Help with number 4 please help with number 4 please An electrical current of 8.7

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Help with number 4 please

help with number 4 please

An electrical current of 8.70 mA exists in a solid cylindrical of the current density in the wire whose diameter is 1.20 mm. Calculate the magnitude of the current density in the wire. 4.34 Times 10^(-2) A. 2.17 Times 10^(4) A/m^2. 7.69 Times 10^(3) A/m^2. 7.17 Times 10^(-3) A/m^2. 8.19 Times 10^(3) A/m^2. Assume that electrons are the charge carriers and the conduction electron density is 2.35 Times 10^29/m^3. (Note that this is the charge carrier density and could also be stated as 2.35 Times 10^29 charge carriers/m^3 or electrons/m^3.) Then, calculate the electron drift speed in the wire. 1.28 Times 10^(-8) m. 3.72 Times 10^(-9) m. 2.26 Times 10^(-3) m/s 9.26 Times 10^(-6) m/s. 2.04 Times 10^(-7) m/s. A wire with a resistance of 0.80 Ohm is drawn out through a die so that its new length is 6 times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are not changed during the drawing process. 28.8 Ohm. 0.23 Ohm. 46.5 Ohm. 12.5 Ohm 1.25 Ohm. A potential difference of 1.8 V is maintained between the ends of a 2700 cm length of wire whose diameter is 0.66 mm. The conductivity of the wire is 5.80 Times 10^7 (ohm.m)^-1. Determine the rate at which energy in the wire is transformed from kinetic to thermal energy. 8.26 W. 5.19J. 3.38J. 2.38W. 5.19W. A solar cell generates a potential difference of 5.62 V when a 1.20 k Ohm external resistance is connected across it. i.e., from one terminal of the cell to the other terminal of it, and a potential difference of 6.84 V when a 2.90 k Ohm external resistance is used. What is the internal resistance r_i of the solar cell? (Give your answer in the form of "abc" Ohm.) 5.25 Times 10^(2) Ohm. 7.67 Times 10^(2) Ohm. 4.67 Times 10^(3) Ohm. 5.93 Times 10^(2) Ohm. 3.93 Times 10^(3) Ohm. What is the emf sigma (V) of the solar cell? 9.93V. 7.67J. 7.07V. 5.93J. 8.08V. For the circuit diagram as Figure 4 of Assignment 7, sigma = 15 kV, R = 4.5 K Ohm, and C = 7800.0 mu F. Determine the time constant for the circuit. 3.51 Times 10^(1) s. 2.58 Times 10^(-1)s. 1.61 Times 10^(-3)s. 7.51 Times 10^(-6)s. 1.54 Times 10^(2)s. Determine the maximum charge that will appear on the capacitor. 1.51 Times 10^(-2)C. 2.58 Times 10^(-5)C. Assuming that the capacitor is initially uncharged, how long will it take for the capacitor to charge to half the maximum charge? 1.51 Times 10^(-1)s. 35.8 s. A point mass consisting of 34 charged particles travels at a velocity of magnitude 4.80 Times 10^7 m/s through a uniform magnetic field of magnitude 4.8 T. Each charged particle has a charge of -12e and a mass of 3.60 Times 10^-29 kg. The angle between

Explanation / Answer

ans the internal resistance is c =4.67 *(10^3)ohm

and the emf of the slar cell is d =5.93 J

We used the formula

V = E - Ir

Where:

V = pd across the external circuit (V)

E = emf of the cell (V)

I = current through the cell (A)

r = value of the internal resistance (Ω)

To caluclate the terminal V used

I= VT/RT

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