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
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
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
Answer
Given,
Reservoir pressure of a supersonic wind tunnel = 5 atm
Local Mach number = ?
Area ration = ?
a) 4 atm.
Pressure ratio =
= 0.8
From Isentropic Flow Tables
M = 0.58 A/A* = 1.213
b) 2.64 atm
Pressure ratio =
= 0.528
From Isentropic Flow Tables
M = 1 A/A* = 1
c) 0.5 atm
Pressure ratio =
= 0.1
From Isentropic Flow Tables
M =2.10 A/A* = 1.8369
Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?
The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.
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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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 .
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’)
A loop is a control structure in programming that allows a block of code to be executed repeatedly.
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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