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UkoKoshka [18]
3 years ago
12

For a steel alloy it has been determined that a carburizing heat treatment of 7 hour duration will raise the carbon concentratio

n to 0.38 wt% at a point 3.8 mm from the surface. Estimate the time (in hours) necessary to achieve the same concentration at a 6.2 mm position for an identical steel and at the same carburizing temperature.

Engineering
1 answer:
Aliun [14]3 years ago
6 0

Answer:

18.6h

Explanation:

To solve this Duck's second law in form of Diffusion will be used.

Also note that since the temperature is constant D (change) will also be constant.

Please go through the attached files for further explanation and how the answer Is gotten.

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What are three types of land reform ​
s344n2d4d5 [400]

Answer:

Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);

5 0
3 years ago
Read 2 more answers
Teresa is emotionally volatile, particularly with friends and boyfriends. She is extremely dramatic about even the smallest disa
Ilia_Sergeevich [38]

Answer: Borderline personality Disorder.

Explanation: This is a type of mental disorder which could affects mood, behavior and relationships.

Its symptoms includes unstable emotions, sense of insecurity, worthlessness, and impulsivity.

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4 0
3 years ago
You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0
3 years ago
A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
Mazyrski [523]

Answer:

P > 142.5 N  (→)

the motion sliding

Explanation:

Given

W = 959 N

μs = 0.3

If we apply

∑ Fy = 0 (+↑)

Ay + By = W

If  Ay = By

2*By = W

By = W / 2

By = 950 N / 2

By = 475 N (↑)

Then  we can get F (the force of friction) as follows

F = μs*N = μs*By

F = 0.3*475 N

F = 142.5 N (←)

we can apply

P - F  > 0

P  > 142.5 N (→)

the motion sliding

6 0
3 years ago
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

6 0
3 years ago
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