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Nat2105 [25]
3 years ago
8

Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load

P for which the normal stress in that portion of BD is 50 .
Engineering
1 answer:
IceJOKER [234]3 years ago
7 0

Answer: 50

Explanation:

You might be interested in
A shipment of rebar that weighs 745 kg would weigh roughly how much in pounds​
Andre45 [30]

Answer:

Dont no but will check

Explanation:

6 0
3 years ago
Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a
guapka [62]

Answer:

K_v=12.34

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo =   Discharge / area of the orifice

or

Vo = \frac{0.01}{\frac{\pi}{4}0.045^2}

or

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, V_v =   Discharge / area of the orifice

or

V_v = \frac{0.01}{\frac{\pi}{4}0.0675^2}

or

V_v = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

or

25 m = K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}

where,

K_v is the loss coefficient for the gate valve

on substituting the values, we get

25 m = 10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}

or

K_v\frac{2.79^2}{2\times9.81} = 4.898

or

K_v=12.34

3 0
4 years ago
____________engineers build and test tools and machines, and they rely heavily on computer, math, and problem-solving skills.
Wittaler [7]

Answer:Mechanical Engineers

Explanation: Mechanical engineering is an engineering branch that combines engineering physics and mathematics principles with materials science to design, analyze, manufacture, and maintain mechanical systems. It is one of the oldest and broadest of the engineering branches.

8 0
3 years ago
Read 2 more answers
Determine the depreciation expense for 2018 and 2019 using the following​ methods: (a)​ Straight-line (SL),​ (b) Units of produc
prohojiy [21]

Answer:

Check Explanation.

Explanation:

(1). The straight-line method: the general clue with this method is that in the two years, depreciation is the same. The formula for Calculating depreciation is given below;

straight-line method = (cost - Residual value)/ useful life in years.

From the question we know that the cost of acquisition is $30,000,000, the residual value of the asset is $4,000,000 and useful life is 7 years. Therefore;

straight-line method = ($30,000,000 - $4,000,000)/ 7.

= $3, 714,285.71 Per year.

That is $3, 714,285.71 for 2018 and 2019.

(2).Units of production​ (UOP) = (cost - Residual value)/ useful life in units.

= ($30,000,000 - $4,000,000)/ 4,375, 000.

Units of production​ (UOP) = $6 per mile.

Hence, the depreciation in 2018 = Depreciation per unit × 2018 year usage.

= 6 × 1,100,000 mile.

= $6,600,000.

depreciation in 2019 = Depreciation per unit × 2019 year usage.

= 6 × 1,200,000.

= $7,200,000.

Double-declining-balance​ (DDB)= (cost - accumulated depreciation) × 2 × 1/(useful life years).

Double-declining-balance​ (DDB) = (30,000,000 - 0)× 2 × (1/7).

= $8,571,428.57 depreciation in 2018.

= $8,571,428.6 depreciation in 2018

Double-declining-balance​ (DDB) = (30,000,000 - 8,571,428.57) × 2 × 1/7.

= $6,122,449.00 depreciation in 2019.

====================================================================

Total depreciation for straight-line method(2018 and 2019) = $7,428,571.42.

Total depreciation for Units of production​ (UOP)(2018 and 2019) = $13,800,000.

Total depreciation for Double-declining-balance (DDB)= $ 14,693,877.6.

5 0
3 years ago
Use the case structure in order to do the following tasks. a) Use a five layer case structure in order to do following for a two
Igoryamba

Answer:

The complete answer along with step by step explanation and output results is provided below.

Explanation:

Task a)

#include<iostream>

using namespace std;

int main()

{

  int op;

  double num1, num2;

   cout<<"Enter num1 and num2"<<endl;

   cin>>num1>>num2;  

  // To provide the option of required 5 cases  

  cout << "Select the operation:"

          "\n1 = Addition"

          "\n2 = Subtraction"

          "\n3 = Multiplication"

          "\n4 = Division"

          "\n5 = Negation\n";

  cin >> op;  // user input the desired operation

  switch(op)  // switch to the corresponding case according to user input

   {

       case 1:

           cout <<"The Addition of num1="<<num1<<" and num2="<<num2<<" is: "<<num1+num2;

           break;

       case 2:

           cout <<"The Subtraction of num1="<<num1<<" and num2="<<num2<<" is: "<<num1-num2;

           break;

       case 3:

           cout <<"The Multiplication of num1="<<num1<<" and num2="<<num2<<" is: "<<num1*num2;

           break;

       case 4:        

        while(num2 == 0) // to check if divisor is zero  

        {

           cout << "\nWrong divisor! Please select the correct divisor again: ";

           cin >> num2; // if divisor is zero then ask user to input num2 again

        }

           cout <<"The division of num1="<<num1<<" and num2="<<num2<<" is: "<<num1/num2;

           break;

       case 5:

           cout <<"The Negation of num1="<<num1<<" and num2="<<num2<<" is: "<<-1*num1<<" "<<-1*num2;

           break;

       default:

           // If the operation is other than listed above then error will be shown

           cout << "Error! The selected operatorion is not correct";

           break;

   }

  return 0;

}

Output:

Test 1:

Enter num1 and num2

2

9

Select the operation:

1 = Addition

2 = Subtraction

3 = Multiplication

4 = Division

5 = Negation

1

The Addition of num1=2 and num2=9 is: 11

Hence the output is correct and working as it was required

Test 2:

Enter num1 and num2

8

0

Select the operation:

1 = Addition

2 = Subtraction

3 = Multiplication

4 = Division

5 = Negation

4

Wrong divisor! Please select the correct divisor again: 2

The Division of num1=8 and num2=2 is: 4

Hence the output is correct and working as it was required

Test 3:

Enter num1 and num2

-2

4

Select the operation:

1 = Addition

2 = Subtraction

3 = Multiplication

4 = Division

5 = Negation

5

The Negation of num1=-2 and num2=4 is: 2 -4

Hence the output is correct and working as it was required

Task b)

#include<iostream>

#include<cmath>   // required to calculate square root

using namespace std;

int main()

{

  int op;

  double num;

   cout<<"Enter a real number > 0"<<endl;

   cin>>num;

  cout << "Press 1 for square root:";

  cin >> op;

  switch(op)  // switch to the corresponding case according to user input

   {

       case 1:

          if (num <= 0) // to check if number is less or equal to zero

        {

           cout << "\nError! number is not valid";

           break;   // if number is not valid then terminate program

        }

           cout <<"The square root of num="<<num<<" is: "<<sqrt(num);

// if number is valid then square root will be calculated

           break;

       default:

           // If the operation is other than listed above then error will be shown

           cout << "Error! The selected operation is not correct";

           break;

   }

  return 0;

}

Output:

Test 1:

Enter a real number > 0

6

Press 1 for square root: 1

The square root of num=6 is: 2.44949

Hence the output is correct and working as it was required

Test 2:

Enter a real number > 0

-4

Press 1 for square root: 1

Error! number is not valid

Hence the output is correct and working as it was required

8 0
3 years ago
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