Answer : The crystal structure of Niobium is, BCC (Z=2)
Explanation :
Nearest neighbor distance, r =
Atomic mass of niobium (Nb) = 92.91 g/mole
Avogadro's number
First we have to calculate the cubing of edge length of unit cell for BCC and FCC crystal lattice.
For BCC lattice :
For FCC lattice :
Now we have to calculate the density of unit cell for BCC and FCC crystal lattice.
Formula used :
.............(1)
where,
= density
Z = number of atom in unit cell (for BCC = 2, for FCC = 4)
M = atomic mass
= Avogadro's number
a = edge length of unit cell
Now put all the values in above formula (1), we get
From this information we conclude that, the given density is approximately equal to the density of BCC unit lattice.
Therefore, the crystal structure of Niobium is, BCC (Z=2)
speed increases when u increase torque
Answer:
a)f=2.25 Hz
b)Time period T=.144 s
c)tex]V_{max}[/tex]=0.42 m/s
d)Phase angle Ф=87.3°
e)
Explanation:
a)
Natural frequency
=14.14 rad/s
w=2πf
⇒f=2.25 Hz
b) Time period
T=
Time period T=.144 s
c)Displacement equation
Boundary condition
t=o,x=0.03 m
t=0,v=.02m/s , V=
Now by using these above conditions
A=0.03,B=0.0014
x=0.03 cos14.14 t+0.0014 sin14.14 t
⇒x=0.03003sin(14.14t+87.3)
=0.42 m/s
d)
Phase angle Ф=87.3°
e)
Maximum acceleration
=6.0041
Answer:
Can you give me a brainlest?? :(
Explanation: