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Ymorist [56]
3 years ago
10

. Carly's Catering provides meals for parties and special events. In Chapter 2, you wrote an application that prompts the user f

or the number of guests attending an event, displays the company motto with a border, and then displays the price of the event and whether the event is a large one. Now modify the program so that the main() method contains only three executable statements that each call a method as follows:
Engineering
1 answer:
klio [65]3 years ago
6 0

Answer:

Explanation:

public class Event

  public final static int PRICE_PER_GUEST = 35;

  public final static int CUT_OFF = 50;

 

  //Attributes

  private String eventNum;

  private int noOfGuest;

  private int price;

 

  /**

  * param eventNum the eventNum to set

  */

  public void setEventNum(String eventNum)

      this.eventNum = eventNum;

 

 

  /**

  * param noOfGuest the noOfGuest to set

  */

  public void setNoOfGuest(int noOfGuest)

      this.noOfGuest = noOfGuest;

      this.price = this.noOfGuest * PRICE_PER_GUEST;

 

 

  /**

  * return the eventNum

  */

  public String getEventNum()

      return eventNum;

 

 

  /**

  * return the noOfGuest

  */

  public int getNoOfGuest()

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Elan Coil [88]

Answer:

d

Explanation:

4 0
4 years ago
A 55-μF capacitor has energy ω (t) = 10 cos2 377t J and consider a positive v(t). Determine the current through the capacitor.
mart [117]

Given :

Capacitor , C = 55 μF .

Energy is given by :

\omega(t)=10cos^2 (377t)\ J .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)

Now , current i is given by :

i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .

6 0
3 years ago
A car crusher has one hydraulic input piston with diameter 2 ft and 6 crushing pistons with diameter 10 feet. If this car needs
IrinaK [193]
Not sure of the answers
4 0
3 years ago
The "view factor" Fij depends on surface emissivity and surface geometry. a) True b) False
Alex

Answer:

(B) FALSE

Explanation:

view factor F_{ij} depends on the surface emissivity and the surface of geometry  view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by  F_{ij}

4 0
3 years ago
Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow
Black_prince [1.1K]

Answer:

\omega_y,\omega_x,\omega_Z  all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function \phi given as follows

 u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}

Rotationality of fluid is given by \omega

\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z

\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x

\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y

So now putting value in the above equations ,we will find

\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},

\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}

So \omega_y=0

Like this all \omega_y,\omega_x,\omega_Z all are zero.

That is why  velocity potential flow is irroational flow.

5 0
4 years ago
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