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3 years ago

EASY MATH PROBLEMS! #8 and #10

Mathematics

1 answer:

xxMikexx [17]3 years ago

4 0

8. volume of cylinder is pi*r^2*h
So V= pi * (radius)^2 * height =720* pi

we know circumference = 2 * pi * r

if we look at first equation. We can edit it like
pi * r * r * h

now we can say it kind of looks like circumference equation but missing 2 so we multiply both sides by 2
2 * pi * r * r * h = 2 * 720 * pi

We can take out the 2 * pi * r and put in circumference (C)

c * r * h = 1440 * pi

We we know h is 20 so

c * r * 20 = 1440 * pi
c * r = 72 * pi

we have to solve for r first so break circumference back up

2 * pi * r * r = 72 * pi

r * r = 36
r = 6

plug into circum equation
C = 2 * pi * 6
C = 12 * pi

10. I assume the cylinders fill uo the box and if radius is 1.5 then diameter is 3. There are 5 in a row some 3 * 5 = 15 box length.
2 in a column so 2 * 3 = 6 box width
the height is 5

So
l * h * w = V

6 * 5 * 15 =V

V = 450

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Answer:

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, $c_{m}$ = The drag coefficient of the prototype, $c_{p}$

The medium of the test for the model, $\rho_m$ = The medium of the test for the prototype, $\rho_p$

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We have;

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Therefore;

$\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2$

$\dfrac{L_p}{L_m} =\dfrac{17}{1}$

$\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2$

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$\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3$

$\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3$ = (1/17)^3 ≈ 0.0002

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Hello!

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