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Vinil7 [7]
3 years ago
14

The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two

seconds later, the propeller is rotating at 10rad/s. Through how many revolutions has the propeller rotated through during the first two seconds?
Physics
2 answers:
UkoKoshka [18]3 years ago
5 0

Answer: The propeller has rotated 5 revolutions

Explanation:

Number of revolutions = angular velocity/ time

Given:

Angular velocity=10rad/s

Time=2seconds

Number of revolutions = 10/2= 5 revolutions.

Maurinko [17]3 years ago
3 0

Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. \omega _0=0 rad/s

after t=10 s

\omega =10 rad/s

using \omega =\omega _0+\alpha t

10=0+\alpha \cdot 10

\alpha =1 rad/s^2

Revolutions turned in 2 s

\theta =\omega _0t+\frac{\alpha t^2}{2}

\theta =0+\frac{1\times 2^2}{2}

\theta =2 rad

To get revolution \frac{\theta }{2\pi }

=\frac{2}{2\pi}=0.318\ revolutions

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12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

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How do you solve for a net force
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Answer:

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8 0
3 years ago
The voltage supplied to a circuit is 17 V and the current running through is 10 A. What is the power generated?
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Answer:

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Explanation:

Applying

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From the question,

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Substitute these values into equation 1

P = (17×10)

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Hence the power generated is 170 W.

The right option is A. 170 W

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