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2 years ago

5

I need help on #9. The help would be grateful appreciated.

1 answer:

ELEN [110]2 years ago

8 0

Original Figure:
Length = 15
Width = 5
Height = 10
Volume = Length*Width*Height
Volume = 15*5*10
Volume = 750

New Figure
Length = 3
Width = 1
Height = 2
Each dimension has been divided by 5 (eg: 15/5 = 3)
Volume = Length*Width*Height
Volume = 3*1*2
Volume = 6

The old volume was 750 and it changes to 6
Notice how 750/6 = 125
Which can be rearranged to 750/125 = 6

Answer: if you divide the old volume by 125, then you get the new volume

Note: the new volume is 125 times smaller than the old volume
Put another way, the old volume is 125 times larger compared to the new volume

The fact that 125 = 5^3 is not a coincidence. If you divide each dimension by some number k, then you divide the volume by k^3

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2 years ago

If the graph of y = lxl is translated so that the point (1, 1) is moved to (1.4), what is the equation of the new graph?

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2 years ago

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3 years ago

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

10 x 200 is how many tens

melisa1 [442]

10 x 200 is 2000
therefore there are 200 tens .

7 0

3 years ago

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