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:


For this case we have that by definition:

Where,
- <em>m: mass of the object
</em>
- <em>a: acceleration of the object
</em>
- <em>F: summation of forces
</em>
We have then:

Then, by clearing the acceleration we have:

Substituting values we have:

Answer:
The acceleration of the box is equal to:

Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Answer:
a)693.821N/m
b)17.5g
Explanation:
We the Period T we can find the constant k,
That is

squaring on both sides,

where,
M=hanging mass, m = spring mass,
k =spring constant
T =time period
a) So for the equation we can compare, that is,

the hanging mass M is x here, so comparing the equation we know that

b) In order to find the mass of the spring we make similar process, so comparing,
