I need last minute answers to these 7 physics questions:(?

Question

21. A 12-V automobile battery is connected to an electric starter motor. The current through the motor is 210 A.
a. How many joules of energy does the battery deliver to the motor each second?
b. What power does the motor use in watts?
c. How much energy does the motor use in 10 seconds?

32. A 20.0- Ω resistor is connected to a 30.0-V battery. What is the current in the resistor?

38. What voltage is applied to a 4- Ω resistor if the current is 1.5 A?

48. What voltage is placed across a motor of 15 Ω operating resistance to deliver 8.0 A of current?

49. A 6- Ω resistor is connected to a 15-V battery.
a. What is the current in the circuit?
b. How much heat is produced in 10 minutes?

56. A lamp consumes 30.0 W when connected to 120 V.
a. What is the current in the lamp?
b. What is the resistance of the lamp?

57. A heating coil has a resistance of 4.0-Ω and operates on 120 V.
a. What is the current in the coil while it is operating?
b. What energy is supplied to the coil in 5.0 min?
c. If the coil is immersed in an insulated container holding 20.0 kg of water, what will be the increase in the temperature of the water? Assume that 100% of the heat is absorbed by the water.

Answer 1

Most of these can be solved with Ohm's Law, P=V*I, and a little algebra....



21a. Remember 1 Watt = 1 Joule/Second. P = V * I = 12V * 210A = 2520 W = 2520 J/s



21b. 2520 Watts (solved above)



21c. 2520 J/s * 10s = 25200 Joules



32. Ohm's Law: V = I * R 30V = I * 20 ohms. Solve for I, the current is 1.5 A



38. V = I * R = 1.5A * 4 ohms = 6 V



48. V = I * R = 8A * 15 ohms = 120 V



49a. V = I * R 15V = I * 6 ohms. Solve for I which is 2.5 A



49b. I do not recall this equation..I think it is part of Joule's Laws. There is a relationship between current, time, and resistance, but I do not recall the equation. Google it and I am sure you will find it.



56a. P = V * I 30W = 120 * I. Solve for I. I is 0.25A



56b. This one you have to use the Power equation and Ohm's law together. I'll let you do the math, but the equation boils down to P = V^2 / R. With a little algebra, R = V^2/P R = 120V * 120V / 30W R = 480 ohms



57a. V = I * R 120V = I * 4. I = 30A



57b. P = V * I = 120 * 30 = 3600W = 3600J/s. Multiply by 300 seconds, and the Energy is 1080000 J



57c. See 49b...don't remember the heat equation.



Good Luck!

Answer 2

All these questions can be answered using ohm's law, V=IR,

Where v=voltage,r=resistance,i=current.



21. V=12,I=210,T=1

a]E=IVT

E=12*210*1=2520Joules

b] P=vi

P=12*210=2520watts

c]E=VIT

E=12*210*10=25200J



32. V=IR

I=V/R

I=30/20

I=1.5AMPERES

38.V=IR

V=1.5*4=6VOLTS

48. V=IR

V=8*15=?

49.

a] I=V/R

I=15/6=?

b] E=ivt

=?*15*6=?

56. P=30,V=120

But, P=IV

a] 30=I * 120

I=30/120=?

b] R=V/I

R=120/?=?

57. R=4, V=120

a] I=V/R

=120/4

=?

b] E=IVT

E=?*120*4

=?



NB*-calculate the answer not given that is in ? And use it to find the remaining answer.I have no clue to the last question its incomplete. . .from where are you?

Answer 3

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