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FrozenT [24]
3 years ago
14

A single-phase inductive load draws a 10 MW at 0.6 power factor lagging. Calculate the reactive power of a capacitor to be conne

cted in parallel with the load to raise the power factor to 0.89. Type your answer on the box, and make sure your answer have two decimal digits and the correct sign.

Engineering
1 answer:
Dafna1 [17]3 years ago
5 0

Answer: The reactive power Q = 9.18 MVAR

Explanation: please find the attached file for the solution

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A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid
dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

3 0
4 years ago
Read 2 more answers
Write a function called largest3 which takes 3 numbers as parameters and returns the largest of the 3. Write a program which tak
tamaranim1 [39]

Answer:

The solution code is written in Python.

  1. def largest3(num1, num2, num3):
  2.    largest = num1
  3.    if(largest < num2):
  4.        largest = num2
  5.    
  6.    if(largest < num3):
  7.        largest = num3
  8.    
  9.    return largest
  10. first_num = int(input("Enter first number: "))
  11. second_num = int(input("Enter second number: "))
  12. third_num = int(input("Enter third number: "))
  13. largest_number = largest3(first_num, second_num, third_num)
  14. print("The largest number is " + str(largest_number))

Explanation:

<u>Create function largest3</u>

  • Firstly, we can create a function <em>largest3 </em>which take 3 numbers (<em>num1, num2, num3</em>) as input. (Line 1).
  • Please note Python uses keyword <em>def </em>to denote a function. The code from Line 2 - 10 are function body of <em>largest3</em>.
  • Within the function body, create a variable,<em> largest</em>, to store the largest number. In the first beginning, just tentatively assign<em> num1 </em>to<em> largest</em>. (Line 2)
  • Next, proceed to check if the current "<em>largest</em>" value smaller than the<em> num2 </em>(Line 4). If so, replace the original value of largest variable with <em>num2</em> (Line 5).
  • Repeat the similar comparison procedure to<em> </em><em>num3</em> (Line 7-8)
  • At the end, return the final value of "<em>largest</em>" as output

<u>Get User Input</u>

  • Prompt use input for three numbers (Line 13 -15) using Python built-in <em>input</em> function.
  • Please note the input parts of codes is done outside of the function <em>largest3</em>.

<u>Call function to get largest number and display</u>

  • We can simply call the function<em> largest </em>by writing the function name <em>largest</em> and passing the three user input into the parenthesis as arguments. (Line 17)
  • The function <em>largest </em>will operate on the three arguments and return the output to the variable <em>largest_number</em>.
  • Lastly, print the output using Python built-in <em>print</em> function. (Line 18)
  • Please note the output parts of codes is also done outside of the function<em> largest3</em>.
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Who wants me to make a device to resurrect Juice Wrld
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I need help with my autos
oksano4ka [1.4K]

Answer:

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Explanation:

6 0
3 years ago
A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turn
Ostrovityanka [42]

Answer: a) 150 rev. b) 2105 rev.

Explanation:

a) Assuming a uniformly accelerated motion, we can use the equivalent kinematic equations, replacing linear variables by angular ones.

In order to get the number of revolutions executed, we can use this:

ωf² - ω₀² = 2 γ Δθ (1)

For the first part, we know that ω₀ = 0 (as it starts from rest).

We can find out the value of angular acceleration γ, just applying the definition of angular acceleration, as the change in angular velocity, regarding time, as follows:

γ = (ωf - ω₀) / Δt (2)

As we would want to use SI units, it is advisable to convert the value of ωf, from rpm to rad/sec.

3600 rev/min . (1min/60 sec) . (2π rad/rev) = 120π rad/sec

Replacing in (2), we get γ:

γ = 120 π / 5 rad/sec² = 24 π rad/sec²

Replacing in (1) and solving for Δθ:

Δθ = 120² π² / 2. 24 π = 300 π rad

As 1 rev = 2π rad, Δθ = 150 rev

b) For the second part, we can use exactly the same equations, taking into account that ω₀ = 120 π rad/sec, and that ωf = 0.

The new value for γ is as follows:

γ = -120π  / 70 rad/sec² = -1.71 rad/sec²

Replacing in (1) and solving for Δθ, we get:

Δθ = -120² π² / 2. (-1.71) π = 4210 π rad

As 1 rev = 2π rad, Δθ = 2105 rev

7 0
4 years ago
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