The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam material has a specific gravity, SG, of 3.1. You may assume that the dam is loosely attached to the ground at its base, though there is significant friction to keep it from sliding.Is the weight of the dam sufficient to prevent it from tipping around its lower right corner?

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

int main(){

//declare arrays

string ItemName[100];

double ItemCost[100]={0.0};

//declare variable

string name="";

double Total=0.0;

int NumItems=0;

cout<<"Enter number of grocery item you will be entering";

cin>>NumItems

cout<<"\n";

if (NumItems>100)

{

  cout<<"Enter grocery items less than 100"

  cin>>NumItems

  cout<<"\n";

}

for (int i=1;i>=NumItems;i++)

{

  cout<<"Please enter the item name in one word only, Example:                                icecream\n";

  cin>>ItemName[i]

  cout<<"Please enter the cost as a decimal number, Example: 2.05\n\n";

  cin>>ItemCost[i];

}

cout<<"Items"<<"   "<<""$""<<"Cost"<<endl

for (int i=1;i>=NumItems;i++)

{

    cout<<ItemName[i]<<"   "<<"$"<<ItemCost[i]<<endl

}

for (int i=1;i>=NumItems;i++)

{

  Total=Total+ItemCost[i];

}

cout<<"Total:$"<<Total<<endl;

system("PAUSE");

return 0;

}

Answer: Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;

COPhp = Th/(Th - Tl)

COPhp = Qh/Wnet

Qh/Wnet = Th/(Th -Tl)

the amount of heat loss is expressed as

Qh = 5400/3600(294 - Tl)

the temperature of sink

( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)

now solving the equation

Tl = 259.7 - 273

Tl = - 13.3°C

so the lowest outdoor temperature is - 13.3°C

Answer:

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       System.out.println(birthMonth+"/"+birthYear);

   }

}

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       // Get input values for birthMonth and birthYear from the user

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       // Output the month, a slash, and the year

       System.out.println(birthMonth + "/" + birthYear);

   }

}

When the program is tested with inputs 1 2000, the output will be:

1/2000

And when tested with inputs 5 1950, the output will be:

5/1950

Know more about java program:

brainly.com/question/34106786

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Answer:

558.1918 kilocalories = 558191.8 calories

Explanation:

Data provided in the question:

Atmospheric pressure = 84.6 KPa

Mass of water, m = 900 g = 0.90 kg

Temperature = 15°C

Now,

Temperature at 84.6 KPa = 94.77°C

Therefore,

Heat energy required = m(CΔT + L)

here,

C is the specific heat of the water = 4.2 KJ/kg.°C

L = Latent heat of water = 2260 KJ/kg

Thus,

Heat energy required = 0.90[ 4.2 × (94.77 - 15) + 2260 ]

= 2335.53 KJ

also,

1 KJ = 0.239  Kilocalories

Therefore,

2335.53 KJ = 0.239 × 2335.53 Kilocalories

= 558.1918 kilocalories = 558191.8 calories

Answer:

Explanation:

The concept of Hooke's law was applied as it relates to deformation.

The detailed steps and appropriate substitution is as shown in the attached file.

Answer:

The air heats up when being compressed and transefers heat to the barrel.

Explanation:

When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.

In an adiabatic transformation:

For air k = 1.4

SO

SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.

After it was compressed the hot air will exchange heat with the barrel heating it up.