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KatRina [158]
3 years ago
12

Help pls .. Which image shows an example of the strong nuclear force in action?

Physics
2 answers:
lora16 [44]3 years ago
8 0

Answer:D

Explanation:

A P E X

Vedmedyk [2.9K]3 years ago
3 0

Answer:

The answer is D.

Explanation:

Just took the quiz. Not sure if u still need help

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A person is wandering in woods and records his movements as a sequence of displacements: d1 is 351 m, 35.0° north of east; d2 is
Deffense [45]

Answer:

18.64

Explanation:

3 0
3 years ago
Read 2 more answers
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
The velocity of sound in air at 300C is approximately :
kumpel [21]

The velocity of sound in at 300C is 511.3 m/s.

Explanation:

The equation that gives the speed of sound in ar as a function of the air temperature is the following:

v=(331.3+0.6T) m/s

where

T is the temperature of the air, measured in Celsius degrees

In this problem, we want to find the speed of sound in ar for a temperature of

T=300^{\circ}C

Substituting into the equation, we find:

v=331.3 + 0.6(300)=511.3 m/s

So, the velocity of sound in at 300C is 511.3 m/s.

Learn more about sound waves:

brainly.com/question/4899681

#LearnwithBrainly

6 0
3 years ago
A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65 m/s in 5 seconds
melisa1 [442]

Answer:

C

Explanation:

65-35 = 30

30 divided by 5 = 6

The answer will be 6 m/s

6 0
2 years ago
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity

\lambda = Wavelength

Our value are given by,

v = 345m/s

\lambda = 63cm

Replacing

f = \frac{345}{0.63}

f = 547.61Hz

Therefore the frequency of the tuning fork is 547.61Hz

6 0
3 years ago
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