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Genrish500 [490]
3 years ago
12

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved

perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 18 foot radius. It is 11 feet tall and has 6 feet of water in it. How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the density of the Kool-Aid contaminated water is σ=63.8lbs/ft^3. Don't forget to enter the correct units.
Mathematics
1 answer:
Rzqust [24]3 years ago
4 0

Answer:

The amount of work required to remove all of the water by pumping it over the side = 3.12 × 10⁶ lbs.ft

Step-by-step explanation:

Work done in moving anything from point A to point B = Fx

For this setup,

If we take an elemental vertical height, dx, The volume would be A.dx

where A = Cross sectional Area = πr² = π(18)² = 1017.88 ft²

dV = 1017.88 dx

The elemental force on that part will be

dF = ρg dV

ρg = 63.8 lbs/ft³

dF = 63.8 × 1017.88 dx = 64940.5 dx

F = ∫dF = 64940.5 dx

W = Fx = (∫dF)x = ∫ 64940.5x dx = 64940.5 ∫ xdx

We'll be integrating from (11 - 6) ft to 11 ft because that's the total height it'll be pumped through

W = 64940.5 (x²/2)¹¹₅ = 64940.5((11² - 5²)/2) = 3.12 × 10⁶ lbs.ft

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2 years ago
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Answer:

k = 4, y = 4x

Step-by-step explanation:

From the question, we get the ratio of :

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In the question, the two variables are :

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7 0
2 years ago
According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var
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Answer:

A)Mean :10.65

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Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

A)

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}

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Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.7

10.9

11.2

11.5

11.8

n=10(even)

Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7

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Standard deviation :\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}

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C)

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Range :(Q1-1.5IQR, Q3-1.5IQR)

Range :(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)

Range :(9.45, 10.15)

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