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Ad libitum [116K]

2 years ago

10

Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is

0.1143 nm, determine the unit cell volume in cubic nanometers. Round off the answer to three significant figures. Do not include the units.

1 answer:

sweet-ann [11.9K]2 years ago

8 0

Answer:

Unit cell volume will be

$4.866*10^{-2}$ $nm^{3}$

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Discuss the nature of materials causing turbidity in

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Answer:

a

Explanation:

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Derive an equation for the work of a mechanically reversible, isothermal compression of 1 mol of a gas from an initial pressure

Lyrx [107]

Answer:

The long derivation for work of a mechanically reversible, isothermal compression was done with detailed steps as shown in the attachment.

Explanation:

what is applied here is a long derivation from Work done in an isobaric process, the expression for the compressibility factor (Z) and the equation of state that was given. The requisite knowledge of Differentiation and Integration was used.

The detailed derivation from firs principle is as shown in the attachment.

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2 years ago

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

8 0

2 years ago

Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.

pantera1 [17]

Answer:

Eye.

Explanation:

Dust, dirt, and metal chips are most unpleasant to get in your eyes. Just experience it and you'll know what I mean.

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3 0

2 years ago

A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a

77julia77 [94]

Answer:

7.65 mm

Explanation:

Stress, $\sigma=\frac {F}{A}$ where F is the force and A is the area

Also, $\sigma=E\times \frac {\triangle L}{L}$

Where E is Young’s modulus, L is the length and $\triangle L$ is the elongation

Therefore,

$\frac {F}{A}= E\times \frac {\triangle L}{L}$

Making A the subject of the formula then

$A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}$

Since $A=\frac {\pi d^{2}}{4}$ then

$d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm$

4 0

2 years ago