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wel
2 years ago
9

The voltage across a 5-uF capacitor is: v (t )equals 10 cos open parentheses 6000 t close parentheses space straight V. What is

the current through this capacitor?
Physics
1 answer:
mamaluj [8]2 years ago
4 0

Answer:

- 0.3sin6000t A

Explanation:

Voltage, v = 10 cos 6000t V

Capacitance = 5-uF

Current flowing through, i(t)

i(t) = c * d/dt (V)

c = 5-uF = 5 * 10^-6 F

i(t) = (5 * 10^-6) * d/dt(10 cos 6000t)

d/dt(10 cos 6000t) = (10 * 6000) * (-sin 6000t)

Hence,

i(t) = (5*10^-6) * (10*6000) * (-sin 6000t)

i(t) = 5*10^-6 * 6*10^4 * - sin6000t

i(t) = 30 * 10^-2 * - sin6000t

i(t) = 0.3*-sin6000t

i(t) = - 0.3sin6000t Ampere

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3 years ago
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

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- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

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drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

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Using the equation for the lift force

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we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

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b) the required engine thrust and power to maintain level flight.

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F_D = T = C_D\frac{1}{2}ρV²A

we substitute

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Therefore, required power is 79.7 kW

8 0
3 years ago
Responsibilities of a class are derived from ________________.
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Answer:

The correct option is a behavioural models of the the to-be system.

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As the list is derived from the behavioural models of the system, thus option a is the correct option.

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