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Shtirlitz [24]
3 years ago
12

An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH

, and C = 23.1 μF. Find (a) the capacitive reactance XC, (b) the impedance Z, and (c) the current.
Physics
1 answer:
Doss [256]3 years ago
3 0

Answer:

Xc= 17.267 Ω,   Z= 415.5 Ω,   I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH,  C = 23.1 μF

a)

Capacitive reactance = Xc=?

Xc= \frac{1}{2\pi fC}

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z=\sqrt{ R^2 + (Xl - Xc)^2}

Xl= 2π * f * L  

Xl= 2π * 399 * 147 * 10^{-3}

Xl= 368.5 Ω

Z=\sqrt{ R^2 + (Xl - Xc)^2} = \sqrt{222^2 + (368.5 - 17.267)^2}

Z= 415.5 Ω  

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

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ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

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Answer:

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B. Vy = -45.73 m/s

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D. θ = -85.46°

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r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

Vy = - 11.46 * 3.99 = -45.73 m/s

C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]

|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

|V| = 45.87 m/s

D. Angle of the velocity relative to the x axis, θ is given as:

tanθ = Vy/Vx

tanθ = -45.73/3.63

tanθ = -12.6

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