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horrorfan [7]
3 years ago
6

Liquid oxygen is stored in a thin-walled, spherical container 0.75 m in diameter, which is enclosed within a second thin-walled,

spherical container 1.1 m in diameter. The opaque, diffuse, gray container surfaces have an emissivity of 0.05 and are separated by an evacuated space.
a. If the outer surface is at 280 K and the inner surface is at 90 K, what is the mass rate of oxygen lost due to evaporation?

Engineering
1 answer:
nadezda [96]3 years ago
3 0

Answer:

Explanation:

the solution is well stated

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When we utilize a visualization on paper/screen, that visualization is limited to exploring: Group of answer choices Relationshi
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As many variables as we can coherently communicate in 2 dimensions

Explanation:

Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.

When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).

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Another focus of effective communication, according to Stephen Covey, is ensuring that:
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3 years ago
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Veronika [31]

Air supplied to a pneumatic system is supplied through the C. Actuator

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3 years ago
Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo
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An AI operated automatic garbage collection system

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a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

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