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hram777 [196]
3 years ago
9

65. A length of wire is bent into a closed loop and a magnet is plunged into it, inducing a voltage and, consequently, a current

in the wire. A second length of wire, twice as long, is bent into two loops of wire, and a magnet is similarly plunged into it. Twice the voltage is induced, but the current is the same as that produced in the single loop. Why
Physics
2 answers:
natulia [17]3 years ago
6 0

Answer:

Resistance of the second wire is twice the first wire.

Explanation:

Let us first see the formula of resistance;

R = pxL/A

Here L is the lenght of the wire, A the area and p is the resistivity of wire.

As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.

Since we have two loop in second case, inducing double voltage but as resistance is doubled so the current would remain same according to ohms law

I = V/R

OLga [1]3 years ago
6 0

Answer:

The resistance of the second wire is two times that of the first wire. Even though the voltage doubles, current still remain the same.

Explanation:

The voltage induced is directly proportional to the number of loops of the wire. Resistance is directly proportional to the length of the wire as long as all other parametersare constant in the formula R=PL/A. The second wire has twice the length and twice the number of turns than the first wire.

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Answer:

The total current supplied by the source of voltage = 10.29 A

Explanation:

We have a 14-Ω coffee maker and a 14-Ω frying pan are connected in series.

Effective resistance = 14 + 14 = 28Ω

Now we have 28Ω and 20Ω in parallel

Effective resistance

             R=\frac{28\times 20}{28+20}=11.67\Omega

So we have resistor with 11.67Ω in a 120 V source of voltage.

We have equation V = IR

Substituting

               120 = I x 11.67

                 I = 10.29 A

The total current supplied by the source of voltage = 10.29 A

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The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig
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Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
  • the velocity of water in the upper channel is zero (v₁ = 0 m/s).
  • y₁ = 2.00 m  (height of the upper channel)
  • y₂ = 0.00 m  (height of the lower channel)
  • g = 9.81 m/s²
  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

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