0 - 1" -20 -15 -10 5 0 1 2 3 0

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

uysha [10]

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.

8 0

3 years ago

An inventor claims to have developed a food freezer that at steady state requires a power input of 0.25 kW to extract energy by

Serga [27]

Answer:

investor claim is acceptable

Explanation:

given data

Win = 0.25 kW

Qc = 3000 J/s = 3kW

Th = 293 K

Tc = 270 K

solution

we get here coefficient of performance of cycle is

coefficient of performance = $\frac{Qc}{Win}$ ..................1

put here value and we get

coefficient of performance = $\frac{3}{0.25}$

coefficient of performance = 1.2

and

coefficient of performance of ideal refrigeration is

coefficient of performance = $\frac{Tc}{Th-Tc}$ ..................2

coefficient of performance = $\frac{270}{293-270}$

coefficient of performance = 11.74

and

we can see here that coefficient of performance of ideal refrigeration is is more than real cycle coefficient of performance

so investor claim is acceptable

6 0

2 years ago

Design roller chain drive. Specify the chain size, the sizes and number of teeth in the sprockets, the number of chain pitches,

Rom4ik [11]

Answer:

The total design can be summarized as follows:

AC motor 55 hp, 750 rpm as a driver

Service factor 1.0

Design power 5 hp

No. 40 chain, 0.50 in pitch, 1 strands

17 teeth, 2.72 in pitch dia, 1 strand small sprocket

39 teeth, 7.46 in pitch dia, 1 strand large sprocket

1 strand of length 54 in

Center distance of 19.9 in

Actual Output speed of 326.9 rpm

Type B lubrication

Explanation:

4 0

3 years ago

A thin flat plate of surface area, 500 cm², is located between two fixed plates such that the gap below the top plate is 20 mm a

kodGreya [7K]

Answer:0.3166N

Explanation:

Given data

Area $\left ( A\right )=500 cm^2$

Gap below top plate $\left ( y_1\right )=20 mm$

Gap above bottom plate $\left ( y_2\right )=30 mm$

SAE 30 oil viscosity = $0.38 N-s/m^2$

Velocity of middle plate $\left ( v\right )=1 m/s$

There will viscous force on middle plate i.e. at above surface and below surface

Viscous force $\left ( F\right )=\mu \frac{Av}{y}$

$Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ]$

$F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ]$

$F=31.66\times 10^{-2}=0.3166 N$

this is force by oil on plate thus we need to apply atleast 0.3166N to move plate

8 0

3 years ago

To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.8 times each persons weight - and t

Yuri [45]

Answer:

μ = 0.55

Explanation:

Given that

Normal weight = 1.8 x weight of person

N= 1.8 mg

We know that friction force Fr

Fr= μ N

μ=Coefficient of friction

N=Normal force

To find μ We have to equate friction and gravity force

Fr= Wt

μ N = m g

μ x 1.8 m g = m g

μ = 0.55

So the coefficient of friction will be 0.55.

5 0

3 years ago