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PolarNik [594]
3 years ago
7

Random Online 1) Using your choice of C# or Java (NOT in Pseudocode), write an interface IMammal. The IMammal contains a method

mammalType(). Then write a concrete class Dog that inherits the IMammal interface and includes the unimplemented mammalType() method with at least one PRINT statement, with whatever you like to display. You don’t need to include the main() method.
Engineering
1 answer:
MAVERICK [17]3 years ago
3 0

Answer:

IMammal:

namespace AnimalInterfaceTest

{

   interface IMammal

   {

       void mammalType();

   }

}

Dog Class:

using System;

namespace AnimalInterfaceTest

{

   class Dog : IMammal

   {

       public void mammalType()

       {

           Console.WriteLine("Mammal type is Dog.");

       }

   }

}

===============main program=================

using System;

namespace AnimalInterfaceTest

{

   class Program

   {

       static void Main(string[] args)

       {

           IMammal m1 = new Dog();

           Dog dog = new Dog();

           m1.mammalType();

           dog.mammalType();

           Console.ReadLine();

       }

   }

}

Explanation:

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The correct answer is An abrasive wheel.
5 0
3 years ago
Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the w
soldi70 [24.7K]

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

4 0
3 years ago
The concrete canoe team does some analysis on their design and calculates that they need a compressive strength of 860 psi. They
vlada-n [284]

Answer:

874 psi

Explanation:

Given a sample mean (x') = 900,

and a standard error (SE) = 10

At 99% confidence, Z(critical) = 2.58

That gives 99% confidence interval as,

x' ± Z(critical) x SE = 900 ± 2.58 x 10

The value of the lower limit is,

900 - 25.8 = 874.2

≈ 874 psi

8 0
3 years ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0
4 years ago
Read 2 more answers
Engineering Questions
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5 is the correct one to choose for this
6 0
3 years ago
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