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jarptica [38.1K]

3 years ago

12

A driverless car collides head on with a stationary sign. The collision brings the car to a stop, but does not move the sign.

2 answers:

Citrus2011 [14]3 years ago

7 0

Oh ghjoihhbbnbv high hj. Hi

neonofarm [45]3 years ago

4 0

Answer:

I'm sorry hindi ko po alam ung sagot sorry po

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How is a vector represented in symbol form?

VikaD [51]

I believe it’s just a “v” with an arrow above it.

6 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 268.75 $MVA$

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

An 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its

nikdorinn [45]

Answer:

The average induced emf in the loop is 0.20 V

Explanation:

Given:

Radius of loop $r = \frac{d}{2} = 9.25 \times 10^{-2}$ m

Magnetic field $B = 1.5$ T

Change in time $\Delta t = 0.20$ sec

According to the faraday's law,

Induced emf is given by

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Where $\phi =$ magnetic flux

$\phi = BA\cos0$ ( here $\theta = 0$ )

Where $A = \pi r^{2}$

We neglect minus sign because it's shows lenz law

$\epsilon = \frac{B \pi r^{2} }{\Delta t}$

$\epsilon = \frac{1.5 \times 3.14 \times (9.25 \times 10^{-2} )^{2} }{0.20}$

$\epsilon = 0.20$ V

Therefore, the average induced emf in the loop is 0.20 V

4 0

4 years ago

FILL IN THE BLANKS here on earth, the pull of gravity on a mass of 1 kg is ......... newtons

Aleks [24]

Answer:

9.8 Newton

Explanation:

At average gravity on earth (conventionally, g=9.80665m/s2),

a kilogram mass exerts a force of about 9.8 newton

I hope this answer helps you

4 0

3 years ago

Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec

yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{3.74\times 9.8}{0.0161}$

k = 2276.52 N/m

The frequency of vibration is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}$

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

8 0

3 years ago