<span>There is an low cost and quickest alternative available for adaptive optics. Name of this technique is wavefront coding. The numerical analysis pretends to show the robustness of the technique under changes in pupil diameter and wavefront shape including intersubject and intrasubject variability, using always the same restoration filter or image decoder .Using this technique it is possible to obtain high resolution images under different ocular aberrations and pupil diameters with the same decoder, opening the possibility of real time high resolution images.</span>
Answer:
1
Explanation:
Its 1 since the shape of it is more downwards which results the reflection going downwards.
Other than for the chemical symbol, the electron dot diagram for silicon would be the same as it was for carbon.
The reason for this is because electron dot diagrams are used to represent the electrons in the outermost, or valence, shell of an atom. In a group of the periodic table, all of the elements have the same number of valence shell electrons. This means that all elements belonging to the same group have the same electron dot diagram, except for the symbol of the element that is within the diagram.
For a car that is pulled by a cable we can say that there are two forces on it
1. force of gravity downwards
2. tension force due to cable upwards
so here if upward force due to cable is more than the weight of the car then we have net upward force on the car
Due to this force the velocity of car will increase with time because here car is also moving upwards and we know that when net force will act in the direction of velocity of car then the velocity will always increase.
So here we can say that velocity of car will increase in this case
Answer:
a) F1 = 1999.8 N
, F2 = 4545 N
, F3 = 2778 N
, c) the cans do not collapse because the pressure is applied on both sides
Explanation:
Let's use the pressure equation
P = F / A
Suppose we have atmospheric pressure 1.01 10⁵ Pa
Let's calculate the area of the can that is a parallelepiped
Length L = 25 cm
width a = 18 cm
high h = 11 cm
Side area A = h a
A = 11 18
A1 = 198 10⁻⁴ m²
Lid area
A2 = L a
A2 = 25 18
A2 = 450 10⁻⁴ m²
Other side area
A3 = L h
A3 = 25 11
A3 = 275 10⁻⁴ m²
Now let's calculate the force on these sides
Side 1
F1 = P * A1
F1 = 1.01 10⁵ 198 10⁻⁴
F1 = 1999.8 N
Side 2
F2 = P A2
F2 = 1.01 10⁵ 450 10⁻⁴
F2 = 4545 N
Side 3
F3 P A3
F3 = 1.01 10⁵ 275 10⁻⁴
F3 = 2778 N
We see that the force is greater on side 2 which is where the can should collapse
b) To compare the previous forces we must use the concept of density, in general the cans are made of aluminum that has a density of 2700 kg / m3
d = m / V
m = d * V
V = L a h
V = 0.25 0.18 0.11
V = 0.00495 m3
m = 2700 0.00495
m = 13.4 kg
This is the maximum weight, because much of the volume we calculate is air that has a much lower density
W = 13.4 * 9.8
W = 131.3 N
Let's make the comparison by saying the two magnitudes
Side 1
F1 / W = 1999.8 / 131.3
F1 / W = 15.2
Side 2
F2 / W = 4545 / 131.3
F2 / W = 34.6
Side 3
F3 / W = 2778 / 131.3
F3 / W = 21.2
c) the cans do not collapse because the pressure is applied on both sides: outside and inside, so the net force is zero on each side.
