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maksim [4K]
3 years ago
10

Coffee is poured at a uniform rate at 20 cm3 / s into a cup whose inside is shaped like a truncated cone. if the upper and lower

radii of the cup are 4 cm and 2 cm and the height of the cup is 6 cm, how fast will the coffee be rising when the coffee is halfway up (hint: extend the cup downward to form a cone.)
Physics
1 answer:
Vladimir [108]3 years ago
5 0

Let the cup is filled to height h after some time

now the total volume of coffee filled in the cup is given as

\frac{2}{y} = \frac{4}{6+y}

2y = 6 + y

y = 6 cm

now volume of the coffee will be

V = \frac{1}{3}\pi r^2(y + 6) - \frac{1}{3}\pi 2^2 (6)

here we know that

\frac{r}{y+6} = \frac{2}{6}

r = \frac{y+6}{3}

V = \frac{1}{3}\pi (\frac{y+6}{3})^2(y+6) - \frac{1}{3}\pi 2^2(6)

now we know that volume flow rate is given as

Q = \frac{dV}{dt}

20 cm^3/s = \frac{1}{3}\pi (\frac{1}{9})(3(y+6)^2)\frac{dy}{dt}

20 \times 9 = \pi (y + 6)^2 v

here y = 3 cm

180 = \pi (9)^2 v

v = 0.71 cm/s

so water will rise up with speed 0.71 cm/s


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