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Marta_Voda [28]

3 years ago

9

The inner surface of a hollow cylinder is subjected to tangential and axial stresses of 40,000 and 24,000 psi, respectively. Det

ermine the maximum shear stress at the inner surface, if the cylinder is pressurized to 10,000 p

1 answer:

Furkat [3]3 years ago

5 0

Answer:

15,000 psi

Explanation:

The solution / solving is attach below.

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A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co

NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx 32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

$V_{hem}=\frac{2}{3}\pi r^3$

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

$V_{cyl}=\pi r^2\cdot h$

3) Conical bottom

Volume of conical bottom of radius'r' and angle $\theta$ is

$V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}$

Applying the given values we obtain the volume of the container up to cylinder is

$V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}$

Hence the capacity in liters is $V=32.355\times 1000=32355Liters$

3 0

4 years ago

Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 85 vol% carbide

velikii [3]

Answer:

Explanation:

k_max = 26.9 w/mk

k_min = 22.33 w/mk

Explanation:

a) the maximum thermal conductivity is given as

K_MAX = k_m v_m + k_p v_p

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.15

v_p = proportion of carbide in the cement = 0.85

$K_MAX = k_m v_m + k_p v_p$

= 66*0.15 + 20*0.85

k_max = 26.9 w/mk

b) the minimum thermal conductivity is given as

$k_min = \frac{ k_{carbide} *k_{metal}}{k_{metal} v_{carbide} +k_{carbide} v_{metal}}$

= \frac{20*66}{20*0.15 +66*0.85}

k_min = 22.33 w/mk

5 0

3 years ago

A motor with an operating resistance of 32 22 is connected to a

Akimi4 [234]

quizlet can help you its very helpful on quizes it some time mostly have the same excact quizes

4 0

3 years ago

On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch

rewona [7]

Answer:

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

Explanation:

Distance of earth from sun = $R_{2} = 147.4 \times 106 Km$

Spacecraft perihelion = $R_{2} = 120\times106Km$

gravitational parameters are now given as

$\mu_{sun} = 132.7\times 10^{9}$

$\mu_{earth} = 398600$

radius of earth = 6378 Km

Heliocentric spacecraft velocity at earth sphere of influence =

$V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }$

$V_{D}^{v} =28.43\frac{km}{s}$

Heliocentric velocity of earth = $v_{earth} = 30.06\frac{km}{sec}$

$V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}$

assume

$r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km$

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

$v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}$

geometric space craft velocity in its circular parking orbit

$v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } } = 7.784 \frac{km}{s}$

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

7 0

3 years ago

The A-36 solid steel shaft is 3.3 m long and has a diameter of 50 mm. It is required to transmit 35 kW of power from the engine

spin [16.1K]

Answer:

Explanation:

Answer is in the following attachment

8 0

3 years ago