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3 years ago

11

it takes a hose 4 minutes to fill a rectangular aquarium 9 inches long, 13 inces wide, and 14 inces tall. how long will it take

the same hose to fill an aquarium measuring 21 inces by 25 inces by 34 inces?

1 answer:

Studentka2010 [4]3 years ago

4 0

Answer: 43.58 min

Explanation:

Knowing the volume of a rectangular object is length x width x height, we have two volumes:

$V_{1}=(9 in)(13 in)(14 in)=1638 in^{3}$

$V_{2}=(21 in)(25 in)(34 in)=17850 in^{3}$

And we know it takes a time of 4 minutes to fill $V_{1}$ .

If we want to know how long will it take the same hose to fill another tank with volume $V_{2}$ , we can use the <u>Rule of three</u>, which is a mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known:

$1638in^{3}$ ---- $4min$

$17850in^{3}$ ---- $?$

$?=\frac{(17850 in^{3})(4min)}{1638 in^{3}}$

Finally:

$?=43.58 min$

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2 years ago

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Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

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