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NeX [460]
3 years ago
5

When mass is the same, what is the relationship between radius and compression strength?

Engineering
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

a gas in a rigid container has a pressure of 632 torrs and a temperature of 45 celsius. The pressure has increased to 842 torrs. What is the new temperature of the gas

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A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?
stiks02 [169]

Answer:

The tax on this property is 3750 dollars

Explanation:

Given

Tax on per $100 is $2.50

Tax on every $1 is \frac{2.5}{100} = 0.025 dollars

Tax on property of value $150,000 is

150,000 * 0.025 = 3750 dollars

The tax on this property is 3750 dollars

7 0
3 years ago
PythonA group of statisticians at a local college has asked you to create a set of functionsthat compute the median and mode of
skelet666 [1.2K]

Answer:

  1. def median(l):
  2.    if(len(l) == 0):
  3.       return 0
  4.    else:
  5.        l.sort()
  6.        if(len(l)%2 == 0):
  7.            index = int(len(l)/2)
  8.            mid = (l[index-1] + l[index]) / 2
  9.        else:
  10.            mid = l[len(l)//2]  
  11.        return mid  
  12. def mode(l):
  13.    if(len(l)==0):
  14.        return 0
  15.    mode = max(set(l), key=l.count)
  16.    return mode  
  17. def mean(l):
  18.    if(len(l)==0):
  19.        return 0
  20.    sum = 0
  21.    for x in l:
  22.        sum += x
  23.    mean = sum / len(l)
  24.    return mean
  25. lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
  26. print(mean(lst))
  27. print(median(lst))
  28. print(mode(lst))

Explanation:

Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).

In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.

In mean function,  after checking the length of list,  we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).

In the main program, we test the three functions using a sample list and we shall get

20.5

12.5

12

3 0
3 years ago
Dear sir i want to ask something about the solution of my question?
Eva8 [605]
No you may not ask the question
3 0
3 years ago
Which of the following is a possible unit of ultimate tensile strength?
levacccp [35]

Answer:

Newton per square meter (N/m2)

Explanation:

Required

Unit of ultimate tensile strength

Ultimate tensile strength (U) is calculated using:

U = \frac{Ultimate\ Force}{Area}

The units of force is N (Newton) and the unit of Area is m^2

So, we have:

U = \frac{N}{m^2}

or

U = N/m^2

<em>Hence: (c) is correct</em>

4 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0
4 years ago
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