Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mechanical resonance frequency is [Equation] = 100 Hz. If the mass is to be made of cube of `silicon, what would its physical dimensions be?

Answers

Answer 1
Answer:

Answer:

Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).

The physical dimension of the silicon is 10kg

Explanation:

Using the formular, Force, F = 1/2π√k/m

At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

Hence, F = 1/2π√mw²/m = f ( f = frequency)

∴ f = F = mg, taking g = 9.8 m/s²

100 Hz = 9.8 m/s² X m

m = 100/9.8 = 10.2kg


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One kind of SS-3xX steel alloy has a melting point of 1450°c. Its specific heat = 0.46 J/g°C, and its heat of fusion 270 J/g. For a 200 kg block of this steel, determine how much heat is required to (a) raise its temperature from 25°C to its melting point and (b) transform it from solid to liquid phase.

Answers

Answer:

a)Q=131.1 MJ

b)Q=54 MJ

Explanation:

Given that

Mass ,m=200 kg

Specific heat Cp=0.46 J/g°C

Cp=0.46 KJ/kg°C

Heat of fusion = 270 J/g

Heat of fusion = 270 KJ/kg

Melting point temperature = 1450°C

a)

Initial temperature = 25°C

Final temperature=1450°C

Heat required to rise temperature from 25°C to 1450°C.

Q= m CpΔT

Q=200 x 0.46 x (1450-25) KJ

Q=131,100 KJ

Q=131.1 MJ

b)

Heat required to transform from solid phase to liquid phase

Q= Mass x heat of fusion

Q=200 x 27 KJ

Q=54,000 KJ

Q=54 MJ

There are many diferent materials available for seal faces . List the following seal face materials in order of hardness. i.e Hardest first, softest last. (a) 316 Stainless Steel (b)-Mild steel (c)- Reaction bonded Silicon carbide (d)- Tungsten carbide

Answers

Answer:

Reaction bonded Silicon carbide: 2500-3500 HV

Tungsten carbide: 1800-2500 HV.

316 Stainless Steel: 152 HV

Mild steel: 130 HV

Explanation:

In order to list those seal face materials by hardness, we look up what are the values of hardness for each material in a hardness scale.

We are going to use Vickers scale, an indentation method of measuring hardness, it measures the deformation left in a sample by a constant compression load from an indenter (a diamond pyramid) with an adequate (to the material) force, as the result is independent from the test force.

1. Reaction bonded Silicon carbide: 2500-3500 HV

2. Tungsten carbide: 1800-2500 HV

3. 316 Stainless Steel: 152 HV

4. Mild steel: 130 HV

The reservoir pressure of a supersonic wind tunnel is 5 atm. A static pressure probe is moved along the centerline of the nozzle, taking measurements at various stations. For these probe measurements, calculate the local Mach number and area ratio: a. 4 atm; b. 2.64 atm; c. 0.5 atm.

Answers

Answer

Given,

Reservoir pressure of a supersonic wind tunnel = 5 atm

Local Mach number = ?

Area ration = ?

a) 4 atm.

  Pressure ratio =(4)/(5)

                         = 0.8

From Isentropic Flow Tables

  M = 0.58   A/A* = 1.213

b) 2.64 atm

  Pressure ratio =(2.64)/(5)

                         = 0.528

From Isentropic Flow Tables

  M = 1  A/A* = 1

c)  0.5 atm

  Pressure ratio =(0.5)/(5)

                         = 0.1

From Isentropic Flow Tables

  M =2.10   A/A* = 1.8369

Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 seconds. In Eager Mode, the app utilizes a faster timer resolution for its computations, so the execution time in Eager Mode is 2 seconds (i.e., Eager Mode execution time is 60% of Lazy Mode execution time).After finishing computation, the app sends some data to the cloud, regardless of the mode it’s in. The data size sent to the cloud is 600 MB. The bandwidth of communication is 15 MBps for WiFi and 5 MBps for 4G. Assume that the communication radio is idle during the computation time. Assume that the communication radio for WiFi has a power consumption of 75 mW when active and 15 mW when idle. Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization. Assume that the power consumption of the CPU is a linear function of its utilization. In other words: P = (Idle Power) + (Utilization)*(Power per unit Utilization). A configuration of the mobile app involves choosing a timer resolution (Lazy or Eager) and choosing a type of radio (WiFi or 4G). For example, faster timer resolution (Eager) and 4G network is a configuration, while slower resolution (Lazy) and WiFi is another. There are four possible configurations in all.

Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?

Answers

The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Why reducing leads to increasing wages?

Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.

Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.

Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.

Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

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An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Answers

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .

Make a python code.Write a function named max that accepts two integer values as arguments and returns the value that is the greater of the two. For example, if 7 and 12 are passed as arguments to the function, the function should return 12. Use the function in a program that prompts the user to enter two integer values. The program should display the value that is the greater of the two. Write the program as a loop that continues to prompt for two numbers, outputs the maximum, and then goes back and prompts again
Here’s an example of program use
Input the first number: 10
Input the second number: 5
The maximum value is 10
Run again? yes
Input the first number: -10
Input the second number: -5
The maximum value is -5
Run again? no
Function max():
Obtain two numbers as input parameters: max(num1, num2):
if num1 > num2 max_val = num1, else max_val = num2
return max_val
Main Program:
Initialize loop control variable (continue = ‘y’)
While continue == ‘y’
Prompt for first number
Prompt for second number
Call function "max," sending it the values of the two numbers, capture result in an assignment statement:
max_value = max (n1, n2)
Display the maximum value returned by the function
print(‘Max =’, max_val)
Ask user for if she/he wants to continue (continue = input(‘Go again? y if yes’)

Answers

A loop is a control structure in programming that allows a block of code to be executed repeatedly.

How to write the python code?

Here is the Python code that implements the function and program described in the question:

def max(num1, num2):

 if num1 > num2:

   max_val = num1

 else:

   max_val = num2

 return max_val

# Main program

continue = 'y'

while continue == 'y':

 n1 = int(input("Input the first number: "))

 n2 = int(input("Input the second number: "))

 max_val = max(n1, n2)

 print("The maximum value is", max_val)

 continue = input("Run again? ")

Loops are an important programming construct that allow you to perform the same task multiple times with minimal code. They are often used to process data, perform calculations, and perform repetitive tasks.

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