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marin [14]
2 years ago
15

Please help extra points!!!!

Physics
1 answer:
Allushta [10]2 years ago
6 0

Answer:

If you increase either mass or velocity, the momentum of the object increases proportionally.

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Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer
SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The High Voltage Rating for Auto - Transformer is 86kV

The  Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75MVA

b

The efficiency is 99.4%

Explanation:

From the question  we are given are given that

 The transformer has Mega Volt Amp rating of 25MVA

                          The frequency is 60-Hz

                           Voltage rating 8.0kV : 78kV

   The short circuit test gives : 453kV,321A,77.5kW

   The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the  Low Voltage Rating for Auto - Transformer is 78kV

 Now to obtain the current flowing through the 8kV  coil in the Auto-transformer we have

             \frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}

The volt will cancel each other

             \frac{25*10^6}{8*10^3} =  3125\ A

 Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

MVA \ rating = (86*10^3)(3125) =268.75

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil I_2 as shown in the circuit diagram can be obtained as

       I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321

Now the efficiency can be obtained as thus

           \frac{(operational \ MVA )*(Power factor \pf))}{(operational\  MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}

             =99.941%

8 0
3 years ago
A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum
Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , m_1= 0.473 kg

mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

Final speed of puck 1 and puck 2 be v_1\, and\, v_2  respectively

Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

=>1.594=0.5775\times v_1+ v_2 -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

u_2-u_1=v_1-v_2

=>0-2.76=v_1-v_2 ------(B)

From equation (A) and (B)

v_1=-0.739\frac{m}{s}

and v_2=2.02\frac{m}{s}

Thus the final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

       

3 0
3 years ago
A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at
Gnom [1K]

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

W=p(V_f -V_i)

where in this case we have:

p = 2 bar = 2\cdot 10^5 Pa is the pressure

V_i = 0.1 m^3 is the initial volume

V_f = 0.12 m^3 is the final volume

Substituting,

W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ

The 1st law of thermodynamics also states that

\Delta U = Q-W

where

\Delta U is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

\Delta U = +0.25 kJ

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ

7 0
4 years ago
Earth that are large enough to be detected by radar
Novosadov [1.4K]

Answer:

I don't now sorry HHHAHAH GOOD LUCK

8 0
3 years ago
An obiect of mass weighing 5,24 k acceleration due to gravity is 9 8 meters/second2 is raised to a height of 1.63 meters. What i
Julli [10]
83.79 J (using significant digits)
7 0
3 years ago
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