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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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Calculate the diameter of pneumatic cylinder to drive a mechanism. The force exerted by the mechanism is 80 N. The return force
sashaice [31]

Answer:

14.27 mm

Explanation:

Force = 80 Newton

Pressure exerted = 5 bar = 5×10⁵ Pa

Pressure exerted = Force/Area

Area = Force/Pressure

Area=\frac{80}{500000}\\\Rightarrow Area=0.00016\ m^2

Area of the cylinder=πd²/4

\Rightarrow \frac{\pi}{4}d^2=0.00016\\\Rightarrow d^2=\frac{4\times 0.00016}{\pi}\\\Rightarrow d^2=0.0002037\\\Rightarrow d=0.01427\ m

Hence diameter of cylinder 14.27 mm

5 0
3 years ago
Which of the following is a way to heat or cool a building without using electricity or another power source?
zzz [600]
I believe the answer is: A. Passive heating and cooling.
8 0
3 years ago
Read 2 more answers
Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i
DedPeter [7]

Answer:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

Explanation:

The internal energy is equal to:

dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV

The entropy is equal to:

dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }

Replacing:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

4 0
3 years ago
A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4
Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0
3 years ago
Air enters a 34 kW electrical heater at a rate of 0.8 kg/s with negligible velocity and a temperature of 60 °C. The air is disch
Flura [38]

Answer:

79 kW.

Explanation:

The equation for enthalpy is:

H2 = H1 + Q - L

Enthalpy is defined as:

H = G*(Cv*T + p*v)

This is specific volume.

The gas state equation is:

p*v = R*T (with specific volume)

The specific gas constant for air is:

287 K/(kg*K)

Then:

T1 = 60 + 273 = 333 K

T2 = 200 + 273 = 473 K

p1*v1 = 287 * 333 = 95.6 kJ/kg

p2*v2 = 287 * 473 = 135.7 kJ/kg

The Cv for air is:

Cv = 720 J/(kg*K)

So the enthalpies are:

H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW

H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW

Ang the heat is:

Q = 34 kW

Then:

H2 = H1 + Q - L

381 = 268 + 34 - L

L = 268 + 34 - 381 = -79 kW

This is the work from the point of view of the air, that's why it is negative.

From the point of view of the machine it is positive.

4 0
4 years ago
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