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Nookie1986 [14]
3 years ago
7

Two positive charges are labeled q Subscript 1 baseline and q Subscript 2 baseline are 0.1 m apart. Two positively charged parti

cles are separated on an axis by 0.1 meters, and q1 has a 6 µC charge and q2 has a 2 µC charge. Recall that k = 8.99 × 109 N • meters squared over Coulombs squared.. What is the force applied between q1 and q2 in N? In which direction does particle q2 want to go?
Physics
1 answer:
Snowcat [4.5K]3 years ago
4 0

Answer:

Explanation:

Let that point be at a distance x from q1

Then Kq1/x^2= Kq2/ (s-x)^2

Taking square roots and simplifying, x =s /[1+(q2/q1)^0.5]

Assuming an identical distance, the rigidity of Q on 2Q is equivalent in value to the rigidity of 2Q on Q. for that reason, had the area R been stored an identical, the two forces could be equivalent. inspite of the shown fact that, via fact the area is being decreased, we could constantly consult with the equation we use to calculate those forces: F = ok(Q1xQ2)/(R^2) because R is squared and is being halved, the final result's that's it being divided by potential of a million/4. for that reason, the rigidity would be expanded by potential of four, and be 4F.

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If I have a picosecond laser how would that be expressed in terms of zeptoseconds? In terms of petaseconds?
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3 years ago
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

V_{R} = 2.05 V

Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

V_{L} = - IwLsin(wt)

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V_{R} = IR

    = (0.044 A) (93 Ω)

V_{R} = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V_{R} = 2.05 V

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