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o-na [289]
2 years ago
14

) A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging proc

ess. Twenty specimens of the polymer were used in an experiment. Ten were assigned randomly to be exposed to an accelerated batch aging process that involved exposure to high temperatures for 10 days. Measurements of tensile strength of the specimens were made, and the following data were recorded on tensile strength in psi:
No aging: 227, 222, 218, 217, 225, 218, 216, 229, 228, 221
Aging: 219, 214, 215, 211, 209, 218, 203, 204, 201, 205

(a) Do a dot plot of the data.
(b) From your plot, does it appear as if the aging process has had an effect on the tensile strength of this polymer? Explain.
(c) Calculate the sample mean tensile strength of the two samples.
(d) Calculate the median for both. Discuss the similarity or lack of similarity between the mean and median of each group.
(e) Calculate the sample variance as well as standard deviation in tensile strength for both samples.
(f) Does there appear to be any evidence that aging affects the variability in tensile strength?
Engineering
1 answer:
bonufazy [111]2 years ago
5 0

Answer:

it will be a scattered plot

Explanation:

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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump eff
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Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

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p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

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Using this formula

Required power input=P/μ

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P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

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