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anastassius [24]
3 years ago
15

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d

ifference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?
Physics
1 answer:
Archy [21]3 years ago
5 0

Answer:

The voltage is   V =   0.993V_b

Explanation:

From the question we are told that

   The time that has passed is  t = \frac{\tau}{2}

 Here \tau is know as the time constant

    The voltage of the  power source is   V_b

Generally the voltage equation for charging a capacitor is mathematically represented as

       V =  V_b  [1 - e^{- \frac{t}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }]

=>   V =  V_b  [1 - e^{- \frac{1}{2} }]

=>   V =   0.993V_b    

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The appropriate expression for the calculation of power by relating the angular energy in a given time.

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P=\tau\omega

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Our values are given by

\tau = 630Nm

\omega = 3200rev/min

The angular velocity must be transformed into radians per second then

\omega = 3200rev/min (\frac{2\pi rad}{60s})

\omega = 335.103rad/s

Replacing,

P=(630)(335.103)

P = 211.11*10^3W

P = 211.1kW

The average power delivered by the engine at this rotation rate is 211.1kW

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A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizonta
shusha [124]

Answer:

19.99 kg m²/s

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L = m r × v.

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where r = lsinθ.

l = 2.4 m

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resolving using newtons second law in the vertical and horizontal components.

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