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Nata [24]
3 years ago
14

Estimate how much power is saved if the voltage is stepped up from 150 VV to 1500 VV and then down again, rather than simply tra

nsmitting at 150 VV . Assume the transformers are each 99%% efficient. Express your ans
Physics
1 answer:
gtnhenbr [62]3 years ago
3 0

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

Suppose 79 kW is to arrive at a town over two 0.115 Ω lines. Estimate how much power is saved if the voltage is stepped up from 150 V to 1500 V and then down again, rather than simply transmitting at 150 V. Assume the transformers are each 99%% efficient. Express your ans up to two decimal points.

Answer:

Power saved = 30772.58 W or 30.77 kW

Explanation:

Since the power loss depends upon current, we need to find the current at each voltage level.

Current at 150 V:

I = P/V = 79,000/150 = 526.67 A

Current at 1500 V:

Considering transformer efficiency of 99%

79,000/0.99 = 79797.97 W

I = P/V = 79,797.97/1500 = 53.19 A

Power loss at 150 V:

P = I²R = (526.67)²*0.115 = 31898.84 W

Power loss at 1500 V:

Here we have to consider both of the losses, due to the transmission line and due to transformer itself

P = I²R = (53.19)²*0.115 = 325.35 W

So the total power at the sending end would be

Receiving end power + losses

79,000 + 325.35 = 79,325 W

Power at the input of the transformer is

79,325/0.99 = 80126.26

Power lost in the transformer is

80,126.26 - 79,000 = 1126.26 W

So the power that can be saved by stepping up the voltage from 150 to 1500 V is

Power saved = 31898.84 - 1126.26

Power saved = 30772.58 W

Power saved = 30.77 kW

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From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle
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Answer:

Part a)

y = 88.5 m

Part b)

v_x = 7.7 m/s

v_y = 2.5 m/s

Part c)

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x = v_x t

equation of position in y direction is given as

y = v_y t + \frac{1}{2}gt^2

Part d)

x = 30.8 m

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

v_x = 8.10 cos18

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Since it took t =4 s to reach the ground

so its initial y coordinate is given as

y = v_y t + \frac{1}{2}a_y t^2

y = 2.5(4) + \frac{1}{2}(9.81)(4^2)

y = 88.5 m

Part b)

components of the velocity is given as

v_x = 8.10 cos18

v_x = 7.7 m/s

v_y = 8.10 sin18

v_y = 2.5 m/s

Part c)

equation of position in x direction is given as

x = v_x t

equation of position in y direction is given as

y = v_y t + \frac{1}{2}gt^2

Part d)

distance where it will strike the floor is given as

x = v_x t

x = 7.7 \times 4

x = 30.8 m

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

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H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

y = v_y t + \frac{1}{2}gt^2

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7 0
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Answer:

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the correct answer is C

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3 years ago
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