An upright image which reduced in size 10 times occurred in a mirror. If the radius of curvature of the mirror is 2 m, bow far is the object from the mirror?
How to solve?​

p = -9 m

Explanation:

For this exercise we use the equation of the geometric optics constructor

where f is the focal length, p and q are the distance to the object and the image, respectively.

The mirrors the focal length is

         f = R / 2

         f = 2/2

         f = 1 m

the magnification is

         m =

indicates that the image was reduced h ’= h/10 implies that m = 1/10  

we write our system of equations

            p = -10q

            1/1 =

we substitute

           1 =

           1 = 1/p  (1 - 10)

           1 = -9 / p

           p = -9 m

c)as speed changes, wavelength and frequency change.

everything changes!

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow