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Taya2010 [7]
3 years ago
10

Let us be two cylindrical conductors connected in parallel, to which a potential difference of V = 170V is applied. The two cond

uctors are made of the same material, but the first is 6 times the length of the second, and the radius of the second. The resistance of the second is R2 = 469Ω. Determine the equivalent resistance.
Physics
1 answer:
PSYCHO15rus [73]3 years ago
3 0

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

<h3>Resistance</h3>

Resistance is a measure of the opposition to flow of electric current. It is measured in ohms.

It is given by the formula:

R=\rho\frac{l}{A} \\\\where\ l=length,A=area,\rho=resistivity

Given that R₂ = 469 ohm, hence:

R_2=\rho\frac{l_2}{A_2} \\\\469=\rho\frac{l_2}{\pi r_2^2}

But l₁ = 6l₂, r₁ = (1/5)r₂, hence:

R_1=\rho \frac{l_1}{A_1}=\rho *\frac{6l_2}{[\pi (1/5)r_2]^2} =150 * \rho \frac{l_2}{[\pi r_2]^2}=30*469=70350\ ohm

The equivalent resistance (R) is:

R=\frac{R_1R_2}{R_1+R_1}=\frac{469*70350}{469+70350}  =466\ ohm

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

Find out more on resistance at: brainly.com/question/17563681

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3 years ago
A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force
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Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

\Sigma F = F - m\cdot g = m\cdot a (1)

Where:

F - Buoyant force, measured in newtons.

m - Mass of the plastic ball, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

a - Net acceleration, measured in meters per square second.

If we know that F = 5\,N, m = 0.408\,kg and g = 9.807\,\frac{m}{s^{2}}, then the net acceleration of the plastic ball is:

a = \frac{F}{m} - g

a= 2.448\,\frac{m}{s^{2}}

The acceleration is 2.448 meters per square second and is vertically upward.

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8 0
4 years ago
If you increase the charge on a parallel-plate capacitor from 3 mu or micro CC to 9 mu or micro CC and increase the plate separa
aliya0001 [1]

Explanation:

The energy stored in a capacitor is given by

U = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}

In the case of a parallel plate capacitor, the capacitance C is given by

C = \dfrac{\epsilon_0A}{d}

so we can rewrite the expression for the energy as

U = \dfrac{Q^2d}{2\epsilon_0 A}

Increasing the charge from 3\:\mu\text{C}\:\text{to}\:9\:\mu\text{C} means that you're tripling the charge. The same thing is true when you increase the distance from 1.8 mm to 5.4 mm, i.e., you triple the separation distance. So the new energy U' is given by

U' = \dfrac{Q'^2d'}{2\epsilon_0 A}

\:\:\:\:\:\:= \dfrac{(3Q)^2(3d)}{2\epsilon_0 A}

\:\:\:\:\:\:= 27\left(\dfrac{Q^2d}{2\epsilon_0 A}\right)

\:\:\:\:\:\:= 27U

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7 0
3 years ago
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