Look at yourself in a plane mirror<span> and </span>you see<span> your </span>image<span> - it is upright. The </span>image <span>is located on the other side of the </span>mirror<span> since reflected rays diverge upon reflection; when </span>mirrors<span> produce </span>images<span> on the the opposite side of the </span>mirror, the images<span> are said to be virtual.</span>
Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
In the absence of a diagram the general answer to the roller coaster scenario is as follows. Energy is always conserved in nature and if we exclude the energy losses due to friction in the moving parts and surfaces of the roller coaster, the sum of kinetic and potential energy of the car will always equate to the same number, called the mechanical energy.
As the rollercoaster moves you can imagine energy flowing between kinetic and potential states. The highest potential energy will be the highest and slowest point on the track, this will also be the lowest kinetic energy state. Similarly, the highest kinetic energy will be the lowest and fastest point on the course, which is also the point of lowest potential energy.
True, They contain old stars and posses little gas or dust