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Rufina [12.5K]
3 years ago
8

The voltage across a parallel-plate capacitor with area A = 820 cm2 and separation d = 5 mm varies sinusoidally as V = (14 mV)co

s(170t), where t is in seconds. Find the displacement current between the plates. (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume that Id is in amperes. Do not include units in your answer.)
Physics
1 answer:
shutvik [7]3 years ago
6 0

Answer:

I_d = -3.454*10^{-10} \ \ (sin ( \ 170 \ t))

Explanation:

The displacement current I_d is given by the expression;

I_d = \epsilon _o \frac{d \phi _ E }{dt}

where

\phi _ E = A.E \\ \\ \\\phi _E = A. \frac{V}{d} \\ \\ \\\phi _E = \frac{A}{d}(14*10^{-3})(cos \ 170 \ t ) \\ \\ \\\phi _E = \frac{820*10^{-4}}{5*10^{-3}} * (14*10^{-3})(cos \ 170 \ t ) \\ \\ \\\phi _E = 2296*10^{-4} (cos \ 170 \ t)

I_d = \epsilon _o \frac{d \phi _ E }{dt}

I_d = 8.85*10^{-12}(2296*10^{-4})(-170)(sin(170 \ t))

I_d = -3.454*10^{-10} \ \ (sin  ( \ 170 \ t))

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a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
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Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
  • Vertical: constant downward acceleration at g = 10\; \rm m \cdot s^{-2}, starting at 0\; \rm m \cdot s^{-1}.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

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3 years ago
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