I need solution for this question

Which option best describes a way engineers can provide maintenance, diagnoses, upgrades, or duplicates even for products they d

guajiro [1.7K]

Answer:

Reverse Engineering

Explanation:

Just took the test

4 0

3 years ago

Which reference source may be consulted to answer questions regarding the Professional Engineers Act?

9966 [12]

Answer: c. The Professional Engineers Act and Board Rules

Explanation:

The reference source may be consulted to answer questions regarding the Professional Engineers Act is the The Professional Engineers Act and Board Rules.

The Professional Engineers Act and Board Rules is an Act that was established in order to regulate the qualifications for professional engineered, register them and also make sure that their conducts and behavior are looked into.

6 0

3 years ago

Implement the function lastChars() that takes a list of strings as a parameter and prints to the screen the last character of ea

Liono4ka [1.6K]

Answer:

The following program is in C++.

#include <bits/stdc++.h>

using namespace std;

void lastChars(string s)

{

int l=s.length();

if(l!=0)

{

cout<<"The last character of the string is: "<<s[l-1];

}

int main() {

string s;//declaring a string..

getline(cin,s);//taking input of the string..

lastChars(s);//calling the function..

return 0;

}

Input:-

Alex is going home

Output:-

The last character of the string is: e

Explanation:

In the function lastChars() there is one argument that is a string.I have declared a integer variable l that stores the length of the string.If the length of the string is not 0.Then printing the last character of the string.In the main function I have called the function lastChars() with the string s that is prompted from the user.

8 0

3 years ago

The heat rate is essentially the reciprocal of the thermal efficiency. a)- True b)- False

jeyben [28]

Answer:

a). TRUE

Explanation:

Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.

Mathematically, thermal efficiency is

η = net work done / heat input

While heat rate is the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.

Mathematically, heat rate is

Heat rate = heat input / net work done

Thus from above we can see that heat rate is the reciprocal of thermal efficiency.

Thus, Heat rate is reciprocal of thermal efficiency.

4 0

3 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago

I need solution for this question ​

I need solution for this question