
(a)
Interpretation:
The reason for which the aluminium is used for airplane bodies is to be explained.
Concept Introduction:
Aluminum is soft, non-magnetic and ductile metal which also forms an alloy by melting with copper silicon.
(b)
Interpretation:
The reason for polyurethane to be used for alignment of teeth is to be explained.
Concept Introduction:
Polyurethane is a polymer made up of organic units by combining units of urethane. The polyurethane is a thermosetting polymer which does not melt on heating.
(c)
Interpretation:
The properties of steel making it useful for ball bearing in the bicycles wheel hub are to be explained.
Concept Introduction:
Steel is one of the important engineering and construction materials. It is an alloy of iron and carbon where, carbon is less than 1%.
(d)
Interpretation:
The reason behind using polyurethane terephthalate for water bottles is to be explained.
Concept Introduction:
Polyurethane terephthalate is the most common thermoplastic polymer used in containers for liquids and foods. It is strong, lightweight plastic used in storage containers for foods and beverages, soft drinks, water, etc.
(e)
Interpretation:
The reason for the wine bottles to be made up of glass is to be explained.
Concept Introduction:
Glass bottles are chemical free, hence, there is no chemical leeching. They are easy to clean. Glass does not react with any other substance.

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Chapter 1 Solutions
Essentials of Materials Science and Engineering, SI Edition
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- Q5: A 5.6 litre V8 engine with a compression ratio of 9.4:1 operates on an air-standard Otto cycle at 2800 RPM, with a volumetric efficiency of 90 % and a stoichiometric air-fuel ratio using gasoline. The exhaust flow undergoes a temperature drop of 44ºC as it passes through the turbine of the supercharger. Calculate (a) mass flow rate of exhaust gas and (b) power available to drive the turbocharger compressor.arrow_forwardFind v(t) for t> 0 in the circuit of Fig. below. Assume the switch has been open for a long time and is closed at t = 0. Calculate v (t) at t = 0.5. 10 V 202 www +21 t=0 60 ww 13 F بلا SVarrow_forwardQ1: determine the area of steel for the slab that rests on brick walls and is shown in the figure. Use the following data: f'c = 25 MPa, fy = 420 MPa, L.L. = 1.5 kN/m², D.L. = 3 kN/m² (without self-weight). Rate from (1 to 4) your ability to solve the problem 7 m 0.3 m 0.3 marrow_forward
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