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AleksandrR [38]

2 years ago

7

You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______

_ before you turn.
The options are as follows:
A. 10 feet
B. 100 feet
C. 75 feet
D. 20 feet

1 answer:

stich3 [128]2 years ago

3 0

I believe it’s D) 20 feet

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Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

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Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

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Write a iterative function that finds the n-th integer of the Fibonacci sequence. Then build a minimal program (main function) t

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Answer:

Codes for each of the problems are explained below

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#include<iostream>

using namespace std;

//fib function that calculate nth integer of the fibonacci sequence.

void fib(int n){

// l and r inital fibonacci values for n=1 and n=2;

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}

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cout << "\n" << c;

}

int main(){

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}

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print("fib({})".format(n), end=' ')

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print(result)

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