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valina [46]
2 years ago
15

WARNING: when people put links in the answer it is a virus DO NOT DOWNLOAD IT

Engineering
2 answers:
inn [45]2 years ago
8 0
Thank youuuu!!!!!!!!!!!!
kobusy [5.1K]2 years ago
6 0

Answer:

Thank you for this!

Explanation:

I was about to click it on a question I saw.

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Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.
Sedbober [7]

Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true

7 0
2 years ago
If anyone knows manufacturing plz help
sveta [45]

Answer:

I don't know ask my dad he would

Explanation:

but I can't ask him because he went to get milk and forgot to come back

8 0
2 years ago
It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown
sertanlavr [38]

Answer with Explanation:

The crown will be pure if it's specific gravity is 19.3

Now by definition of specific gravity it is the ratio between the weight of an object to the weight of water of equal volume

Since it is given that the weight of the crown is 11.8 N we need to find it's volume

Now According to Archimedes principle when the crown is immersed into water the water shall exert a force in upwards direction on the crown with a magnitude equaling to weight of the water displaced by the crown

Mathematically this is the difference between the weight of the crown in air and weight when immersed in water

Thus Buoyant force is F_{B}=11.8-10.9=0.9N

Now by Archimedes principle This force equals in magnitude to the weight of water of same volume as of the crown

Thus the specific gravity of the crown equals

S.G=\frac{11.8}{0.9}=13.11

As we see that the specific gravity of the crown material is less than that of pure gold hence we conclude that it is impure.  

4 0
3 years ago
Prepare deflection angles for staking the curves in problem 2-2. In every example assume that Sta. P.I. = 37+53.21. Carry the re
weeeeeb [17]

Answer:

ya mom

Explanation:

4 0
2 years ago
A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness
Bingel [31]

Answer:

a) heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

Explanation:

Assumptions:

  1. Constant properties
  2. Steady state conditions
  3. Negligible effect of radiation
  4. Negligible constant resistance between tube and insulation
  5. one dimensional radial conduction

a) What is the heat gain per unit tube length

R_{conv,i}'=\frac{1}{2\pi r_1h_i}

d_1=36mm Therefore r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}

r_2=2mm=2*10^{-3}m

k_{st}=14.2W/m.k

h_o=6W/m^2

h_i=400W/m^2

R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W

R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W

R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W

R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W

heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

r_3=r_1+r_2+10mm=30mm=0.03m

R_{conv,i}' and R_{cond,st}' are the same, but R_{conv,o}' changes.

Therefore:

R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W

R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W

The total resistance R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W

heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

8 0
3 years ago
Read 2 more answers
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