Question

A 6.0 cm object is located 20 cm in front of
a convex mirror with a radius of curvature
equal to 120 cm. What is the size
of the image produced by the mirror?

Answer 1

@IrishBoy123

Answer 2

i get 0.45 and u ?

Answer 3

@IrishBoy123

Answer 4

4.5

Answer 5

how did u do it?

Answer 6

how did you do it?

you first....

Answer 7

k i found -1.5 for di

Answer 8

in the end it did (1.5)(6)/20

Answer 9

to get 0.45

Answer 10

what is f in this question??!!

Answer 11

\(R_c = 120\)

Answer 12

so f = ??

Answer 13

60

Answer 14

-60 since it convex

Answer 15

seems we are in agreement - ish (___), cool.

for me:

\(\large \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)

\(f = -60\)

\(d_o = 20 \implies d_i = -15\)

\( \large \frac{h_i}{h_0} = -\frac{d_i}{d_o} \)

\(h_o = 6\)

follows that \(h_i = 4.5\)

Answer 16

-1/60-1/20=1/di???

Answer 17

@IrishBoy123

Answer 18

-1/60 = 1/20 +1/di

Answer 19

yea so when the 1/20 come to the other side it becomes negative

Answer 20

ohh wait its 1200 not 120

Answer 21

omg

Answer 22

these question are really easy just these silly mistakes ruin them

Answer 23

btw do u come to site often ? i will be needing lots of help in the coming days since i have excam

Answer 24

\(\large \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)

\(f = - 60\)

\(\large -\frac{1}{60} = \frac{1}{20} + \frac{1}{d_i}\)

\(\large \frac{1}{d_i} = -\frac{1}{60} - \frac{1}{20} = -\frac{1}{20} (\frac{1}{3} + 1) = -\frac{1}{15}\)

\(\large d_i = -15\)

\(\large \frac{h_i}{h_o} = -\frac{d_i}{d_o}\)

etc etc

Answer 25

plus, for "lots of help" on Physics, be sure to tag @Michele_Laino

Qualified Helper, and brilliant in every sense of the word

Answer 26

thanks!! @IrishBoy123