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Artemon [7]
3 years ago
14

An RC circuit takes t = 2.5 ms to charge to 35% of its full charge after it weas connected to a battery, What is the total time

it takes for the capacitor of this curcuit to charge to 95% of the full charge?
Physics
1 answer:
marin [14]3 years ago
4 0

Answer: time taken to charge to 95%

t = -5.80[ln(1-0.95)]

t = 17.38ms

Explanation:

For an RC Charging circuit

Where Vs

Vc = Vs (1 - e^(-t/RC))

Vc/Vs = 1 - e^(-t/RC)

-t/RC = ln(1 - Vc/Vs)

t = -RC[ln(1 - Vc/Vs)] and RC = k = -t/ln(1 - Vc/Vs)

Where ;

Vc = voltage across the capacitor

Vs = voltage supply

t = charging time = 2.5ms

k = RC = time constant.

Vc/Vs = 0.35

To calculate the time constant k;

k = -t/ln(1- Vc/Vs)

k = -2.5/ln(1-0.35)

k = 5.80ms

time taken to charge to 95%

t = -5.80[ln(1-0.95)]

t = 17.38ms

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                           =  (30,000 / 30)  sec

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Jazz is a 172 lb athlete who exercised at 7.6 METs. At this workload, what is his energy expenditure in kcals/min.? Round to the
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Answer:

Energy expenditure in K cals/min = 10 K cals /min (approximately)

Explanation:

As we know

Energy expenditure in Kcal/min=  METs x 3.5 x Body weight (kg) / 200

Given is METs=7.6

Weight of Jazz= 172lb=78.02kg

putting the values in formula,

Energy expenditure in K cals/min=  7.6 x 3.5 x 78.02 / 200

                                                       =10.38 K cals /min

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Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min

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Answer:

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Explanation:

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q₂ is the charge on the second object

r is the distance between the two objects

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q₂ = 2q₂

Force between the two charged objects will become

F_2 = \frac{K2q_12q_2}{r^2} =  \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1

Therefore, the force between them quadruples

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