1. If the size of the charge value is changed 6X for both of two point charges a
ID: 306799 • Letter: 1
Question
1. If the size of the charge value is changed 6X for both of two point charges and their separation is changed by a factor of 4X, the mutua force between them will be changed by what factor? A) 1.1868 B) 225 C) 1.7 D) 7.2 E) 5.16 2.If a 600 w A) 80 oh heater carriss a current of 3 A, what is the resistance of the heating element? 12 B) 2 00ohm (C)6667 ohm D) 277778 hm E) 20 ohm 3. A RL series then closed, fran circuit with harer o 30olsas R25 ohms and L-6.5 Hemries. When the cireuit is nd the time for the voltage across the resistor to reach 25 volts. sec C) 1.893204 sec D) 1.725 sec E) 8.385 sec A) 11.5 sec B) 4.659 all metal carr isms. What is the w resistance of 4 f rails 0.5 m apart. The rails are connected to a 28 volt battery and have a Tesla is pointing vertically? art is What is the velocity of the cart when the current in the rails is 2.8 A and the B-field of 1. A) 3.22N B) 22.4 m/s C) 3.5N D) 2N E) 6N resistanc to a battery. It you replace the resistor with one that has 3.5x A) 0.285714 N B) 3.5Nc) 42.875 N D) 0.08163 N E) 1225 dissipated in the circuit change? 6. If I1 goes to the right through R Ia 80 v goes to the right through R. and Is goes to the right through Rs, what is the resulting equation resulting from applying Kirchhoff's Loop rule for a clockwise loop around the perimeter of the circuit? 90 V B) 8.0 V- 11R, + 13R3-0 1.1 T will follow which of the paths described below (g-1.6 x 10.1, c and np-1.67 x 10-r kg) A) a circular path ot 1.138636 cn radius D) a circular path of 1.45 cn radius B) a circular path of 3.036 cm radius C) a circular path of 3.6036 cn radius . A proton moving with a speed of 12 x 10m/s perpendicular to a uniform magnetic tield of E) a straight line path 8. Ifa 12 V battery, a 2uF capacitor and a 4,F capacitior were wired in parallel, what would be the charge leaving the battery A) 40 C B) 120 uC C) 16.67 C D) 1.6C E) 72Explanation / Answer
1)
Finitial = K*Q1*Q2/R^2
Ffinal = K*6Q1*6Q2/(4R)^2
= 36*K*Q1*Q2/16*R^2
So,
Ffinal / Finitial = 36/16
= 2.25
Answer: B
2)
Use:
P = i^2*R
600 = 3^2 * R
R = 66.67 ohm
Answer: 66.67 ohm
I am allowed to answer only 1 question at a time
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