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Nadya [2.5K]
3 years ago
13

A resistor is made out of a long wire having a length LLL. Each end of the wire is attached to a terminal of a battery providing

a constant voltage VVV0. A current III flows through the wire. If the wire were cut in half, making two wires of length LLL/2, and both wires were attached to the battery (the end of both wires attached to one terminal, and the other ends attached to the other terminal), what would be the total current flowing through the two wires?
Physics
1 answer:
Reika [66]3 years ago
6 0

Answer:

Explanation:

length of the wire = L

Potential difference = Vo

Current = I

new length of the wire, L' = L/2

As we know that the resistance of the wire is directly proportional to the length of the wire.

As the length is halved so that the resistance is also halved.

Resistance = Voltage / Current

V = I R

So, Voltage is constant and the resistance is halved so that the current is doubled.

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Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E =  Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west,  q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west

4 0
3 years ago
What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h
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The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
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The sample of silver of our problem has a mass of m=24.7 kg. Its specific heat capacity is C_s = 236 J/g^{\circ}C and the increase in temperature is
\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C

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8 0
4 years ago
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A softball of mass 0.220 kg that is moving with a speed of 5.5 m/s (in the positive direction) collides head-on and elastically
Elanso [62]

Answer:

The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

Explanation:

Given that,

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Using conservation of momentum

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The velocity approach is equal to the separation of velocity

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(b). We need to calculate the mass of the target ball

Now, Put the value of v₂ in equation (I)

m_{2}\times1.6=2.068

m_{2}=\dfrac{2.068}{1.6}

m_{2}=1.29\ kg

Hence, The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

3 0
3 years ago
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