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2 years ago

Might a synchronous motor equally well be called a synchronous inductor

1 answer:

3 0

In Engineering, a synchronous motor is sometimes referred to as synchronous capacitor but not a synchronous inductor.

<h3>What is a synchronous motor?</h3>

A synchronous motor can be defined as an alternating current (AC) electric motor in which the rotation speed of the shaft is directly proportional to the frequency of the supply current, especially at steady state.

Basically, a synchronous motor is sometimes referred to as a synchronous capacitor because it can be used to correct the power factor (PF) of a lagging load such as an induction motor.

However, a synchronous motor cannot be called a synchronous inductor because the rotation speed of the shaft is not proportional to the synchronous speed of the motor. Thus, they are called an synchronous motor.

Read more on synchronous motor here: brainly.com/question/12975042

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3 years ago

A steel bolt has a modulus of 207 GPa. It holds two rigid plates together at a high temperature under conditions where the creep

VikaD [51]

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14.36((14MPa) approximately

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2 years ago

A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni

arlik [135]

Answer:

The population size would be $p' = 5000$

The yield would be $MaxYield = 2082 \ fishes \ per \ year$

Explanation:

So in this problem we are going to be examining the application of a population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield, so basically we would be applying an engineering solution to fishing

So the current yield which is mathematically represented as

$\frac{dN}{dt} = \frac{2000}{1 \ year }$

Where dN is the change in the number of fish

and dt is the change in time

So in order to obtain the solution we need to obtain the rate of growth

For this we would be making use of the growth rate equation which is

$r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}$

Where N is the population of the fish which is given as 4,000 fishes

and K is the carrying capacity which is given as 10,000 fishes

r is the growth rate

Substituting these values into the equation

$r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]} =0.833$

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

$Max Yield = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year$

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by $p'$ would be

$p' = \frac{K}{2} = \frac{10,000}{2} = 5000\ fishes$

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3 years ago

Read 2 more answers

The sum ofall microscopic forms of energy of a system is quantified as flow energy. a)True b) False

Sliva [168]

Answer: b) False

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3 years ago