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Ierofanga [76]
2 years ago
7

At resonance, a driven rlcrlc circuit has vcvc = 5.0 vv , vrvr = 8.0 vv , and vlvl = 5.0 vv . part a what is the peak voltage ac

ross the entire circuit?
Physics
1 answer:
Simora [160]2 years ago
4 0

The real peak voltage is 120/0.707 = 170 V.

The peak voltage is the highest point or voltage value in any voltage waveform. A power quality issue arises when Pulse Width Modulation (PWM) devices, such as variable frequency drives, are added to a power system with a peak voltage equal to the square root of two times the RMS voltage. The peak voltage, for instance, if the RMS voltage is 85 V. The average voltage and maximum voltage of AC power coming from the wall are both about 110 V. Therefore, the real peak voltage is 120/0.707 = 170 V. The sinusoid's amplitude is divided in half by this. Peak-to-peak voltage (Vp-p), also known as the total amplitude, is 340 V, or twice the peak voltage.

To learn more about peak voltage here:-

brainly.com/question/24334542

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anyanavicka [17]

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

8 0
3 years ago
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An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte
telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2

a. The angular acceleration is Torque divided by moments of inertia:

\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2

b. 5 revolution would be equals to 10\pi rad, or 31.4 rad. Since the engine just got started

\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5

\omega = \sqrt{1893.5} = 43.5 rad/s

c. Work done during the first 5 revolution would be torque times angular displacement:

W = T*\theta = 1930 * 31.4 = 60633 J

d. The time it takes to spin the first 5 revolutions is

t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s

The average power output is work per unit time

P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

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7 0
3 years ago
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Answer:

410.4J

Explanation:

Step one:

given

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h=12.49m

Required

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Step two:

the work done is given  as

WD= force* distance

WD= 32.86*12.49

WD= 410.4J

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A sprinter who is running a 200-m race travels the second 100 m in much less time than the first 100m because
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Answer:

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Answer:

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