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STatiana [176]
2 years ago
7

If you wanted to

Physics
1 answer:
VLD [36.1K]2 years ago
5 0

Answer:

Option C

Explanation:

According to the formula

  • \\ \boxed{\sf R=\rho\dfrac{\ell}{A}}

So

If we use wide wire we increase the area of cross section so resistance decreases

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A cylinder contains 250 L of hydrogen gas (H2) at 0.0^∘Cand a pressure of 10.0 atm. How much energy is required to raise the tem
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Answer:

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Explanation:

The step by step calculation can be found in the attachment below. Thank you.

8 0
3 years ago
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
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Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

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Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

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Pressure
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Pressure
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Decomposition breaks down organic material into simpler units and:
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d.) provides proteins for plants​

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