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3 years ago

Write all the different ways you can think of that describe what it means to be healthy

Physics

1 answer:

My name is Ann [436]3 years ago

3 0

Answer:

Eat more healthy foods. Workout and build your immune system.

Explanation:

Eat Healthy foods like Carrots, Apples, Bannas, Pears, and anything that deals with not much of any sugar. An example of unhealthy foods is Cakes, Chocolates, Candy, and more. Drink a lot of water.

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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:

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

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

Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What i

frutty [35]

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

$v = \sqrt{\frac{F}{\mu} }$

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ = m / l

where m = mass of the string

l = length of the string

This implies that:

$v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }$

Let us make mass, m, the subject of the formula:

$v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}$

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

$m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg$

5 0

3 years ago

If you doubled the load resistor in a Wheatstone bridge, the load current would not be half as much. Why not?

Fofino [41]

Doubling the size of a load resistor increases the load current. Increasing the load resistance, in turn the total circuit resistance is reduced. The load current would not be half as much since when you increase the size of load resistor then load current increases.

6 0

3 years ago

How are humans potentially vulnerable to some of these same chemicals? How are humans potentially vulnerable to some of these sa

Svetradugi [14.3K]

The 5 major unit processes include chemical coagulation, flocculation, sedimentation, filtration, and disinfection (described below). There are chemicals added to the water as it enters the various treatment processes.

6 0

2 years ago

Planet Nine is speculated to be on average 20 times farther away from the Sun than Neptune (on average distance from the Sun). H

saveliy_v [14]

Answer:

The distance is 55.636 billion miles, or 528.2 AU.

Explanation:

Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

$d=(20)(2781800000\ miles)=55636000000\ miles$

or 55.636 billion miles.

Since 1 astronomical unit (AU) is 93 million miles, that distance is also:

$d=(55636000000\ miles)(\frac{1AU}{93000000\ miles})=598.2\ AU$

6 0

3 years ago