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liubo4ka [24]
3 years ago
7

Basic output with variables (Java) This zyLab activity is intended for students to prepare for a larger programming assignment.

Warm up exercises are typically simpler and worth fewer points than a full programming assignment. Warm up exercises are ideally suited for an in-person scheduled lab meeting or as self-practice. The last section provides a full programming assignment. A variable like userNum can store a value like an integer. Extend the given program to print userNum values as indicated. (Submit for 2 points) (1) Output the user's input. Enter integer: 4 You entered: 4 (2) Extend to output the input squared and cubed. Hint: Compute squared as userNum userNum. (Submit for 2 points, so 4 points total). Enter integer: 4 You entered: 4 4 squared is 16 And 4 cubed is 64!! (3) Extend to get a second user input into userNum2. Output sum and product. (Submit for 1 point, so 5 points total). Enter integer: 4 You entered: 4 4 squared is 16 And 4 cubed is 64!! Enter another integer: 5 4 5 is 9 45 is 20 LAB 1.16.1: Basic output with variables (Java) 0/5 ACTIVITY OutputWithVars.java Load default template... 1 import java.util.Scanner; 2 3 public class OutputWithVars public static void main(String [] args) Scanner scnr new Scanner(System. in); int userNum - 0 6 7 System.out.println ("Enter integer: ") userNum scnr. next Int ( ) ; 9 10 return; 11 12 13
Engineering
1 answer:
9966 [12]3 years ago
3 0

Answer:

1

Explanation:

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an object of mass 2kg is released from a top of inclined plane 30° and height 6m. The coefficient of kinetic friction of the sur
mel-nik [20]

Explanation:

1) Work done = force x distance x cos(θ)

= 0.15 x 6 x cos(30)

= 0.779

2) Ek = ½mv²

v = acceleration due to gravity so 9.81

Ek = ½(2)(9.81)²

Ek = 96.2361

3) v = (√(2em)) / m

= (√(2(96.2361)(2)) / 2

= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.

5 0
2 years ago
Gold and silver rings can receive an arc and turn molten. True or False
liubo4ka [24]
The answer is False!
The answer is false
8 0
3 years ago
Read 2 more answers
KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown
kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

i_x+50mA=i_y\\i_x-i_y=0.05A

Also

V_{\Delta}=1K*i_y

Now applying KVL on loop 1 as indicated in the attached figure

1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0

Similarly for loop 2

2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0

So the system of equations become

i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

8 0
4 years ago
Which of the following uses pressure and flow to transmit power from one location to another?
lord [1]

Answer:

fluid power

Explanation:

fluids commonly used in fluid power are Oil, Water, Air, CO², and Nitrogen gas, fluid power is commonly confused with hydraulic power, which only uses liquids, fluid power uses either liquids or gases

5 0
2 years ago
A 50 kg diver stands motionless at the top of 10.0 meter high dive. What is her potential energy?
brilliants [131]

Answer:

Potential energy=4905 Joule

Explanation:

Potential energy is defined as the energy between two bodies due to its relative positions. Here one body is the diver and the other body is the earth. So acceleration will be the constant 9.81\ m/s^2

P.E.=m\times g\times h\\where\ m=mass\ of\ body=50\ kg\\g=acceleration\ due\ to\ gravity=9.81\ m/s^2\\h=height\ at\ which\ the\ body\ is\ from\ the\ ground=10.0\ m\\\Rightarrow P.E.=50\times 9.81\times 10.0\\\therefore Potential\ Energy=4905\ Joule

7 0
4 years ago
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