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jonny [76]
3 years ago
10

The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03

Engineering
1 answer:
kykrilka [37]3 years ago
7 0

Answer:

Answer is c 0.500

Explanation:

SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}

We know that \rho_{water}=62.42lb/ft^{3}

Applying values we get

SpecificGravity=\frac{31.2}{62.4}=0.5

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When mining diamonds with a stone pick what will be the outcome
Soloha48 [4]

Answer:

The diamond ore will break and you won't get any diamonds.

Explanation:

4 0
3 years ago
Read 2 more answers
Strain gage is a device that senses the strain of the structure. The property of the strain gage that is used to correlate with
Anni [7]

Answer:

  resistance

Explanation:

A strain gauge changes resistance with applied strain.

8 0
3 years ago
Reference Parameters (returning multiple values): Write a C++ function that converts standard time to military time. Inputs incl
valkas [14]

Answer:

Code is given as below:

Explanation:

#include <iostream>

using namespace std;

//function prototype declaration

void MilitaryTime(int, int, char, int &, int &);

int main()

{

    //declare required variables

    int SHour, SMin, MHour, MMin;

    char AorP;

    //promt and read the hours from the user

    cout<<"Enter hours in standard time : ";

    cin>>SHour;

    //check the hours are valid are not

    while(SHour<0 || SHour>12)

    {

         cout<<"Invalid hours for standard time. "

             <<"Try again..."<<endl;

         cout<<"Enter hours in standard time : ";

         cin>>SHour;

    }

    //promt and read the minutes from the user

    cout<<"Enter minutes in standard time : ";

    cin>>SMin;

    //check the minutes are valid are not

    while(SMin<0 || SMin>59)

    {

         cout<<"Invalid minutes for standard time. "

             <<"Try again..."<<endl;

         cout<<"Enter minutes in standard time : ";

         cin>>SMin;

    }

    //promt and read the am or pm from the user

    cout<<"Enter standard time meridiem (a for AM p for PM): ";

    cin>>AorP;

    //check the meridiem is valid are not

    while(!(AorP=='a' || AorP=='p' || AorP=='A' || AorP=='P'))

    {

         cout<<"Invalid meridiem for standard time. "

             <<"Try again..."<<endl;

         cout<<"Enter standard time meridiem (a for AM p for PM): ";

         cin>>AorP;

    }

    //call function to calculate the military time

    MilitaryTime(SHour, SMin, AorP, MHour, MMin);

    //fill zeros and display standard time

    cout.width(2);

    cout.fill('0');

    cout<<SHour<<":";

    cout.width(2);

    cout.fill('0');

    cout<<SMin;

    if(AorP=='a' || AorP=='A')

         cout<<" am = ";

    else

         cout<<" pm = ";

    //fill zeros and display military time

    cout.width(2);

    cout.fill('0');

    cout<<MHour;

    cout.width(2);

    cout.fill('0');

    cout<<MMin<<endl;

    system("PAUSE");

    return 0;

}

//function to calculate the military time with reference parameters

void MilitaryTime(int SHour, int SMin, char AorP, int &MHour, int &MMin)

{

    //check the meredium is am or pm

    //and calculate hours

    if(AorP=='a' || AorP=='A')

    {

         if(SHour==12)

             MHour = 0;

         else

             MHour = SHour;

    }

    else

         MHour = SHour+12;

    MMin = SMin;

5 0
3 years ago
4. If a hot wire is shorted to ground, what will usually happen?
barxatty [35]

Answer:

nothing will happen

Explanation:

except for whatever that wire feeds will lose power

help support our cause by spreading this flag all over brainly

5 0
2 years ago
S1.1 The Acre-Foot. Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as
Mars2501 [29]

Answer:

1072 acre foot

1331424000 kg

Explanation:

1 feet has 12 inches, so 2 in is 0.167 feet.

1 km^2 has 1 million m^2.

1 acre is 4074 m^2.

So, 1 km is 247 acres.

Then 26 km^2 is 6422 acres.

So, the volume of water is

6422 * 0.167 = 1072 acre-foot

Since one cubic meter of water has 1000 kg

One inch is 25.4 mm = 0.0254 m

One feet is 12 * 0.0254 = 0.3048 m

An acre-feet has a volume of

4074*0.3048 = 1242 m^3

And that is a mass of water of

1242 * 1000 = 1242000 kg/acre-feet

Therefore the mass of rainwater in the town is of

1072 * 1242000 = 1331424000 kg = 1331424 tons

4 0
3 years ago
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