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forsale [732]
3 years ago
11

Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig

ht as positive and forces pointing to the left as negative. Rank positive forces as larger than negative forces.
1. q1=+1nC
q2=-1nC
q3 =-1nC
2. q1= -1nC
+ q2 = + 1nC
q3= +1nC
3. q1= +1nC
q2= +1nC
q3= +1nC
4. q1= +1nC
q2= + 1nC
q3= -1nC
5. q1= -1nC
q2= - 1nC
q3= -1nC
6. q1=+1nC
q2=-1nC
q3 =+1nC

Physics
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Answer:

Plss see attached file

Explanation:

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To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

L= 1.5*10^{-2}H

R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

i= 0.0133A

i_1 = 0.01052A

3 0
3 years ago
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