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strojnjashka [21]

3 years ago

6

The assignment is to modify the class implementation without changing the functionality. Remove the feet and inches data members

and replace them with one data member named total_inches. Re-implement all the class methods so they have he same functionality as originally. This means that any program written using the public members of the new class must produce exactly the same results as the original implementation.

1 answer:

Lorico [155]3 years ago

4 0

Answer: The solution can be found in the attached

Explanation:

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Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

Ann [662]

The number of tubs that each of them sold is; 24 tubs each

The number of days it will take for both of them to sell same amount of tubs is; 4 days

Number of cookies that Nicole had already sold = 4 tubs

Number of cookies sold by Josie before counting = 0 cookies

Nicole now sells 5 tubs per day and

Josie now sells 6 tubs per day.

Let the number of days it will take for them to have sold the same amount of cookies be d.

Thus;

5d + 4 = 6d + 0

6d - 5d = 4

d = 4 days

Thus, total number of cookies for both are;

Total for Nicole = 4 + 5(4) = 24 cookies

Total for Josie = 6(4) = 24 cookies

Read more about proportion at; brainly.com/question/870035

6 0

3 years ago

What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin

DochEvi [55]

Answer:

$m_{LP}=0.45\,kg$

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

$Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]$

$Q_{water} = 3599.435\,kJ$

The heat liberated by the LP gas is:

$Q_{LP} = \frac{3599.435\,kJ}{0.16}$

$Q_{LP} = 22496.469\,kJ$

A kilogram of LP gas has a minimum combustion power of $50028\,kJ$ . Then, the required mass is:

$m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }$

$m_{LP}=0.45\,kg$

6 0

3 years ago

Assume you have four fins, each with a mass of 8.0 grams. What is the total weight of these four fins? (Hint: watch your units!)

soldier1979 [14.2K]

The answer is 32.0 grams

7 0

3 years ago

A tool used to put a concave edge on a plane iron is

tigry1 [53]

Answer:

C) grinder

Explanation:

7 0

2 years ago

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