Consider the mirror from the last question. an object 4cm tall stands 10cm in front of a converging mirror of focal length 5cm.

This question is incomplete, here is the complete question;

"Consider the mirror shown here. An object 4 cm tall stands 10 cm in front of a converging mirror of focal length 5 cm.

The image formed by this mirror is: (Select all that apply.)

real

virtual

enlarged

the same size

diminished

Inverted

erect"

Virtual, Upright and Enlarged

Explanation:

The formula for image formed by a converging mirror is given by;

------------------- eqn 1

where f = focal length = 5 cm

u = object height = 4 cm

v = image height = ?

substitute for u and f in eqn 1 to find v.

v = -20 cm

From the value of the image formed i.e -20 cm

The negative sign means the image is behind the mirror and this means image is;

1. Virtual

2. Upright

3. Enlarged

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2