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3 years ago

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You recall an algorithm from elementary school for factoring a number N: Divide out all factors of 2, then of 3, then of 4, then

of 5, then of 6, then of 7, etc. Finally, divide out all factors of N (of which there can be at most one). (a) Write pseudo-code for this algorithm (and print the prime factors)

1 answer:

Contact [7]3 years ago

4 0

Answer:

let number = 0

while number < 1

begin

print "Enter a positive integer: "

read number

end

end_while

find and print number's factors:

let prime = TRUE

let currentFactor = 2

let lastFactor = the square root of number truncated

to an integer value

while currentFactor <= lastFactor

begin

if number is evenly divisible by currentFactor

begin

print currentFactor

let number = number / currentFactor

end

else

let currentFactor = currentFactor + 1

end_if

end

end_while

print a message if number is prime:

if prime == TRUE

print "Your number is prime"

end_if

Explanation:

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The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm

oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

$m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s$

The exit temperature is:

$T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25$ °C

b) The mean fluid velocity is:

$V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s$

The hydraulic diameter is:

$D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m$

The Reynold´s number is:

$Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36$

Assuming fully developed turbulent flow, the Nusselt number is:

$Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69$

$h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C$

The highest component surface temperature is:

$T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22$ °C

6 0

4 years ago

Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

4 0

3 years ago

Open the"stateData3.c" program and try to understand how the tokenization works. If you open the input file "stateData.txt", you

babymother [125]

Answer:

Kindly see explaination

Explanation:

Code

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#define size 200

int main(void)

{

int const numStates = 50;

char tempBuffer[size];

char tmp[size];

char fileName[] = "stateData.txt"; // Name of the text file (input file) which contains states and its populations

char outFile[] = "stateDataOutput1.txt"; // Output file name

// Open the input file, quit if it fails...

FILE *instream = fopen(fileName, "r");

/* Output File variable */

FILE *opstream;

if(instream == NULL) {

fprintf(stderr, "Unable to open file: %s\n", fileName);

exit(1);

}

//TODO: Open the output file in write ("w") mode

/* Opening output file in write mode */

opstream = fopen(outFile, "w");

//TODO: Read the file, line by line and write each line into the output file

//Reading data from file

while(fgets(tmp, size, instream) != NULL)

{

//Writing data to file

fputs(tmp, opstream);

}

// Close the input file

fclose(instream);

//TODO: Close the output file

/* Closing output file */

fclose(opstream);

return 0;

}

5 0

3 years ago

The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

slamgirl [31]

Answer:

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2 ⇒X=4 ft

A=8 x 10=80 $ft^2$

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

4 0

4 years ago

A Class A fire extingisher is for use on general combustibles such as:

Tatiana [17]

Answer:

used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.

Explanation:

7 0

3 years ago