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2 years ago

What are the four categories of engineering materials used in manufacturing?

Engineering

1 answer:

alexgriva [62]2 years ago

8 0

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They have low weight and are poor conductors of electricity and heat

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Water at 20oC, with a free-stream velocity of 1.5 m/s, flows over a circular pipe with diameter of 2.0 cm and surface temperatur

Lisa [10]

Answer:

Average heat transfer coefficient = 31 kw/m^2 k

Heat transfer rate per meter length of pipe = 116.808 KW

Explanation:

water temperature = 20⁰c,

free-stream velocity = 1.5 m/s

circular pipe diameter = 2.0 cm = 0.02 m

surface temperature = 80⁰c

A) calculate average heat transfer coefficient

we apply the formula below :

m = αAv

A (area) = $\pi /4 (d)^2$

m = 10^3 * $\pi / 4 ( 0.02)^2$ * 1.5

= 10^3 * 0.7857( 0.0004) * 1.5

= 0.4714 kg/s

Average heat transfer coefficient

h = $\frac{m(cp)}{A}$ , A = $\pi DL$

L = 1 m , m = 0.4714 kgs , cp = 4.18

back to equation

h = $\frac{0.4714*4.18}{\pi * 0.02 }$ = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k

B) Heat transfer rate per meter length of pipe

Q = ha( ΔT ), a = $\pi DL$

= 31 * 0.0628 * ( 80 - 20 )

= 31 * 0.0628 * 60 = 116.808 KW

3 0

2 years ago

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

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2 years ago

Turning a screw with a screwdriver pushes the screw into the wood. The force is __________ over many turns

IrinaK [193]

Answer:

repeated?

Explanation:

not really sure what type of answer choices you have

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Answer:

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