Answer:
The distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Explanation:
From the image uploaded, a Face centered cubic structure (100) plane, there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell.
In terms of the atomic radius, R, we determine the distance between the centers of adjacent atoms.
Let this distance = AC
the two adjacent sides = AB and BC
AB = a = 2R
BC = a = 2R
Using Pythagoras theorem
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = √2a²
AC = a√2
But a = 2R
AC = 2R√2
Therefore, the distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Answer:
its a view point for auto cad
Explanation:
from my knowlege in IED we learned about it as a way of sing how an object would look in inventor or auto CAD
Answer:
i)ω=3600 rad/s
ii)V=7059.44 m/s
iii)F=1245.8 N
Explanation:
i)
We know that angular speed given as
We know that for one revolution
θ=2π
Given that time t= 2 hr
So
ω=θ/t
ω=2π/2 = π rad/hr
ω=3600 rad/s
ii)
Average speed V
Where M is the mass of earth.
R is the distance
G is the constant.
Now by putting the values
V=7059.44 m/s
iii)
We know that centripetal fore given as
Here given that m= 200 kg
R= 8000 km
so now by putting the values
F=1245.8 N
Answer:
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Answer:
Tech A is correct
Explanation:
Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.
Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.
