\ V$

$V=270\ V$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=0.285\times10^{-10}\times270$

$Q=76.95\times10^{-10}\ C$

Here, the charge on both the positive and negative plates

$Q=+76.95\times10^{-10}\ C$

$Q=-76.95\times10^{-10}\ C$

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

$n=\dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}$

$n=4.80\times10^{10}\ electrons$

Hence, The number of excess electrons are on the negative surface is $4.80\times10^{10}\ electrons$

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3 years ago

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tresset_1 [31]

Answer:

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Andrej [43]

Answer:

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