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Lesechka [4]
3 years ago
14

6–3 The current through a 0.1‐μF capacitor is a rectangular pulse with an amplitude of 2 mA and a duration of 5 ms. Find the cap

acitor voltage at the end of the pulse when the capacitor voltage at the beginning of the pulse is –1 V.
Physics
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

999 V

Explanation:

Charge deposited on the capacitor

= current x time  

= 2 x 10⁻³ A x 5 x 10⁻³

=  10⁻⁵ C

increase in volt = charge deposited / capacitance

10⁻⁵ / .1 x 10⁻⁶

= 10² V

= 100 V

voltage in the beginning = -1 V

voltage at the end

= -1 + 100

99 V .

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Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity (\omega_o) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity  (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

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\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\

b)

\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)}  =112.4\ rev

θ = 323 rad

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W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J

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