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schepotkina [342]
3 years ago
15

Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour

s.
Engineering
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by  \frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}

So availability of system =\frac{900}{900+100}=\frac{900}{1000}=0.9

So the availability of system will be 0.9

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A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the
muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, \L_e = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

P =\frac{\pi^2EI}{L_e^2}

where, I is the moment of inertia

on substituting the respective values, we get

10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}

or

I = 13.13 in⁴

also for circular cross-section

I = \frac{\pi}{64}\times d^4

thus,

13.13 = \frac{\pi}{64}\times d^4

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0
3 years ago
Consider the following list. list = {24, 20, 10, 75, 70, 18, 60, 35} Suppose that list is sorted using the insertion sort algori
Greeley [361]

Answer:

Option B

10,20,24,75,70,18,60,35

Explanation:

The first, second and third iteration of the loop will be as follows

insertion sort iteration 1: 20,24,10,75,70,18,60,35

insertion sort iteration 2:10,20,24,75,70,18,60,35

insertion sort iteration 3: 10,20,24,75,70,18,60,35

8 0
3 years ago
Part A Engineering stress and strain are calculated using the actual cross-sectional area and length of the specimen. True or fa
Galina-37 [17]

Answer: True

Explanation:

Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.

While Engineering strain is the amount that a material deforms per unit length in a tensile test.  It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.

5 0
3 years ago
Read 2 more answers
A liquid phase reaction, A → B, is to be carried out in an isothermal, well mixed batch reactor with a volume of 1L. Initially t
Archy [21]

Answer:

the time, in hours = 4.07hrs

Explanation:

The detailed step by step derivation and appropriate integration is as shown in the attached files.

4 0
3 years ago
Assume you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all
garik1379 [7]

Answer:

Explanation:

Given data;

  • In line 1, v1 = 30mi/hr
  • in line 2, v2 = 45mi/hr
  • in line 3, v3 = 60mi/hr
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  • = VT = 45mi/hr

  • space mean speed ; Vs
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  • Hence Vs = V x n = 3 x 13.85 = 41.55mi/hr
5 0
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