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grandymaker [24]
3 years ago
7

I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground. C

alculate the velocity in m/s after 1.10 seconds of freefall.
Physics
1 answer:
alex41 [277]3 years ago
6 0
Assuming that air resistance is negligible you can use suvat.
you don't know the distance, the initial velocity is 0, the velocity at the point you are trying to work out is v, the acceleration is gravity (9.8ms^-2) and the time is 1.1s.
you use the equation v=u+at which gives v=0+(9.8x1.1)=10.78ms^-1
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A defibrillator passes 14.0 A of current through the torso of a person for 0.0300 s. How much charge moves in coulombs?
krek1111 [17]
<h2>Answer:</h2>

4.2 C

<h2>Explanation:</h2>

The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.

i.e

Q = I x t

Where;

I = current = 14.0A

t = time taken  = 0.0300s

Substituting the values of I and t into the equation above gives

Q = 14.0 x 0.0300

Q = 4.2 C

Therefore quantity of charge moving is 4.2 C

4 0
3 years ago
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%

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Any collection of things that have some influence on one another can be thought of as a system. When we talk about a system, we
Katen [24]

Answer:

B

Explanation:

Its B

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