? What is the difference between the Primitive cell and convectional cell
1 answer:
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Answer:
- This means that the integral of the square modulus over the space is dimensionless.
Explanation:
We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral
must be dimensionless, as represents a probability.
As the differentials has units of length
for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:
taking the square root this gives us :
Answer:
A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?
The focal length is of 10 cm, object distance is 30 cm and magnification is -2.
Explanation:
Given:
A concave mirror:
Radius of curvature of the mirror, as C = 20 cm
Object distance in-front of the mirror = 15 cm
a.
Focal length:
Focal length is half of the radius of curvature.
Focal length of the mirror = = 10 cm
According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.
Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.
b.
We have to find the object distance:
Formula to be use:
⇒
⇒ Plugging the values.
⇒
⇒
⇒
⇒
⇒
⇒
Image will be formed towards negative x-axis 30 cm away from the pole.
c.
Magnification (m) is the negative ratio of mage distance and object distance:
⇒
⇒
⇒
The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.
Answer: 230.50 m
Explanation:
We have the following information:
the barometric reading at the top of the building
the barometric reading at the bottom of the building
density of air
density of mercury
gravity
And we need to find the height of the building.
In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building and the perssure at the bottom
:
(1)
(2)
From (1): (3)
From (2): (4)
Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure :
(5)
(6)
On the other hand, we have a column of air with a cross-section area and the same height of the building, lets name it
.
As pressure is defined as the force exerted on a specific area
, we can write:
(7)
If we isolate we have:
(8)
Also, the force gravity exerts on this column of air (its weight) is:
(9)
Knowing the density of air is: (10)
where the volume of air can be written as: (11)
Substituting (1) in (10):
(12)
Isolating :
(13)
Substituting (13) in (9):
(14)
Matching (8) and (14)
(15)
Isolating :
(16)
Substituting the known and calculated values:
(17)
Finally:
This is the height of the building