The question is missing a diagram of the ray reflection. I attached a diagram which comes from a similar question in the answer section. The full question should be as follows:
Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cmfrom their point of intersection, as shown in the figure. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long) after reflecting from the first mirror?
Answer:
34.6°
Explanation:
To strike the midpoint of the second mirror, the ray light will have to travel half of the distance vertically
i.e. 29/2 = 14.5
We can solve this through trigonometry.
Let the angle between the ray and the vertical plane mirror is known as α
tan α = 10/14.5
α = = 34.6°
The angle of incidence is the angle between the ray and the normal line of the mirror.
Let angle of incidence of first mirror be β
β = α = 34.6
An observable phenomenon that is commonly associated with a spinning object moving through a fluid.
Answer:
-4*10⁴ units.
Explanation:
As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must still remain to be zero.
So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.
The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.
Segment D ...
Walking AWAY from home; distance increases as time increases.
Segment B ...
Not walking; distance doesn't change as time increases.
Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.
Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.
I think it is c density and temperature