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3 years ago

9

Two technicians are discussing the need for the history of the vehicle. Technician A says that an accident could cause faults du

e to hidden damage. Technician B says that some faults could be related to a previous repair. Which technician is correct?a. Tech A onlyb. Tech B onlyc. Both a and bd. Neither a and b

1 answer:

Zigmanuir [339]3 years ago

3 0

Answer:

(C) both tech A and tech B

Explanation:

both technicians are correct because an accident could cause faults due to hidden damages which might not have been detected at the time of the accident and some faults could be related to previous repair probable due to human error from the repair man or even improper repair due to lack of adequate knowledge on the vehicle.

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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 268.75 $MVA$

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

What conventions are used in SI to indicate units

klemol [59]

Answer:

<u>Conventions used in SI to indicate units are as follows:</u>

Only singular form of units are used. for example: use kg and not kgs.
Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
Units named after a scientist should be written in small letters. for example: newton, henry.
Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k

5 0

3 years ago

What is matter made of ? Will mark as brainliest to answer

Masja [62]

Answer:

Explanation:

All matter is made up of atoms, which are turned up of protons, neutrons and electrons.

They bond together to make up matter

8 0

3 years ago

Read 2 more answers

A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo

m_a_m_a [10]

Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

The radius of the circular loop is $r = 9.50 \ cm = 0.095 \ m$

The intensity of the wave is $I = 0.0215 \ W/m^2$

The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

B is the magnetic field which can be mathematically represented from the equation as

$B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }$

substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

$B = 1.342 *10^{-8} \ T$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

So

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

$\epsilon = 0.1041 \ V$

7 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

kumpel [21]

Answer: $\frac{D_A}{v_B-v_A}$

Explanation:

Given

car A had a head start of $D_A$

and it starts at x=0 and t=0

Car B has to travel a distance of $D_A and d_a$

where $d_a$ is the distance travel by car A in time t

distance travel by car A is

$d_a=v_A\times t$

For car B with speed $v_B$

$d_B=D_A+d_a$

$v_B\times t=D_A+v_A\times t$

$t=\frac{D_A}{v_B-v_A}$

7 0

3 years ago