Answer:
(a) A = 
(b) 
(c) 
(d) 
Solution:
As per the question:
Radius of atom, r = 1.95 
Now,
(a) For a simple cubic lattice, lattice constant A:
A = 2r
A = 
(b) For body centered cubic lattice:
(c) For face centered cubic lattice:
(d) For diamond lattice:
Answer:
b = 0.6487 kg / s
Explanation:
In an oscillatory motion, friction is proportional to speed,
fr = - b v
where b is the coefficient of friction
when solving the equation the angular velocity has the form
w² = k / m - (b / 2m)²
In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction
let's call
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Let's find the angular velocities
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
we subtitute
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
When the metals touch
together, half the charge of the charged metal flows to the other because the
electrons all repel each other. Therefore this also means that each metal ball
contains the same amount of electrons. Each ball has 5^10 electrons, this is
equivalent to a total charge of:
Q1 = Q2 = (1.602 * 10^-19
coulombs / electron) 5^10 electrons = Q
Q = 1.564 * 10^-12 C
Now using the Coulombs
law to find for the electric force:
F = k q1 q2 / r^2 = k (Q)^2
/ r^2
where k is a contant = 9
* 10^9 N m^2 / C^2
r = the distance of the
two metals = 0.2 m
So,
F = (9 * 10^9 N m^2 /
C^2) (1.564 * 10^-12 C)^2 / (0.2 m)^2
F = 5.51 * 10^-13 N
Since the two metals
repel therefore they are the one which exerts the force hence the magnitude
must be negative:
<span>F = - 5.51 * 10^-13 N</span>
