Answer:
(a) Increases
(b) Increases
(c) Increases
(d) Increases
(e) Decreases
Explanation:
The tensile modulus of a semi-crystalline polymer depends on the given factors as:
(a) Molecular Weight:
It increases with the increase in the molecular weight of the polymer.
(b) Degree of crystallinity:
Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.
(c) Deformation by drawing:
The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.
(d) Annealing of an undeformed material:
This also results in an increase in the tensile strength of the material.
(e) Annealing of a drawn material:
A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.
The following scenarios are pertinent to driving conditions that one may encounter. See the following rules of driving.
<h3>What do you do when the car is forced into the guardrail?</h3>
Best response:
- I'll keep my hands on the wheel and slow down gradually.
- The reason I keep my hands on the steering wheel is to avoid losing control.
- This will allow me to slowly back away from the guard rail.
- The next phase is to gradually return to the fast lane.
- Slamming on the brakes at this moment would result in a collision with the car behind.
Scenario 2: When driving on a wet road and the car begins to slide
Best response:
- It is not advised to accelerate.
- Pumping the brakes is not recommended.
- Even lightly depressing and holding down the brake pedal is not recommended.
- The best thing to do is take one foot off the gas pedal.
- There should be no severe twists at this time.
Scenario 3: When you are in slow traffic and you hear the siren of an ambulance behind
Best response:
- The best thing to do at this moment is to go to the right side of the lane and come to a complete stop.
- This helps to keep the patient in the ambulance alive.
- It also provide a clear path for the ambulance.
- Moving to the left is NOT recommended.
- This will exacerbate the situation. If there is no place to park on the right shoulder of the road, it is preferable to stay in the lane.
Learn more about rules of driving. at;
brainly.com/question/8384066
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Answer:FALSE
Explanation: A negative pressure respirator is a respiratory system which is known to have a low air pressure inside the mask when compared to the air pressure on the outside during Inhalation.
Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
= 0.75 < 25.84°
attached below is the remaining part of the solution
<u>B) Find the input current on the primary side in real units </u>
load current in primary = 31.38 < 25.84 A
<u>C) find the input power factor </u>
power factor = 0.9323 leading
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<em>attached below is the detailed solution </em>