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Finger [1]
2 years ago
6

Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance

+/- 0.000005 m^3/kg)?
Engineering
1 answer:
Svetradugi [14.3K]2 years ago
6 0

Answer:

specific\ volume=0.00097\ m^3/kg

Explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000 kg/m^3

Density of sea water = 1025  kg/m^3

We know that

Density=\dfrac{mass}{Volume}   ---1

Specific volume

specific\ volume=\dfrac{Volume}{mass}    ---2

From equation 1 and 2

We can say that

specific\ volume=\dfrac{1}{density}\ m^3/kg

specific\ volume=\dfrac{1}{1025}\ m^3/kg

specific\ volume=0.00097\ m^3/kg

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A flexural member is fabricated from two flange plates 7-1/2 x ½ and a web plate 17 x 3/8. The yield stress of steel is 50 ksi.
MissTica

Answer:

(a)

Plastic section modulus Z =92.72 in^3

Plastic moment Mp = 4636 kip-in

(b)

Elastic section modulus S = 80.88 in^3

Yield moment My = 4044 kip-in

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached files.i hope my explanation will help you in understanding this particular question.

7 0
2 years ago
NEEDS TO BE IN PYTHON:ISBN-13 is a new standard for indentifying books. It uses 13 digits d1d2d3d4d5d6d7d8d910d11d12d13 . The la
kondor19780726 [428]

Answer:

Follows are the code to this question:

n=input("Enter the first 12 digits of an ISBN-13 as a string:")#defining a varaible isbn for input value

if len(n)!=12: #use if block to check input value is equal to 12 digits

   print("incorrect input") #print error message

elif n.isdigit()==False: #use else if that check input is equal to digit

   print("incorrect input") #print error message

else:# defining else block

   s=0 #defining integer vaiable s to 0

   for i in range(12):#defining for loop to calculate sum of digit

       if i%2==0: #defining if block to check even value

           s=s+int(n[i])#add even numbers in s vaiable  

       else: #use else block for odd numbers

           s=s+int(n[i])*3 #multiply the digit with 3 and add into s vaiable

   s=s%10#calculate the remainder value  

   s=10-s#subtract the remainder value with 10 and hold its value

   if s==10: #use if to check s variable value equal to 10  

       s=0#use s variable to assign the value 0

   n=n+s.__str__() #u

Output:

please find attached file.

Explanation:In the above Python code, the "n" variable is used for input the number into the string format uses multiple conditional statements for a check input value, which can be defined as follows:

  • In if block, it checks the length isn't equal to 12, if the condition true, it will print an error message.
  • In the else, if the block it checks input value does not digit, if the condition is true, it will print an error message.
  • In the else block, it uses the for loop, in which it calculates the even and odd number sum, and in the odd number, we multiply by 3 then add into s variable.
  • In this, the s variable is used to calculate its remainder and subtract from the value and use the if block to check, its value is not equal to 10 if it's true, it adds 0 into the last of n variable, otherwise, it adds its calculated value.    

4 0
2 years ago
Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in
LenKa [72]

The given question is incomplete, the complete question is as follows:

Our text describes a trade-off that we must make as engineers between our confidence in the value of a parameter versus the precision with which we know the value of that parameter. That trade-off might be affected by whether we are looking at a two-sided or bounded (one-sided) interval.

Question: Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in one direction or the other in an engineering situation.

Answer: A balancing point is required to be reached to obtain a better confidence level in the predicted values.

Explanation:

The confidence interval and precision are the two terms that aims at providing the accurate estimation of the measurability of an object. If the precision increases, we can compromise on the confidence level and if the confidence level increases, then the precision of the predicted value also dilutes.

Thus a balance point is required to be reached between these two variables so that we get better confidence in the values being predicted without losing the correct estimation on precision. Ensuring that both the confidence and precision are maintained.

8 0
2 years ago
The rigid bar CDE is attached to a pin support at E and rests on the 30 mm diameter brass cylinder BD. A 22 mm diameter steel ro
Leno4ka [110]

Answer:

stress = 38.84 MPa

Explanation:

S_{D} = \alpha _{brass} * (delta T) *(L_{BD} )\\= (18.8 * 10^(-6) )*(30)*(0.3)\\= 0.0001692 m\\\\E_{BD} = stress / strain\\stress = E_{BD} * S_{D}\\stress = (200 *10^9) * (0.0001692)\\stress = 33.84 MPa

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2 years ago
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