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Taya2010 [7]
3 years ago
5

In a creep test, increasing the temperature will (choose the best answer) A. increase the instantaneous initial deformation B. i

ncrease the steady-state creep rate
Engineering
1 answer:
Hitman42 [59]3 years ago
8 0

Answer:

All of the above

Explanation:

firstly, a creep can be explained as the gradual deformation of a material over a time period. This occurs at a fixed load with the temperature the same or more than the recrystallization temperature.

Once the material gets loaded, the instantaneous creep would start off and it is close to electric strain. in the primary creep area, the rate of the strain falls as the material hardens. in the secondary area, a balance between the hardening and recrystallization occurs. The material would get to be fractured hen recrstallization happens.  As temperature is raised the recrystallization gets to be more.

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A network address of 172.16.0.0 /12 has been given. Which of the following accurately describes this network? (select one or mor
ludmilkaskok [199]

Answer:

B and E is correct.

Explanation:

Given that network address

172.16.0.0/12

This is class B network type.

The ending of this network will be 172.31.255.255

In IP version 4 there are four following type of classes

1)Class A (0-127)

2)Class B (128-191)

3)Class C (191-223)

4)Class D(224-239)

5)Class E (240-255)

Generally class A,B,C and D are used.

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3 0
3 years ago
Motion is defined as a change in an object's position when compared to other objects around it. Mary Ann was watching a slug cra
lara [203]

Answer:

Explanation:

Do it yourself bum

4 0
4 years ago
Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5
pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

Aa.Va=Ab.Vb=Q

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3

Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3

Using the volume rate:

Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s

Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s

Assuming no losses, the energy equation for fluids can be written as:

Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m

Pb = 70000\ Pa-27303\ Pa - 14715\ Pa

Pb = 27,996\ Pa = 28\ kPa

6 0
3 years ago
What should I do I if a patient I’s bleeding th our his penis
Brilliant_brown [7]

Answer:

YOU SHOULD HELP HIM

Explanation:

6 0
3 years ago
Read 2 more answers
Select the correct answer.
Rina8888 [55]

A: Inexhaustible resources are limited

is true

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