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Vadim26 [7]
3 years ago
6

Write using about 10-15 lines for each of the six materials (metals, ceramics, glasses, polymers, composites, and semiconductors

). For each, list at least three most important properties, two drawbacks, three well-known examples, and three well-known applications.
Engineering
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer:

See Answer below- Explanation is the entire answer

Explanation:

Metals:

Properties: Ductile, good heat conductivity, good electrical conductivity, high strength;

Drawbacks: Relatively high weight, reactive with oxygen to create oxides- corrosion is presented;

Examples: steel, aluminum alloys, brass, copper, titanium

Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

Ceramics:

Properties: Brittle, poor heat conductors, poor electrical conductors, high wear resistance, corrosion resistance;

Drawbacks: Deforms by fracturing, shock resistance is low, no conductivity of electricity;

Examples: concrete, tungsten carbide, diamond

Applications: bricks for constructions, clay pots to keep heat, cutting tools for metals;

Glasses:

Properties: amorphous, transparent, high weight

Drawbacks: poor conductors of heat and electricity; brittle; low shock resistance;

Examples: Silica, lead glass, glaze;

Applications: windows, protection screens;

Polymers:

Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

Drawbacks: low strength, low operating temperatures;

Examples: polyethylene, nylon, ABS-plastic, rubber;

Applications: toys, tires, insulation covers for the wires.

Composites:

Properties: high strength to weight ratio, can get combination of properties from the used materials, rarely conductive, good shock resistance;

Drawbacks: high cost, hard to recycle, expensive;

Examples: steel-reinforced concrete, carbon fiber, fiber glass, Nomex, sandwich roof panels;

Applications: buildings, bullet proof vests, body of the Formula 1 cars, rockets, roof panels.

Semiconductors:

Properties: brittle, change conductive behavior under certain scenario, poor heat conductors;

Drawbacks: hard to manufacture, expensive;

Examples: Silicon-based semiconductors, Germanium-based semiconductors, Ga-based semiconductors;

Applications: chips, LED, diodes, transistors, op-amps, microprocessors.

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
kompoz [17]

Answer:

a. The mass flow rate (in lbm/s) is 135lbm/s

b. The temperature (in o F) is 200.8°F

Explanation:

We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

a. The mass flow rate (in lbm/s) is 135lbm/s

b.

m1 = Rate at inlet 1 = 125lbm/s

m2 = Rate at inlet 2 = 10lbm/s

The mass flow rate (in lbm/s) is calculated as m1 + m2

Mass flow rate = 125lbm/s + 10lbm/s

Mass flow rate = 135lbm/s

Hence, the mass flow rate (in lbm/s) is 135lbm/s

b. To calculate the temperature.

First we need to determine the enthalpy h1 at 14.7psia

Using table A-3E (thermodynamics)

h1 = 180.15 Btu/Ibm

h2 at 14.7psia and 60°F = 28.08 Btu/Ibm

Calculating h3 using the following formula

h3 = (h1m1 + h2m2) / M3

h3 = (180.15 * 125 + 28.08 * 10)/135

h3 = 168.8855555555555

h3 = 168.89 Btu/Ibm

To get the final temperature; we make use of table A-2E of thermodynamics.

Because h3 < h1, it means the liquid is at a compressed state.

The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

6 0
3 years ago
A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round
Harlamova29_29 [7]

Answer:

\frac{Q_{square}}{Q_{circle}} =  0.785  

Explanation:

given data

types of drinking straws

  1. square cross-sectional shape
  2. round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square  = perimeter of circle  

4 × S = π × D

here S is length and D is diameter

S = \frac{\pi D}{4}        ....................1

and

ratio of  flow rate through the square and circle is here

\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}  

\frac{Q_{square}}{Q_{circle}} =  0.785  

4 0
3 years ago
Hey, I have a question, I was thinking that if you have engineering skills or drawing skill you could help me to start a project
gayaneshka [121]

Answer

Dont have one;(

Explanation:

7 0
2 years ago
A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum
antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

  • Density = 61 veh/km.
  • Speed = 59 km/hr.
  • Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

q_{max}=\frac{V_f \times K_i}{4}

<u>Where:</u>

  • V_f is the free flow speed.
  • K_i is the Jam density.

In order to calculate the free flow speed, we would use this formula:

V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr

Substituting the parameters into the model, we have:

q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0
2 years ago
Match the test to the property it measures.
Vinvika [58]

Answer:

a. Rockwell              3. hardness

b. Instron                 2. stress vs strain

c. Charpy                 1. impact strength

d. Fatigue                4. Endurance Limit

e. Brinell                  3. hardness

f. Izod                      1. impact strength

Explanation:

Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.

Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.

Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.

Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.

7 0
3 years ago
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