Aconcave mirror has a focal length of 30.0 cm. the distance between an object and its image is 45.0 cm. find the object and image dis- tances, assuming that (a) the object lies beyond the center of curvature and (b) the object lies between the focal point and the mirror.

object and image distances when object lies beyond center of curvature is 90 cm

object and image distances when object lies between focal point and mirror is 15 cm

Explanation:

given data

focal length f = 30.0 cm

distance between object and image = 45.0 cm

to find out

object and image distances

solution

we will apply here lens formula that is

(1/v) + (1/u) = (1/f)   1

and here v - u = 45

so v = 45 + u       2

so from equation 1

(1/(45+u)) + (1/u) = (1/30)

solve it we get u

u = 45 cm or -30 cm

so from equation 2

v = 45 + u

v = 45 + 45

v = 90 cm

object and image distances when object lies beyond center of curvature is 90 cm

and

v = 45 -30

v = 15 cm

object and image distances when object lies between focal point and mirror is 15 cm

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

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.