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emmainna [20.7K]
3 years ago
9

Calculate the charge that flows through the cell in 1 minute. Each filament lamp has a power of 3 W and a resistance of 12 Ω

Physics
1 answer:
lapo4ka [179]3 years ago
5 0

Answer:

24 Coulumbs

Explanation:

Given data

time= 1 minute= 6 seconds

P=2 W

R= 12 ohm

We know that

P= I^2R

P/R= I^2

2/12= I^2

I^2= 0.166

I= √0.166

I= 0.4 amps

We know also that

Q= It

substitute

Q= 0.4*60

Q= 24 Columbs

Hence the charge is 24 Coulumbs

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Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

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refractive index (n) = \frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}

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S_{medium}  = \frac{S_{vacuum} }{n_{medium} }

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S_{water} = \frac{1 }{1.33}

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