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Ivahew [28]
3 years ago
7

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y

-axis.
y = 5x2, y = 30x − 10x2
Engineering
1 answer:
kow [346]3 years ago
3 0

Answer:

40π

Explanation:

First, find the limits (intersections).

5x² = 30x − 10x²

15x² − 30x = 0

x² − 2x = 0

x (x − 2) = 0

x = 0 or 2

Within this interval, 30x − 10x² is greater than 5x².

Dividing the volume into cylindrical shells, the volume of each shell is:

dV = 2π r h t

dV = 2π x (30x − 10x² − 5x²) dx

dV = 2π x (30x − 15x²) dx

dV = 30π (2x² − x³) dx

The total volume is the sum (integral):

V = ∫ dV

V = ∫₀² 30π (2x² − x³) dx

V = 30π ∫₀² (2x² − x³) dx

V = 30π (⅔ x³ − ¼ x⁴)|₀²

V = 30π (⅔ 8 − ¼ 16)

V = 30π (16/3 − 4)

V = 10π (16 − 12)

V = 40π

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Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

so we substitute

e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

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A thin 20-cm*20-cm flat plate is pulled at 1m/s horizontally through a 4-mm thick oil layer sandwiched between two stationary pl
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2 years ago
The diffusion coefficients for species A in metal B are given at two temperatures:
Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

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I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

R = 8.314\ J/mol*K

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

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5 0
3 years ago
A DOHC V-6 has how many camshafts?
viktelen [127]

Answer:

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4 0
3 years ago
A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
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Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

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L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m

The Biot number is given as;

B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952

B_i < 0.1,  thus apply lumped system approximation to determine the constant time for the process;

\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s

The time for the heating process is given as;

t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

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Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0
3 years ago
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