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hram777 [196]
3 years ago
5

An autotransformer is used to reduce the voltage of a 100-kilovolt amp, 480-volt secondary of an isolated type transformer, to s

upply a 100-kilovolt amp load with 277 volts. What size autotransformer is needed? (Use the co-ratio to determine the size.)
Physics
1 answer:
tigry1 [53]3 years ago
5 0

Answer:

42KVA

Explanation:

Given data

High Voltage (HV)= 480V

Low Voltage (LV)= 277V

Fo find

Size of transformer=?

Solution

To find the size of transformer here we use the co-ratio.The Co-ratio is given as:

Co-Ratio= (HV - LV)/HV

where

HV is High Voltage

LV is Low Voltage

Now put the values we get

Co- Ratio=(480-277)/480=.42

So the size of transformer is 42KVA

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Why does the density of a substance remain the same for different amount of the substance
tresset_1 [31]

Think of it this way: 
-- Any time you have something that means (some number) PER UNIT,
it doesn't matter how many units there are on the table or in the bucket,
because that amount doesn't change the (number) PER UNIT.

-- If oranges cost $1 PER POUND, it doesn't matter how many pounds
you buy, the whole bagful is still $1 PER POUND.

-- If a certain salad dressing has 40 calories PER Tablespoon, it doesn't
matter whether you eat a drop of it or drink the whole jar.  You still get
40 calories PER Tablespoon.

-- Density means '(mass) PER unit of volume'.  Whether you have a tiny
chip of the substance or a whole truckload of it, there's still the same
amount of mass IN EACH unit of volume.

6 0
3 years ago
Is static electricity more harmful than it is helpful or vice versa
Deffense [45]

Answer:

You might even see a spark if the discharge of electrons is large enough. The good news is that static electricity can't seriously harm you. Your body is composed largely of water and water is an inefficient conductor of electricity, especially in amounts this small. Not that electricity can't hurt or kill you.

Explanation:

You might even see a spark if the discharge of electrons is large enough. The good news is that static electricity can't seriously harm you. Your body is composed largely of water and water is an inefficient conductor of electricity, especially in amounts this small. Not that electricity can't hurt or kill you.

6 0
3 years ago
Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
3 years ago
On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around
Mekhanik [1.2K]

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

7 0
4 years ago
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
konstantin123 [22]

Answer:

I=2.71\times 10^{-5}\ A

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.  

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}, r is radius

Let I is the displacement current. It is given by :

I=C\dfrac{dV}{dt}

Here, \dfrac{dV}{dt} is rate of increasing potential difference

So

I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A

So, the value of displacement current is 2.71\times 10^{-5}\ A.

4 0
3 years ago
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