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3 years ago

15

given the classes above, what output is produced by the following code? meg[] elements ={new Lois(), new Stewie(), new Meg(), ne

w Brian()}; for (int i = 0; i

1 answer:

LuckyWell [14K]3 years ago

6 0

Answer:

<u> Inheritance Mystery </u>

<u />

Lois a Meg a

Lois b

Meg

Lois a Meg a Stewie a

Lois a Meg a Stewie a Brian b

Brian Stewie

Meg a

Meg b

Meg

Lois a Meg a

Lois a Meg a Brian b

Brian

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Which one is dependent variable?

GREYUIT [131]

Answer:

The dependent variable is MEDV - Median value of owner-occupied homes in $1000's

Explanation:

The median value of the house has to be predicted, based on its properties and neighborhood properties, this can be done by using a linear regression model.

The dependent variable in Machine Learning is the output variable that we want to predict.

Therefore, according to the question given "MEDV" is the dependent variable.

8 0

3 years ago

Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.

zepelin [54]

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

$T_{sat}=179.88\°c$

$H_{fg} = 2014.6kJ/kg$

$c_p=4.18 kJkg^{-1}{K^{-1}$

$\nu_g = 0.19436m^3/kg$

Replacing,

$\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})$

$\Delta h = 2014.6+4.18(179.88-24)$

$\Delta h=2666.17kJ/kg$

With the specific volume we know can calculate the mass flow, that is

$\dot{m}=\frac{\frac{15000}{3600}}{0.19436}$

$\dot{m} = 21.4378kg/s$

Then the heat required in input is,

$Q=\dot{m}\Delta h$

$Q=21.4378*2666.17$

$Q=57157.036kW$

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

$V= \frac{\dotV}{A}$

$V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}$

$V=235.79m/s$

Finally we can apply the steady flow energy equation, that is

$\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2$

Re-arrange for Q,

$Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})$

$Q=\dot{m}(\Delta h-\frac{V^2}{2000})$

$Q= (21.4378)(2666.17-\frac{235.79^2}{2000})$

$Q= 56560.88kW$

We can note that consider the Kinetic Energy will decrease the heat input.

4 0

3 years ago

How many types of arcs do we have in AutoCad? O a.9 O b. 10 Oc. 12 O d. 11

Dmitry [639]

Answer:

Letter D

Explanation:

AutoCAD provides eleven different ways to create arcs. The different options are used based on the geometry conditions of the design. To create an arc, you can specify various combinations of center, endpoint, start point, radius, angle, chord length, and direction values.

8 0

3 years ago

Read 2 more answers

For a bronze alloy, the stress at which plastic deformation begins is 266 MPa and the modulus of elasticity is105 GPa.

pentagon [3]

Answer:

88750 N

Explanation:

given data:

plastic deformation σy=266 MPa=266*10^6 N/m^2

cross-sectional area Ao=333 mm^2=333*10^-6 m^2

solution:

To determine the maximum load that can be applied without

plastic deformation (Fy).

Fy=σy*Ao

=88750 N

7 0

3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

Free_Kalibri [48]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

then $\frac{x_1^2}{t_1}=\frac{x_2^2}{t_2}$ we have given $x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm$ and we have to find $t_2$ putting all these value in equation

$\frac{2^2}{15}=\frac{6^2}{t_2}$

so $t_2=135\ hour$

5 0

3 years ago