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daser333 [38]
3 years ago
6

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

w many time constants (a decimal number) must elapse before a capacitor in a series RC circuit is charged to 39.0% of its equilibrium charge
Physics
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer: t = 0.492τ

Explanation: In a circuit where there is a resistor and a capacitor, the equation for a charging capacitor is given by:

q = q_0(1 - e^{\frac{t}{RC} })

where:

q_0 is the equilibrium charge

q is the charge at time t

RC is time constant also called τ (tau)

For this problem, the circuit is charged to 39%, which means: q = 0.39 q_0

0.39q_0 = q_0 (1 - e^\frac{-t}{RC} )

0.39 = 1 - e^\frac{-t}{RC}

e^\frac{-t}{RC} = 0.61

\frac{-t}{RC} = ln(0.61)

-t = ln(0.61)τ

t = 0.492τ

For the condition to be met it is needed 0.492 time constants must elapse.

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Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

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By solving the above equation for "t" we have:

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