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denis-greek [22]
3 years ago
9

Find the True statement

Engineering
2 answers:
dmitriy555 [2]3 years ago
6 0

Answer:

Option A is correct ( Stress relaxation is time- and temperature-dependent).

Explanation:

Luda [366]3 years ago
6 0

Answer:

Option A is correct.

The only completely true statement in all of the available options is that "Stress relaxation is time- and temperature-dependent"

Explanation:

Stress relaxation is the constant decrease of stress with time in a material which is exposed to a constant strain at a constant temperature.

It is similar in concept to vaccination; in that, the material is being prepared to be able to undergo a particular amount of stress at a specific temperature in order to make the stress levels in the material lower than it usually should be.

It is evident that stress relaxation, (the extent), in a material completely depends on the amount of time that the constant strain is applied and the temperature at which this whole process occurs.

Polymers always have a melting temperature.

While it is true that almost all materials can melt. For Polymers, it is not necessarily true that the melting takes place at a constant temperature. In very complex Polymers, melting occurs over an appreciable range of temperature, not sharply as in some simple polymers.

Hence, polymers don't always have a melting temperature, some polymers melt over a range of temperature. So, this statement isn't completely true.

Polyethylene becomes weaker if H is replaced with other radicals inside branches.

This also is a statement that isn't always true. Replacing the H in some of the branches of polyethylene with radicals such as the carboxylic group, the hydroxyl group etc., have been shown to seriously strengthen the polyethylene. Polymers formed by replacing the H with ion-Like radicals are known as ionomers.

These groups have a polarity, hence, they are drawn together by their charges, come together in microdomains, toughening and strengthening the polyethylene without taking its ability to be cast to permanent shapes away.

This statement isn't always true too.

Crazing indicates cracks inside the polymer.

This is also false. Crazing doesn't indicate cracks in materials. What it does is that Crazing precedes the cracks.

Crazing is a phenomenon that entails the formation of microvoids in a material in response to excessive tensile stress being applied to the material. The microvoids (which forms in a plane normal to the tensile stress that caused them) often leads to cracks or fracture later.

So, Crazing doesn't indicate cracks, rather, it precedes them in most cases.

Hope this Helps!!!

You might be interested in
4.71 A full-wave rectifier circuit with a 1-kΩ operates from a 120-V (rms) 60-Hz household supply through a 6-to-1 transformer h
brilliants [131]

Answer:

V_{p (load)} = 28,3 V - 0,7 V = 27,6 V

V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA

Explanation:

The peak voltage after the 6 to 1 step down is V_{p} = \frac{120}{6} \sqrt{2} =  28,3V. Then, the peak voltage of the rectified output is V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi}  = \frac{55,2 V}{\pi } = 17,6 VV_{d}[/tex] and according to the statement, the diodes can be modeled to be V_{d} = 0,7 V. Then, the peak voltage in the load is V_{p (load)} = 28,3 V - 0,7 V = 27,6 V.

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

The average output voltage is calculated as:

V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V

The average current in the load is calculated as:

I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA

4 0
4 years ago
The application of technology results in human-made things called
Sergio039 [100]

Answer:

Internet of things

Explanation:

This is a good example where the application of technology results are applied to human made things.

Internet of things (IOT), involves the application of one technology results–the internet, embedded into devices such as refrigerator, television etc so as to send and receive data (digital instructions). Such applications of technology results has revolutionized the way we use "human made things".

8 0
4 years ago
11. Technicians A and B are discussing
algol [13]

Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

5 0
3 years ago
A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?
IceJOKER [234]

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V <em>(Here 208 V is three phase voltage and 120 V is single phase voltage) </em>

<em>Since,</em>

<em />

<em />V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\<em />

V_{1ph} = 120 V

The formula for apparent power in three phase system is given as:

S = \sqrt{3} VI

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A

 

4 0
3 years ago
A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is
jonny [76]

Answer:

a) t = 277.477\,s\,(4.625\min), b) v_{f} = 0\,\frac{mi}{h}, c) a = -0.128\,\frac{ft}{s^{2}}

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}

a = -0.185\,\frac{ft}{s^{2}}

The time required to travel is:

t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }

t = 277.477\,s\,(4.625\min)

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be v_{f} = 0\,\frac{mi}{h}.

c) The final constant deceleration is:

a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}

a = -0.128\,\frac{ft}{s^{2}}

7 0
4 years ago
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