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3 years ago

Write down one metal or alloy that is best suited for each of the following applications:

Engineering

1 answer:

777dan777 [17]3 years ago

6 0

Answer:

Explanation:

a. Cast iron or Aluminium alloy are typically used. Aluminium is much lighter in weight and it can transfer heat better to the coolant. While Cast Iron is typically stronger and is thus still used by the manufacturers.

b. Copper can be used as a condensing heat exchanger for hot steam due to its optimal thermal properties and its ability to resist corrosion.

c. high-speed steel are perfect for producing drill bits because of its hardness and resistance to heat to an extent. Drill bits tend to produce heat as a result of the friction between it and the material to be drilled.

d. lead can be used as a container for strong acids because of its anti-corrosive properties

e.zinc and copper can be used as fuel in pyrotechnics mainly due to the fact that burn with refreshing colours. Aluminium can also be used.

f. Platinum is the metal that best suits this purpose because of its high melting point and resistivity to oxidation.

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Integrated circuits typically are mounted on ________, which are then plugged into the system board.

MrRa [10]

Answer:

chip carriers

Explanation:

The components of a transistor or an integrated circuit are contained on a chip carrier. It is also frequently referred to as a chip container or chip package. With the help of this packaging, the chips can be connected or plugged into a circuit board without risking damage to their delicate components. As chip carriers have shrunk in size to accommodate new technologies, the procedure of installing them has grown more complicated.

5 0

1 year ago

Compute the number of kilo- grams of hydrogen that pass per hour through a 6-mm-thick sheet of palladium having an area of 0.25

nydimaria [60]

Answer:

The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is $4.1 * 10^{-3} \frac{kg}{h}$

Explanation:

Given

x1 = 0 mm

x2 = 6 mm = 6 * $10^{-3}$ m

c1 = 2 kg/ $m^{3}$

c2 = 0.4 kg/ $m^{3}$

T = 600 °C

Area = 0.25 $m^{2}$

D = 1.7 * $10^{8} m^{2}/s$

First equation

J = - D $\frac{c1 - c2}{x1 - x2}$

Second equation

J = $\frac{M}{A*t}$

To find the J (flux) use the First equation

J = - 1.7 * $10^{8} m^{2}/s$ * $\frac{2 kg/m^{3} - 0.4 kg/m^{3}}{0 - 6 * 10^{-3} } = 4.53 * 10^{-6} \frac{kg}{m^{2}s }$

To find M use the Second equation

$4.53 * 10^{-6} \frac{kg}{m^{2}s}$ = $\frac{M}{0.25 m^{2} * 3600s/h}$

M = $4.1 * 10^{-3} \frac{kg}{h}$

4 0

2 years ago

(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of

solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg. Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0

2 years ago

This problem demonstrates aliasing. Generate a 512-point waveform consisting of 2 sinusoids at 200 and 400-Hz. Assume a sampling

aalyn [17]

Answer and Explanation:

clear all; close all;

N=512;

t=(1:N)/N;

fs=1000;

f=(1:N)*fs/N;

x= sin(2*pi*200*t) + sin(2*pi*400*t);

y= sin(2*pi*200*t) + sin(2*pi*900*t);

for n = 1:20

a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))

b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))

c(n) = sqrt(a(n).^2+b(n).^2)

theta(n) =-(360/(2*pi))*atan(b(n)./a(n));

end

plot(f(1:20),c(1:20),'rd');

disp([a(1:4),b(1:4),c(1:4),theta(1:4)])

8 0

3 years ago

The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg

kiruha [24]

Answer:

percentage change in volume = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × $10^{11}$ N/m²

bulk stress = $10^{7}$ N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = $\frac{bulk\ stress}{bulk\ strain}$ ...............1

put here value and we get

3.5 × $10^{11}$ = $\frac{10^7}{bulk\ strain}$

solve it we get

bulk strain = 2.8571 × $10^{-5}$

and

bulk strain = $\frac{change\ volume}{original\ volume}$

so that percentage change in volume is = 2.8571 × $10^{-5}$ × 100

percentage change in volume = 0.00285 %

6 0

3 years ago