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3 years ago

Advantages of using metal

Engineering

1 answer:

Contact [7]3 years ago

4 0

Answer:

Metals have high melting points thus unlikely to degrade when temperatures increase, they can be fabricated and are cost effective due to availability.

Explanation:

Aluminum is the most abundant in the Earth's crust with good thermal and electric properties. It is soft, malleable ,ductile and lighter making it a vital metal in construction industry. An alloy of copper and tin, bronze is a better connector of heat and electricity ,commonly used in automobile industry for bearings and springs production. Steel a carbon alloy has applications in forging and automotive.

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A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball

faltersainse [42]

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

$\sigma = \frac{p*r}{2t}$

<u>Where:</u>

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm

Hence, we need to find r, as follows:

$r_{inner} = r_{outer} - t$

$r_{inner} = \frac{d}{2} - t$

<u>Where:</u>

d: is the outer diameter = 300 mm

$r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm$

Now, we can find the normal stress (σ) in the wall of the basketball:

$\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa$

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!

3 0

3 years ago

1. Discuss the benefits of observing good safety measures in relation to an increase in

Aneli [31]

The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the spills of chemicals and other kinds of accidents, as well as reduce the damage to the environment that is outside of the lab.

<h3>What are the benefits of practicing safety in the laboratory?</h3>

A laboratory is known to be one that is known to have a lot of potential risks that is said to often arise due to a person's exposure to chemicals that are corrosive and toxic, flammable solvents, high pressure gases and others.

Therefore, A little care and working in line to all the prescribed safety guidelines will help a person to be able to avoid laboratory mishaps.

Therefore, The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the spills of chemicals and other kinds of accidents, as well as reduce the damage to the environment that is outside of the lab.

Learn more about safety measures from

brainly.com/question/26264740

#SPJ1

8 0

2 years ago

1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g

Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x $10^{22}$ $m^{-3}$

atomic weight = 58.69 g/mol

density = 8.8 g/cm³

solution

we get here N that is

N = $\frac{N_A \times \rho}{A}$ ...........1

N = $\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}$

N = $9.030 \times 10^{28}$

and here no of vacancy will be

Nv = $N \times e^{\frac{-Qv}{kT}}$ .................2

put here value

$4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}$

$10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}$

take ln both side

$ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})$

$-14.468 = \frac{-Qv}{0.0968}$

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0

3 years ago

an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required

ASHA 777 [7]

Answer:

The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

Explanation:

Given;

Resistance, R₁ = 50Ω

Resistance, R₂ = 75Ω

Total resistance, R = (R₁R₂)/(R₁ + R₂)

Total resistance, R = (50 x 75)/(125)

Total resistance, R = 30 Ω

According to ohms law, sum of current in a parallel circuit is given as

I = I₁ + I₂

$I = \frac{V}{R_1} + \frac{V}{R_2}$

Voltage across each resistor is the same

$1.6 = \frac{V}{R_2}$

V = 1.6 x R₂

V = 1.6 x 75

V = 120 V

Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.

3 0

3 years ago

Read 2 more answers

What are the three different types of cabling recognized by TIA/EIA as acceptable for horizontal wiring

snow_tiger [21]

Answer:

UTP, STP, or fiber optic cable

Explanation:

The three cabling types are

1. UTP is unshielded twisted pair cables. It contains two unshielded wires that are wrapped around each other. Mostly used in telephone and LANs wires.

2. STP is shielded twisted pair. This can be compared to UTP but each if the pair has an extra copper braid jacket protecting it. It is used mainly for ethernet networks.

3. Fibre optics cables. These are network cables with strands iv glass fibers that are in an insulated casing.

This kind of cable can provide greater bandwidth as well as data transmission that goes over a long distance.

4 0

3 years ago

Read 2 more answers