Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement
so when we wanted to move a 100kg a distance of 1m
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m
by the same way:-
------------------------
work in = 100kg * 2m * g (m/
)= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m
That depends on what quantity is graphed.
It also depends on what kind of acceleration is taking place ...
continuous change of speed or continuous change of direction.
-- If the graph shows speed vs time, and the acceleration is a change
in speed, then the graph is a connected series of straight-line pieces.
Each straight piece slopes up if speed is increasing, or down if speed
is decreasing.
-- If the graph shows speed vs time, and the acceleration is a change in
direction only, then the graph is a straight horizontal line, since speed is
constant.
-- If the graph shows direction vs time, and the acceleration is a change
in speed only, then the graph is a straight horizontal line, since direction
is constant.
-- If the graph shows direction vs time, and the acceleration is a change
in direction, then the graph is a connected series of pieces of line.
Each piece may be straight if the direction is changing at a constant rate,
or curved if the direction is changing at a rate which grows or shrinks.
Each piece may slope up if the angle that defines the direction is growing,
or may slope down if the angle that defines the direction is decreasing.
-- If the graph shows distance vs time, and the acceleration is a
change in speed, then the graph is a connected series of pieces
of curves. Each piece curves up if speed is increasing, or down if
speed is decreasing.
-- If the graph shows distance vs time, and the acceleration is a change
in direction only, then the graph is a straight line sloping up, since speed
is constant.
Answer:
It can occur only when light is incident on an interface where the index of refraction on the other side is less.
Explanation:
When the light passes from a denser medium, with refractive index n1, to another less dense medium, with refractive index n2, the incident light beam is refracted in such a way that it is not able to cross the surface between both media, the light beam is fully reflected and completely confining in the optically denser medium through which it propagates. For this phenomenon to occur, it is necessary that the angle of the incident light beam with respect to the normal be greater than or equal to the critical incidence angle θc. The critical angle can be calculated as :
The answer would probably be B.
