A 24-kW electric furnace is connected to a 240-V line, what is the 0.42 ohm 1000

Question

A 24-kW electric furnace is connected to a 240-V line, what is the 0.42 ohm 1000 ohm 10 ohm 2.4 ohm 100 ohm Two coins carry identical charges. These two coins are lying 1.50 m apart on a table. If each of these coins experiences a force of 2.00 N, how large is the charge on each coin? 1.12 mu C 11.2 mu C 6.67 mu C 2.24 mu C 22.4 mu C A carbon resistor has a resistance of 18 ohm at a temperature of 20 degree C. What is its resistance at a temperature of 120 degree C? (The temperature coefficient of resistivity for carbon is -5.0 times 10^-4/C degree.) 15 ohm 14 ohm 17 ohm 16 ohm 18 ohm What is the voltage across a 5.0-ohm resistor if the current through it is 5.0 A? 100V 4.0V 25V 1.0V The magnitude of the charge on each plate of a parallel plate capacitor is 4 mu C and the potential difference between the plates is 80 V. What is the capacitance of this capacitor. 100 times 10^-6F 0.1 times 10"6 F 300 times 10"6 F 5 times 10^-6 F 20 times 10^-6 F Two point charges of +50.0 mu C and -11.0 mu C are separated by a distance of 20.0 cm. A +7.00 mu C charge is placed midway between these two charges. What is the electric force acting on this charge because of the other two charges? 3.84 N directed towards the negative charge 3.84 N directed towards the positive charge 384 N directed towards the negative charge 384 N directed towards the positive charge 0.453 N directed towards the negative charge How much charge must pass by a point in 1.5 s for the current to be 2.0 A? 1.3 C 4.3 C 3.0 C 0.75 C A charge of 12 C passes through an electroplating apparatus in 2.0 min. What is the average current? 1.0 A 0.60 A 0.24A 6.0 A 0.10 A

Explanation / Answer

4.

given data:
Coulomb's law, force of attraction/repulsion
F = kQQ/r²
Q and Q are the charges in coulombs
r is separation in meters
k = 8.99*10^9 Nm²/C²
r=1.50m
F = 2N

2 = (9*10^9)(Q^2) / (1.50)^2
Q^2 = 0.5*10^-9
Q =2.24*10^-5 C

Q = 22.4 micro C

5.given
temperature T1 = 20 degree celcius = 20+273=293 K
temperature T2 =120 degree celcius = 120+273=393 K
coefficient of resistivity alfa = -5*10^-4 /C0
R0 = 18 ohms
R = Ro*(1+alfa * (delta T))
R = 18*(1- 5*10^-4*(393-293))
R = 18*(1-0.05)
R = 17 ohms

6.given
R = 5 ohms
I = 5 A
ohms law: V = IR
V=5*5 = 25 volts

7. given
Q= 4*10^-6 C
V =80 V
C = Q/V
C = 4*10^-6 /80 = 0.5*10^-7 =5*10^-8 F

8.given :
Let's say the +50*10^-6 C charge is on the left and the
-11*10^-6 C charge on the right.
The +50*10^-6 C charge will repel the +7*10^-6 C charge to the right and the -11*10^-6 C charge will attract the
+7*10^-6 C charge to the right.
r = 20cm = 0.2m
k = 9*10^9

Using Coulomb's law:
F1 = 9*10^9*(7*10^-6)(50*10^-6)/(0.1)^2=315 N
F2 = 9*10^9*(7*10^-6)(11*10^-6)/(0.1)^2=69.3N

Adding:
F1 + F2 = 384 N directed towards possitive charge

9.given:
t = 1.5 s
I = 2 A
chage Q = I*t = 2*1.5=3 C

10. given:

Q = 12 C
t = 2min = 2*60 sec = 120 s
I = ?
Q = I*t
12=I*120
I = 0.1 A

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