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11111nata11111 [884]
3 years ago
10

A capacitor is connected to an ac power supply. The maximum current is 4.0 A and the reactance of the capacitor is Xc = 8.0 Ω. W

hat is the average power dissipated by the capacitor? Select one: a. 32 W b.0 W c. 128 W d. 64 W
Physics
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

option (b)

Explanation:

Io = 4 A

Xc = 8 ohm

The formula for the average power dissipated is given by

P = Vrms x Irms x cos Ф

Where, Ф be the phase difference between Vrms and Irms .

As we know that in case of capacitor, the current leads the voltage by 90 degree, it means the value of Ф is 90 degree.

Thus, Average power = Vrms x Irms x Cos 90

Average power = 0 Watt

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The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply
kykrilka [37]

Answer: Got It!

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v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

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0  0

4 0
3 years ago
What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
Marianna [84]

Answer:

The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

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Using Bernoulli effect

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

When two point are at same height so ,

P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

Q=A_{1}v_{1}

v_{1}=\dfrac{Q}{A_{1}}

v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}

v_{1}=56.58\ m/s

For output velocity,

v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}

v_{2}=6.28\ m/s

Put the value into the formula

P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

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(b). We need to calculate the maximum height

Using formula of height

\Delta P=\rho g h

Put the value into the formula

1.58\times10^{6}=1000\times9.8\times h

h=\dfrac{1.58\times10^{6}}{1000\times9.8}

h=161.22\ m

Hence, The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

8 0
3 years ago
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
NARA [144]

Answer:

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P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

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The time taken to lift the load is 141.26626 seconds

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4 0
3 years ago
How much work does it take to move a 50 μC charge<br> against a 12 V potential difference?
lukranit [14]
<span>work =V*Q =12*50*10^-6

The total work done will be equal to 

work = V.Q

which means 

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>
8 0
3 years ago
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