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jeka94
3 years ago
13

Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

L 2, how do L 1 and L 2 compare if wire 1 is made from copper and wire 2 is made from aluminum? The resistivity of copper is 1.7 × 10 –5 Ω·m and the resistivity of aluminum is 2.82 × 10 –5 Ω·m.
Physics
1 answer:
SVEN [57.7K]3 years ago
7 0

Answer with Explanation:

We are given that

Length of wire 1=L_1

Length of wire 2=L_2

Resistivity of copper wire=\rho_1=1.7\times 10^{-5}\Omega-m

Resistivity of aluminum wire=\rho_2=2.82\times 10^{-5}\Omega-m

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance =\frac{\rho l}{A}

When diameter of wires are same then their cross section area are also same .

l=\frac{RA}{\rho}

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity  is small then the length of wire will be large.

\rho_1

Therefore, L_1>L_2

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66

L_1=1.66L_2

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A gumdrop is released from rest at the top of the empire state building, which is 381 m tall. disregarding air resistance, calcu
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<u>Answer:</u>

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<u>Explanation:</u>

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

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 Displacement after 1 second = 0*1+\frac{1}{2}*9.8*1^2=4.9m

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We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

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1. energy and air

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8 0
3 years ago
Consider two oppositely charged, parallel metal plates. The plates are square with sides L and carry charges Q and -Q. What is t
antoniya [11.8K]

Answer:

 E = \frac{Q}{L^2 \epsilon_o}

Explanation:

For this exercise we use that the electric field is a vector, so the resulting field is

          E_total = E₁ + E₂                      (1)

since the field has the same direction in the space between the planes

Let's use Gauss's law for the electric field of each plate

Let's use a Gaussian surface that is a cylinder with the base parallel to the plate, therefore the normal to the surface and the field lines are parallel and the angle is zero so cos 0 = 1

          Ф  = ∫ .dA = q_{int} /ε₀

if we assume that the charge is uniformly distributed on the plate we can define a charge density

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as the field exists on both sides of the plate on the inside

          E A = A σ / 2ε₀

          E = σ / 2ε₀

           

we substitute in equation 1

         E = σ /ε₀

for the complete plate

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we substitute

         E = \frac{Q}{L^2 \epsilon_o}

7 0
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