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Ierofanga [76]
3 years ago
5

A 325 N force applied to an object for 5.5 s. What’s the impulse?

Physics
1 answer:
vazorg [7]3 years ago
4 0
According to Newton second law of motion, the resultant force is directly proportional to the rate of change in momentum while maintaining other factors constant. Therefore, F = (mv-mu)/t where F is the resultant force , m is the mass of the object, v is the final velocity and u is the initial velocity.
Hence, Ft = mv-mu, but impulse is given by force multiplied by time, thus, impulse is equivalent to the change in momentum.
Impulse = Ft
              = 325 × 2.2 sec
              = 715 Ns
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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
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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

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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%

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