Discharging a capacitor: a capacitor of capacitance c =4.50 uf is connected in series with a resistor r = 8.30 k2 to a (a) what is the time constant? (b) what is the maximum charge stored in the capacitor? (c) what is the time (in seconds) to reach 14% of qmax? (d) what is the current value at % qmax?

The answer is b hope this

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

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