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zhuklara [117]
2 years ago
13

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in t

he unit cell and diameter of the metal atom.

Engineering
1 answer:
uysha [10]2 years ago
8 0

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

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A copper-nickel alloy of composition 60 wt% Ni-40 wt% Cu is slowly heated from a temperature of 1250°C (2280 °F). (a) At what te
makkiz [27]

Answer:

a. The very first liquid process, when heated from 1250 degree Celsius, is expected to form at the temperature by which the vertical line crosses the phase boundary (a -(a + L)) which is about <em>1310 degree Celsius. </em>

b. The structure of that first liquid is identified by the intersection with ((a+ L)-L) phase boundary; <em>47wt %of Ni</em> is of a tie line formed across the (a+ L) phase area <em>at 1310 degrees.</em>

c. To find the alloy's full melting, it is determined that the intersection of the same vertical line at 60 wt percent Ni with (a -(a+L)) phase boundary is around <em>1350 degrees.</em>

c. The structure of the last remaining solid before full melting correlates to the intersection with the phase boundary (a -(a + L), of the tie line built at 1350 degrees across the (a + L) phase area, <em>being 72wt % of Ni.</em>

4 0
3 years ago
Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.
atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage v_p=120volt

(i) We know that peak voltage is give by v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt

(ii) We know that for transformer \frac{v_p}{v_s}=\frac{n_p}{n_s}

So \frac{169.08}{v_s}=\frac{10}{1}

v_s=16.90volt

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt

(v) Now dc current is given by i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A

4 0
2 years ago
If a system of pulleys results in a force of 25% of the load, how far will the rope need to move to pull the load a distance of
GaryK [48]

Answer:

  40 ft

Explanation:

Assuming no loss of energy in the system of pulleys, the work done is the same whether you move the load directly or through the pulleys.

  W = Fd . . . . . . . . work is the product of force and distance

  F(10 ft) = (0.25F)(d) . . . . . where d is the distance we want to find

  d = 10F/(0.25F) = 40

The rope will need to move 40 feet.

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3 years ago
Air is compressed by an adiabatic compressor from 100 kPa and 20°C to 1.8 MPa and 400°C. Air enters the compressor through a 0.1
Tamiku [17]

Answer:

(a) the mass flow rate of air is 5.351 kg/s

(b) the input power required is 2090.786 kW

Explanation:

Given;

initial pressure, P₁ = 100 kPa

initial temperature, T₁ = 20 °C

Final pressure, P₂ = 1.8 MPa

Final temperature, T₂ = 400 °C

Inlet area of the compressor = 0.15 m²

outlet area of compressor = 0.078 m²

velocity of air = 30 m/s

Part (a) mass flow rate of air through the inlet

Mass flow rate = Area x velocity = density x volumetric rate

m = Av = ρV

from ideal gas law, PV = nRT and ρ = m/V

substitute these values in the above equations, we will have;

m = \frac{PAv}{RT}

m = \frac{100000*0.15*30}{287*(20+273)}= 5.351 \ kg/s

Part (b) the required power input

W + m(h_1+\frac{v_1{^2}}{2}) = mh_2

where;

W is the input power

m is the mass flow rate

h₁ is the initial enthalpy

h₂ is the final enthalpy

initial and final enthalpy are obtained from steam table using interpolation;

h₁ = 293.166 kJ

h₂ = 684.344 kJ

W + m(h_1+\frac{v_1{^2}}{2}) = mh_2\\\\W = mh_2 - m(h_1+\frac{v_1{^2}}{2})\\\\W = 5.351 ( 684.344) - 5.351 (293.166 + \frac{30^2}{2000}) \\\\W = 3661.925 \ kW -1571.139 \ kW\\\\W = 2090.786 \ kW

7 0
2 years ago
Read 2 more answers
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