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2 years ago

11

The filament of an electric bulb draws a current of 0.4 ampere. Calculate the amount of charge flowing through the filament if t

he bulb glows for 2 hours. Also calculate the number of electrons passed.

1 answer:

Marina86 [1]2 years ago

3 0

Here I =0.4 amp and t=2hr or 2×60×60 sec
We known I=nQ/t where Q is charge and n is no of charge.
n=It/Q
0.4x(2x60x60)/1.6x10^-19

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Explanation:

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Initial angular speed of the ball is $\omega = 0$

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Due to the presence of frictional force, ball moves with

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(a)

velocity $v_{com0}$ in terms of $\omega$ is

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(b)

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(c)

During sliding, ball's angular acceleration is calculated as

$\alpha=-\frac{\tau}{I}\\\\-\frac{\mu mgR}{(\frac{2}{5}mR^2)}\\\\-\frac{2}{5}\frac{\mu g}{R}\\\\-\frac{2}{5}\frac{(0.35)(9.8)}{0.11}\\\\-77.95rad/s^2$

(d)

The time for which the ball slides is calculated from the

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$V_{cm}= V_{cm0} + at\\\\V_{cm} = V_{cm0} + (-\mu g )t\\\\-0.11\omega=6.0m/s -(3.43m/s^2 )t\\\\-(0.11) (\alpha t) =6.0 m/s - (3.43 m/s^2)\\\\- (0.11)(-77.95 rad/s^2)t = 6.0m/s - (3.43 m/s^2 )t\\\\8.5745t + 3.43t= 6.0\\\\12.0045t = 6.0\\\\t= 0.4998s$

(e)

Distance traveled by the ball is

$X= V_{com,0}+ \frac{1}{2}at^2\\\\= (6.0m/s)(0.4998 s)+ 0.5(-3.43m/s^2) (0.4998 s)^2\\\\=2.57m$

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The speed of the ball when smooth rolling begins is

$V_{cm} = V_{com, 0}+ at\\\\=6.0 m/s +(-3.43m/s^2 )(0.4998 s)\\\\= 4.29m/s$

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