An object 9.00 mm tall is placed 12.0 cm to the left of the vertex of a convex mirror whose radius of curvature has a magnitude of 20.0 cm.(a) calculate the position, size, orientation (erect or inverted), and nature (real or virtual) of the image. (b) draw a principal-ray diagram showing the formation of the image (how would i do that.. i know you can't draw it here)..

Explanation:

a )

for convex mirror, focal length

f = + 20 / 2 = 10 cm

object distance u = 12 cm ( negative )

Using mirror formula

Putting the values

v = 5.45 cm

Size of image

=

= 4.08 mm .

It will be erect because only erect image is formed in convex mirror .

It will be virtual .

b )

Draw ray from tip of object parallel to principal axis and after reflection appears to be coming from  focus . Second ray falls on the pole and reflects back making equal angle with principal axis . Draw the reflected ray backwards to cut the other ray . They cut each other where image is formed .

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.