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patriot [66]
2 years ago
13

A_____ transducer is a device that can convert an electronic controller output signal into a standard pneumatic output. A. pneum

atic-to-current (P/I) B. voltage-to-pneumatic (V/P) C. current-to-pneumatic (I/P) D. pneumatic-to-voltage (P/V)
Engineering
1 answer:
makkiz [27]2 years ago
6 0

Answer:

The correct answer is

option C. current to pneumatic (V/P)

Explanation:

A current to pneumatic controller is  basically used to receive an electronic signal from a controller and converts it further into a standard pneumatic output signal which is further used to operate a positioner or control valve. These devices are reliable, robust and accurate.

Though Voltage and current to pressure transducers are collectively called as electro pneumatic tranducers and the only electronic feature to control output pressure in them is the coil.

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What is the linear distance traveled in one revolution of a 36-inch wheel
Afina-wow [57]

Answer:

  36π inches ≈ 113.0973 inches

Explanation:

The circumference of the wheel is pi times its diameter.

  C = πd

  C = π(36 in) = 36π in ≈ 113.0973 in

The distance around the wheel is 36π inches, about 113.0973 inches.

7 0
2 years ago
The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13
prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0
2 years ago
A group of statisticians at a local college has asked you to create a set of functions that compute the median and mode of a set
iVinArrow [24]

Answer:

Functions to create a median and mode of a set of numbers

Explanation:

def median(list):

   if len(list) == 0:

       return 0

   list.sort()

   midIndex = len(list) / 2

   if len(list) % 2 == 1:

       return list[midIndex]

   else:

       return (list[midIndex] + list[midIndex - 1]) / 2

def mean(list):

   if len(list) == 0:

       return 0

   list.sort()

   total = 0

   for number in list:

       total += number

   return total / len(list)

def mode(list):

   numberDictionary = {}

   for digit in list:

       number = numberDictionary.get(digit, None)

       if number == None:

           numberDictionary[digit] = 1

       else:

           numberDictionary[digit] = number + 1

   maxValue = max(numberDictionary.values())

   modeList = []

   for key in numberDictionary:

       if numberDictionary[key] == maxValue:

           modeList.append(key)

   return modeList

def main():

   print "Mean of [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]: ", mean(range(1, 11))

   print "Mode of [1, 1, 1, 1, 4, 4]:", mode([1, 1, 1, 1, 4, 4])

   print "Median of [1, 2, 3, 4]:", median([1, 2, 3, 4])

main()

3 0
2 years ago
Read 2 more answers
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz
nikitadnepr [17]

Answer:

25 - \sqrt[4]{26.66*10^{-8} }  mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2\pif = 2 * \pi * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter  using the relation

  \frac{J}{c_{2}  } = \frac{T}{t_{max} }  = 467.92 / (60 * 10^6) =  7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

\frac{J}{c_{2} } = \frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} ) =  \frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1}  ) ]

therefore; 0.025^4 - c^{4} _{1} = 0.050 / \pi (7.8 *10^-6)

c^{4} _{1} = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

    39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

c_{1} = \sqrt[4]{26.66 * 10^{-8} }  =

THE TUBE THICKNESS

c_{2} - c_{1} = 25 - \sqrt[4]{26.66*10^{-8} }  mm

4 0
2 years ago
The air contained in a room loses heat to the surroundings at a rate of 50 kJ/min while work is supplied to the room by computer
Artemon [7]

Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

4 0
2 years ago
Read 2 more answers
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