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nadezda [96]
3 years ago
5

What is the lowest Temperature in degrees C?, In degrees K? in degrees F? in degrees R

Engineering
1 answer:
Colt1911 [192]3 years ago
3 0

Answer:

-273.16 °C

-459.677 °F

0 °K

0 °R

Explanation:

The lowest temperature is the absolute zero.

Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.

Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.

Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.

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Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

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- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

                            C = W*log_2 ( 1 + P/N)

- Plug the values in:

                            C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

                            C = (20*10^6)*log_2 (25001)

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3 years ago
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Answer:

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3 0
3 years ago
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/* Function findBestVacation * duration: number of vacation days * prefs: prefs[k] indicates the rate specified for game k * pla
alexira [117]

Answer:

This is the C++ code for the above problem:

#include<bits/stdc++.h>

using namespace std;

int computeFunLevel(int start, int duration, int prefs[], int ngames, int plan[]) {

if(start + duration > 365) { //this is to check wether duration is more than total no. of vaccation days

return -1;

}

int funLevel = 0;

for(int i=start; i<start+duration; i++) { //this loop runs from starting point till

//start + duration to sum all the funlevel in plan.

funLevel = funLevel + prefs[plan[i]];

}

return funLevel;

}

int findBestVacation(int duration, int prefs[], int ngames, int plan[]) {

int max = 0, index = 0, sum = 0 ;

for(int i=1; i<11; i++){ //this loop is to run through whole plan arry

sum = 0; //sum is initialized with zero for every call in plan ,

//in this case loop should run to 366,but for example it is 11

//as my size of plan array is 11

for(int j=0; j<duration; j++) { // this loop is for that index to index+duration to calc

//fun from that index

sum = sum + prefs[plan[i]];

}

if(sum>max) { //this is to check max funlevel and update the index from which max fun can be achieved

max = sum;

index = i;

}

}

return index;

}

int main() {

int ngames = 5;

int prefs[] = { 1,2,0,5,2 };

int plan[] = { 0,2,0,3,3,4,0,1,2,3,3 };

int start = 1;

int duration = 4;

cout << computeFunLevel(start, duration, prefs, ngames, plan) << endl;

cout << computeFunLevel(start, 555, prefs, ngames, plan) << endl;

cout << findBestVacation(4, prefs, ngames, plan) << endl;

}

The screen of the program is given below.

3 0
3 years ago
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Explanation:

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3 years ago
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Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

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we can apply the following expression to find T_{2}

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T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

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To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

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