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Kobotan [32]
3 years ago
13

Please help on two I will give brainiest​

Engineering
2 answers:
Readme [11.4K]3 years ago
7 0

Answer:

Prototype I believe

Explanation:

Delicious77 [7]3 years ago
4 0
Prototype correct me if i am wrong
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A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete
Alex

Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

d is grain diameter to be achieved after heating in mm

d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

t₁= 500 min = 500/60 =25/3 hrs

t₂=?

n=2.2

First find K

K=(d²-d²₀ )/t₁

K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3

K=(0.051²-0.024²) ÷25/2

K=0.000243 mm²/h

Re-arrange equation for K ,to get the equation for d as;

d=√(d₀²+ Kt)  where now t=t₂

d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min

4 0
3 years ago
HfrrghhJbfrfefefft nbgyjjiutrwdgju
Tasya [4]

Answer:

hshdhriwjajaldh skshdjdywuusg

Explanation:

null

4 0
3 years ago
Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______
Alina [70]

Answer:

Multiplying impulse response by t  ( option D )

Explanation:

We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .

When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .

5 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete
Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

\sigma_c=\frac {K}{Y \sqrt {a\pi}} and making Y the subject

Y=\frac {K}{\sigma_c \sqrt {a\pi}}

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, \sigma_c  is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

\sigma_c=\frac {K}{Y \sqrt {a\pi}}  and making K the subject

K=\sigma_c Y \sqrt {a\pi}  and substituting 260 MPa for \sigma_c  while a is taken as 0.003m and Y is already known

K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0
3 years ago
The settlement of foundations is typically the result of three separate occurrences that take place in the soil which provides s
evablogger [386]

Answer:

The differences are listed below

Explanation:

The differences between consolidation and compaction are as follows:

In compaction the mechanical pressure is used to compress the soil. In consolidation, there is an application of stead pressure.

In compaction, there is a dynamic load by rapid mechanical methods like tamping, rolling, etc. In consolidation, there is static and sustained pressure applied for a long time.

In compaction, the soil volume is reduced by removing air from the void. In consolidation, the soil volume is reduced by squeezing out water from the pores.

Compaction is used for sandy soil, consolidation on the other hand, is used for clay soil.

7 0
3 years ago
Read 2 more answers
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