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Dmitry_Shevchenko [17]
3 years ago
6

An electric winch operates on 140 volts​ [V] and draws 6 amperes​ [A] of current. The winch has an efficiency of 64​%. The winch

is used to lift an object that weighs 510 ​pound-force ​[lb Subscript f​] for 11 seconds​ [s]. Determine the height the object is lifted​ to, in units of meters​ [m].
Engineering
1 answer:
Elis [28]3 years ago
3 0

Answer:

Explanation:

efficiency = (energy output / energy input)

energy input = ivt where i is current = 6 A, v is volts = 140 and t time = 11 seconds

energy input = 6 × 140 × 11 = 9240 J

energy output = 0.64 × 9240 = 5913.6 J

510 pound-force = 2268.593 N

at height h, potential energy = mgh = 5913.6 J

2268.593 × h = 5913.6

h = 5913.6 / ( 2268.593) = 2.61 m

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Sketch the asymptotes of the Bode plot magnitude and phase for the open-loop transfer ()=100(S+1)/((S+10)(S+100)) Use MATLAB to
Salsk061 [2.6K]

The asymptotes of the open loop transfer are:

  • Horizontal: y = 0
  • Vertical: x = -10 and x = -100

<h3>How to plot the asymptotes?</h3>

The open loop transfer function is given as:

f(s) = 100(s + 1)/((s + 10)(s + 100))

Set the numerator of the function to 0.

So, we have:

f(s) = 0/((s + 10)(s + 100))

Evaluate

f(s) = 0

This means that, the vertical asymptote is y = 0

Set the denominator of the function to 0.

(s + 10)(s + 100)  0

Split

s + 10 = 0 and s + 100 = 0

Solve for s

s = -10 and s = -100

This means that, the horizontal asymptotes are s = -10 and s = -100

See attachment for the graph of the asymptotes

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brainly.com/question/4084552

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2 years ago
Find the resultant of the force system on the body OABC as shown .find the points where the resultant will cut the X and Y axis?
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Explanation:

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\sqrt{}  {x}^{2}  +  {y}^{2}

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3 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
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Answer:

119.1°c

Explanation:

Temperature can be referred to as a measure of the warmth or coldness of an object or substance with reference to some standard value. The temperature of two systems is the same when the systems are in thermal equilibrium.

To determine the tube surface temperature immediately after installation and after prolonged service.

We will apply to appropriate formula, but first lets define the terms in the question.

Please kindly check attachment.

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Esma and hasan are putting the finishing touches on their model for a tiny chip-based energy source to power their lighting syst
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The phase of the engineering design process which should be completed next is to test their work and is denoted as option C.

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These are the series of steps and techniques which are done by individuals in the making of functional product and services.This employs the use of scientific methods and also ensures an easier living for different individuals.

The first stage involves identifying the problem and then building a prototype through the use of different materials. This is then tested before the final finishing work is done to ensure the parts are properly placed before they are moved for evaluation by other people.

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7 0
2 years ago
Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and
jonny [76]

Answer:

- Work done is 2.39 kJ

- heat transfer is 20.23 kJ/kg

Explanation:    

Given the data in the question;

First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;

Specific Volume v₁ =  0.02547 m³/kg

Specific enthalpy u₁ = 243.78 kJ/kg

Specific Volume V₂ = 0.02846 m³/kg

Specific enthalpy u₂ = 261.62 kJ/kg

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Any changes in kinetic and potential energy are negligible.

So we determine the work done by using the equation at constant pressure

]Work done W = p( v₂ - v₁ )

we substitute

W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )

W = 800 kPa( 0.00299 m³/kg )

W = 2.39 kJ

Therefore, Work done is 2.39 kJ

Heat transfer;

using equation at constant pressure

Heat transfer Q = W + ( u₂ - u₁ )

so we substitute

Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )

Q = 2.392 kJ +  17.84 kJ/kg )

Q = 20.23 kJ/kg

Therefore, heat transfer is 20.23 kJ/kg

3 0
3 years ago
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