<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>
First question:
The magnitude of current flowing in a circuit is described in
units of Amperes. The device used to measure it is an Ampmeter,
or Ammeter.
Second question:
This question is so absurd that it should not be dignified with an
answer. Although 'E' is often used as the symbol for Electromotive
force, potential difference, and voltage, there's certainly no rule.
Anyone is free to use 'M', 'Q', 'Θ', or 'Щ' to denote voltage when
they write electrical formulas, just as long as they make sure to
explain the meaning of whatever symbols they use.
Answer:

Explanation:
Given:
Gravity of Mars = 0.38 times the gravity at Earth
Gravity of Earth is, 
Radius of Mars (R) = 3400 km
Mass of mars (M) = ?
We know that, the acceleration due to gravity of a planet of mass 'M' and radius 'R' is given as:

Now, as per question:

Plug in 9.8 for
and solve for
. This gives,

Now, plug in this value in the above equation and solve for 'M'. This gives,

Therefore, the mass of Mars is
.
Answer:
The new energy is 1/16 of the original energy
Explanation:
A capacitor is an electric device which is able to store charge when it is connected to a power supply.
The energy stored in a capacitor is given by the equation

where:
Q is the charge stored on the capacitor
C is the capacitance of the capacitor
For the capacitor in this problem, initially we have
E = 7 J (energy) when the charge stored is Q and the capacitance is C
Later, the charge is decreased to 1/4 of its original value, so the new charge is

Since the capacitance remains the same, the new energy is

Therefore, the new energy is 1/16 of the original energy.
800 going upwards
..........