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Marizza181 [45]
3 years ago
6

32 gal/min = what qt/h

Mathematics
2 answers:
Irina18 [472]3 years ago
6 0

Answer: 7529.4 qt/hr

Step-by-step explanation:

The conversion used from gal to qt is:

1 gal = 4 qt

32 gal= \frac{4}{1}\times 32=128

The conversion used from min to hr is:

60 min = 1 hr

1 min =\frac{1}{60}\times 1=0.017hr

We are asked: 32 gal/min = ? qt/hr

32gal/min=\frac{128}{0.017}qt/hr=7529.4qt/hr

Therefore, the conversion 32 gal/min = 7529.4 qt/hr

mafiozo [28]3 years ago
3 0

32g/m = 32*60 = 1920 gallons per hour

1 gallon = 4 quarts

1920 *4 = 7680 quarts per hour

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A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.8kg/L
Scorpion4ik [409]

Answer:

1. \dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}

2. y(40) = 110.873 \ kg

Step-by-step explanation:

Given that:

A very large tank initially contains 100 L of pure water.

Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min.

. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

As 5L/min is entering and 3L/min is drained out, there is a 2L increase per minute. Therefore, the amount of water at any given time t = (100 +2t) L

t = (50 + t ) L

Since it is given that we should  consider y(t) to be the  amount of salt (in kilograms) in the tank after t minutes.

Then , the differential equation that  y satisfies can be computed as follows:

\dfrac{dy}{dt}=rate_{in} - rate_{out}

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\dfrac{dy}{dt}=(0.8)(5) -\dfrac{3y}{100+2t}

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How much salt is in the tank after 40 minutes?

So,

suppose : e^{\int \dfrac{3}{100+2t} \ dt} = (t+50)^{3/2}

Then ,

( t + 50)^{3/2} y' + \dfrac{3}{2}(t+50)^{1/2} y = 4(t+50)^{3/2}

( t + 50)^{1.5} y' + \dfrac{3}{2}(t+50)^{0.5} y = 4(t+50)^{3/2}

[y\ (t + 50)^{1.5}]' = 4(t+ 50)^{1.5}

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C= -1.6(50)^{2.5}

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y(40) = 144  - 1.6 \times 17677.66953 (90)^{-1.5}

y(40) = 144  - 1.6 \times 17677.66953 \times 0.001171213948

y(40) = 144  - 33.12693299

y(40) = 110.873 \ kg

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4 years ago
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