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Lesechka [4]
3 years ago
13

A wye-connected load has a voltage of 480v applied to it. What is the voltage drop across each phase

Physics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

Y_A=277.128 \angle 30v

Y_B=277.128 \angle (-150)v

Y_C=277.128 \angle (90)v

Explanation:

From the question we are told that

Voltage V_L_L =480v

Generally in a case of Y_connection V_p_ h is mathematical represented as

V_p_h=\frac{V_l_l}{\sqrt{3}} \angle (\phi-30)v

Generally voltage drop across phase A

Y_A=\frac{408}{\sqrt{3}} \angle -(0-30)

Y_A=277.128 \angle 30v

Generally voltage drop across phase B

Y_B=277.128 \angle (-30-120)

Y_B=277.128 \angle (-150)v

Generally voltage drop across phase C

Y_C=277.128 \angle (-30+120)

Y_C=277.128 \angle (90)v

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3 years ago
At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until
Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

  • Initial angular velocity,  \omega_0 = 30.0 \text{ rad/s}
  • Angular acceleration, \alpha =35.0 \text{ rad/s}^2

(A)

At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by

\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

\omega = \omega_0+\alpha  t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t

Here, \theta_2=440 (the angular displacement during deceleration)

The subscripts, <em>i</em> and <em>f</em>, on <em>ω</em> denote the initial and final angular velocities during deceleration.

\omega_i = 100

\omega_f = 0

t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}

This is the time for deceleration. The deceleration began at <em>t</em> = 2 s.

Hence, the wheel stops at <em>t</em> = 2 + 8 = 10 s.

(C)

The deceleration is given by

\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2

The negative sign appears because it is a deceleration.

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3 years ago
A car is moving at 14 m/s. After 30 s, its speed inncreased to 20 m/s. What is the acceleration over time
Juli2301 [7.4K]

Answer:

0.2 m/s^2

Explanation:

initial speed 14m/s

final speed 20m/s

acceleration:

(20m/s - 14m /s) /30s = (6m/s)/30s = 0.2 m/s^2

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