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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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A parachutist jumps out of an airplane and accelerates with gravity to a maximum velocity of 58.8 m/s in 6.00 seconds. She then
MrRissso [65]

Answer:

914m

Explanation:

We must divide the parachutist motion into three parts:

First we have free fall motion:

y_1=\frac{v_f^2-v_0^2}{2g}\\y_2=\frac{(58.8\frac{m}{s})^2-0^2}{2(9.8\frac{m}{s^2})}=176.4m

Then, we have a uniformly accelerated motion. The initial speed is this part will be the same final speed as the previous part.

a=\frac{v_f-v_0}{t}\\a=\frac{10.0\frac{m}{s}-58.8\frac{m}{s}}{4s}=-12.2\frac{m}{s^2}\\y_2=\frac{v_f^2-v_0^2}{2a}\\y_2=\frac{(10.0\frac{m}{s})^2-(58.8\frac{m}{s})^2}{2(-12.2\frac{m}{s^2})}=137.6m

Finally, we have a uniform linear motion:

y_3=v*t\\y_3=10\frac{m}{s}*60s=600m

The total heigh will be the sum of all heights:

y=y_1+y_2+y_3\\y=176.4m+137.6m+600m=914m

7 0
3 years ago
Please take 50 points please help me <br>please ​
ch4aika [34]

I am using a calculator for this question so hopefully this helps + I added working out just in case.

a) mean - 124 (add them all together then divide by 25)

b) standard deviation - 7.776888838 (7.76 to d.p)

c) the person might have eaten foods that increase or decrease the BP so every time they took their blood pressure, it might not have been done as the same routine as others (if this makes sense). Also, some blood pressures are lower than others like for example some might go down to 110 but others go all the way up to like 132, does why I said that they might have eaten food or done activites that increases the blood pressure.

6 0
2 years ago
A human eye is an example of a simple machine found in the human body. True or false
Lisa [10]

Answer:

The answer to this is True!

6 0
3 years ago
With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release
Dmitry [639]

THIS IS THE COMPLETE QUESTION BELOW

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground

And what speed does the ball hit the ground? Solve this problem using energy.

Answer

a)minimum speed must you toss a 130 g is 15.9090m/s

b)speed the ball hit the ground is 16.57m/s

Explanation:

a)We know that For any closed/isolated system, the total energy is CONSERVED.

K.E. lost by the ball=The change in P.E of the ball at 1.1m above the ground as well as the P.E. of the ball at 14 m-high roof

This statement can be expressed as the expression below from K.E and P.E energy formula

P.E. = mgh

K.E. = (1/2)mv^2

Therefore,

(mgh1 - mgh2)=(1/2)mv^2

Where h2=the ball height above the ground=1.1m

h1=ball height at roof of the gymnasium= 14m

Then if we substitute we have

[(10) x (0.14) x (9.81)] - [(1.1) x (0.14) x (9.81)] = (1/2)(0.14)(v^2)

16.45137=0.065V^2

V=15.9090m/s

minimum speed must you toss a 130 g is 15.9090m/s

b)To calculate the speed the ball hit the ground?

This is the highest point (14m-high roof),and the type of the energy the ball possesses is Po.tential energy only.

At the lowest point (ground), the energy the ball possesses is K.E. only.

P.E at 10m-high roof = K.E. at ground.

(14) x (0.13) x (9.81) = (1/2) x (0.13) x v^2

17.8542= 0.065V^2

V= 16.57

Therefore,And speed the ball hit the ground is 16.57m/s

6 0
3 years ago
A 1.7m long barbell has a 20kg weight on its left and a 35kg weight on its right. (a) If you ignore the weight of the bar itself
Levart [38]

Answer:

(a) 1.08 m

(b) 1.06 m

Explanation:

<u>Step 1:</u> calculate the center of gravity from 20kg mass

Let the center of gravity from 20kg mass = X

Applying the principle of moment; clockwise moment = ant-clockwise moment

20*X = 35*(1.7-X)

20X = 59.5 - 35X

55X = 59.5

X = 59.5/55

X = 1.08 m

Ignoring the weight of the bar, the center of gravity is 1.08m from left end of the barbell.

<u>Step 2:</u> calculate the center of gravity from 20kg mass, if the 8.0kg mass of the barbell in considered.

Applying the principle of moment

(20*X)+\frac{X}{2}(\frac{8}{1.7}) = 35(1.7-X) + (\frac{1.7-X}{2})(\frac{8}{1.7})

20X + 2.353X = 59.5 -35X +2.353(1.7-X)

20X +2.353X = 59.5 -35X + 4 - 2.353X

59.706X = 63.5

X = 63.5/59.706

X = 1.06 m

considering the weight of the bar, the center of gravity is 1.06m from left end of the barbell.

4 0
3 years ago
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