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Nostrana [21]
3 years ago
13

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside thi

s metal resistor?
Physics
1 answer:
Zanzabum3 years ago
7 0

Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

ΔV = EL

where;

ΔV is change in electric potential

E = ΔV / L

E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

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The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate o
WINSTONCH [101]

The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which \mu k = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant \theta = 30^{o}.

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

           \sum F_{y} = 0

         N - W = 0

           N = W

or,      N = 75 lb

Again, equilibrium condition in the vertical direction is  as follows.

        \sum F_{x} = 0

       T_{2} - F_{k} = 0

         T_{2} = \mu_{k} N

                  = 0.4 \times 75 lb

                  = 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

         \sum F_{x} = 0

       T Cos (30^{o}) + T Cos (30^{o}) = T_{2}

          2T Cos (30^{o}) = T_{2}

    or,             T = \frac{T_{2}}{2 Cos (30^{o})}

                        = \frac{30}{2 Cos (30^{o})}

                        = \frac{30}{1.732}

                        = 17.32 lb

Now, we will calculate the output power of the motor as follows.

             P = Tv

                = 17.32 lb \times 6

                = 103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}

                = 0.189 hp

or,             = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

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