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Rudik [331]
3 years ago
5

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

Engineering
1 answer:
marta [7]3 years ago
8 0

Answer:

(a) 3.185*10^{21} cells

(b) 6.37*10^{21} atoms

Explanation:

(a)

Volume, V of unit cell

V=(2.866*10^{-8})^{3}=2.354*10^{-23}

Number of unit cells, N

N=\frac {W_{mat}}{V\rho_{mat}} Where W_{mat} is weight of material and \rho_{mat} is density of material

N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron=3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms

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Answer:

2.6 miles

Explanation:

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A ruptured desiccant bag in a reciever-driver is usually caused by what?​
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2 years ago
What are the rigging devices used to move loads such as steel plates and sheet piles without the use of slings, but grip the loa
scoundrel [369]

The rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

<h3>What are the rigging devices?</h3>

The rigging devices are used to lift the objects and items when the safety is required. This device is used in the industries.

Types of rigging devices

  • Rigging hooks-These rigging device is used when the heavy load need to be lift.
  • Lifting clamps-Lifting clamp are used to lift the device with jaw tension to secure the load. In this, there is no use of slings.
  • Pulley and blocks-In the load is lifts with the help of block and pulley arrangement. This is a widely used rigging device.

Thus, the rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

Learn more about the rigging devices here;

brainly.com/question/8430576

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4 0
2 years ago
Do you understand entropy? Why the concept of entropy is difficult to engineering students?
Darya [45]

Answer:

Entropy:

  Entropy is the measure of randomness of system.In other words the entropy is the measurement of tendency of system towards the disorder.

The concept of entropy arise from second law of thermodynamics.It is given as follows

ds=\int \dfrac{dQ}{T}

Entropy is a extensive property of system .

Entropy of universe = Entropy of system + Entropy of surrounding.

The entropy of the system can be zero,positive and negative.But entropy of the surrounding can not be negative,but it can be zero or positive.

Actually the concept of entropy is difficult to understand because we can not visualize because it is not like beam and like rods.Only we have to  realize that there is entropy.

4 0
3 years ago
One cylinder in the diesel engine of a truck has an initial volume of 650 cm3 . Air is admitted to the cylinder at 35 ∘C and a p
kupik [55]

Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

P*V=n*R*T

since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

P*V^k = constant = C, k= adiabatic coefficient for air = 1.4

then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)  

T2 = (1-k)W* T1/(P1*V1) +T1

replacing values (W=-450 J since it is the work done by the gas to the piston)

T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

therefore

T2/T1= (V2/V1)^(1-k)

V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³

3 0
3 years ago
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