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zvonat [6]
3 years ago
9

Em uma estação elevatória de água tratada eleva 144 m³/h de água para um reservatório o através de

Advanced Placement (AP)
2 answers:
Genrish500 [490]3 years ago
8 0

Explanation:

The frictional resistance of the various pipes are given by the

K value in the table which may be used with the formula

hf = KQ2

to relate the magnitude of head loss hf in the pipeline

to the volumetric flow rate, Q. Water is drawn at the constant flow

rates from the network at nodes C and D. The static heads

(elevation + pressure head) at nodes B, C and D are 100m, 65m

and 61m respectively above the local datum. Calculate the

discharges at C and D and the water level in reservoir A. (The

data has been added to the diagram to aid the solution)

Use no more than 3 iterations and 3 significant figures.

Daniel [21]3 years ago
8 0

Explanation:

that is porchugis i think mass

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1. The rate at which people enter a movie theater on a given day is modeled by the function S defined by S(t) = 80 -12 cos 6 The
Arlecino [84]

Hi there!

a.

To find the total amount of people that have ENTERED by t = 20, we must take the integral of the appropriate function.

\text{Amount that entered} = \int\limits^{20}_{10} {S(t)} \, dt \\\\ = \int\limits^{20}_{10} {80 - 12cos(\frac{t}{5})} \, dt

Evaluate using a calculator:

= 899.97 \approx \boxed{900\text{ people}}

b.

To solve, we can find the total amount of people that have entered of the interval and subtract the total amount of people that have left from this value.

In other terms:
\text{Amount of people} = \int\limits^{20}_{10} {S(t)} \, dt - \int\limits^{20}_{10} {R(t)} \, dt

We can evaluate using a calculator (math-9 on T1-84):


\text{\# of people} = \int\limits^{20}_{10} {80-12cos(\frac{t}{5})} \, dt - \int\limits^{20}_{10} {12e^{\frac{t}{10}}+20} \, dt

= 899.97 - 760.49 = 139.47 \approx \boxed{139 \text{ people}}

c.

If:
P(t) = \int\limits^t_{10} {S(t) - R(t)} \, dt

Then:

\frac{dP}{dt}  = P'(t)= \frac{d}{dt}\int\limits^t_{10} {S(t) - R(t)} \, dt  = S(t) - R(t)

Evaluate at t = 20:


S(20) = 80 - 12cos(\frac{20}{5}) = 87.844\\\\R(20) = 12e^{\frac{20}{10}} + 20 = 108.669

S(20) - R(20) = 87.844 - 108.669 = -20.823

This means that at t = 20, there is a <u>NET DECREASE</u> of people at the movie theater of around 20.823 (21) people per hour.

d.

To find the maximum, we must use the first-derivative test.

Set S(t) - R(t) equal to 0:

80 - 12cos(\frac{t}{5}) - 12e^{\frac{t}{10}} - 20 = 0\\\\60 - 12(cos(\frac{t}{5}) + e^{\frac{t}{10}})= 0

Graph the function with a graphing calculator and set the function equal to y = 0:

According to the graph, the graph of the first derivative changes from POSITIVE to NEGATIVE at t ≈ 17.78 hours, so there is a MAXIMUM at this value.

<u>Thus, at t = 17.78 hours, the amount of people at the movie theater is a MAXIMUM.</u>

8 0
2 years ago
Which hormones regulates the usage of sugars in the bloodstream, and is insensitive in Type II Diabetics?
Rainbow [258]

Answer:

Insulin

Explanation:

3 0
3 years ago
There is a very low chance that buying a single lottery ticket will win you a jackpot. However a relative gives you a lottery ti
zheka24 [161]

Getting hit by lightning is almost 4 times more likely than winning the lottery. The chances of being fatally struck is 1 in 10 million, the Mirror claims. You're 45 times more likely to die from flesh eating bacteria than securing the jackpot. The odds are 1 in 1 million, the Daily Beast reports.

6 0
3 years ago
Which of the following jobs is in the agriculture career cluster ?
Romashka [77]
Hello there.

<span>Which of the following jobs is in the agriculture career cluster ? 

</span><span>C. Animal scientist 
</span>
4 0
3 years ago
Suppose a fox in a food web feeds only on rabbits, squirrels, mice and birds, all of which feed exclusively on plant material. A
Illusion [34]

Answer:

square metres of plant materials required to support the fox = 3650 m²

Explanation:

From the given information:

the daily caloric requirement of the fox = 800  

In a year, we have = 800 kcal × 365 /year = 292,000 kcal/year

Also, only 10% of the energy at a particular trophic level can be passed onto the next.

the net productivity of the plant material = 8000 kcal/m²/yr

So, using 10% of energy at a particular level, the fox only need 80 kcal/m²/yr

The objective is to determine in square meters, how many materials are required to support the fox.

square metres of plant materials required to support the fox = \dfrac{292,000}{80}

= 3650 m²

7 0
3 years ago
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