Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved mirror than the centre of curvature.</em>
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During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:


Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore





The value of gravity is canceled because it is a constant



The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%
You can see the Stud Multipliers right away in your Holoprojector menu under the Extras tab.
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
The horizontal velocity was constant, so:



it traveled 90meters