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3 years ago

Which dimension of wellness focuses on finding enjoyment and satisfaction in daily endeavours?

Engineering

1 answer:

UNO [17]3 years ago

5 0

Answer:

I believe its emotional because it says satisfaction and most of the time that has to deal with emotions running through the body.

Explanation:

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A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles

tensa zangetsu [6.8K]

Answer:

(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

Explanation:

We are given:

L 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

= 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

= 8480/69

Mean = 122.9

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

=7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

∑fx² = 1104600

Var = ∑fx²/∑f - (Mean)²

= 1104600/69 - (122.9)²

= 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

P(X<115) = P(z<(115-122.9)/30.071)

= P(z<-0.26)

Using the normal distribution probability table, we can compute the value of P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

(c) σ = 30.071

P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

z = (x-μ)/σ

2.525 = (115 - μ)/30.071

75.93 = 115 - μ

μ = 115 - 75.93

μ = 39.07

4 0

3 years ago

A 36 ft simply supported beam is loaded with concentrated loads 16 ft inwards from each support. On the left side, the dead load

lana66690 [7]

Answer:

1st part: Section W18X76 is adequate

2nd part: Section W21X62 is adequate

Explanation:

See the attached file for the calculation

8 0

3 years ago

Which is not required when working in a manufacturing facility?

Artyom0805 [142]

Flip flops are not required

5 0

3 years ago

Explain 3 ways that people in sports use engineering to increase their performance?

LenKa [72]

Designing systems for manufacturing, motion analysis or impact testing;
building and testing prototypes;
analyzing the human body to prevent injury;
developing or designing new light weight materials that will be more comfortable and withstand greater impacts or forces;

7 0

3 years ago

A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni

arlik [135]

Answer:

The population size would be $p' = 5000$

The yield would be $MaxYield = 2082 \ fishes \ per \ year$

Explanation:

So in this problem we are going to be examining the application of a population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield, so basically we would be applying an engineering solution to fishing

So the current yield which is mathematically represented as

$\frac{dN}{dt} = \frac{2000}{1 \ year }$

Where dN is the change in the number of fish

and dt is the change in time

So in order to obtain the solution we need to obtain the rate of growth

For this we would be making use of the growth rate equation which is

$r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}$

Where N is the population of the fish which is given as 4,000 fishes

and K is the carrying capacity which is given as 10,000 fishes

r is the growth rate

Substituting these values into the equation

$r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]} =0.833$

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

$Max Yield = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year$

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by $p'$ would be

$p' = \frac{K}{2} = \frac{10,000}{2} = 5000\ fishes$

7 0

3 years ago

Read 2 more answers