A object 1.5cm high produces a real image 2cm high. placed at a distance of 20cm from a concave mirror calculate: [a] the position of the image [b] focal lenght of the concave mirror

a. 26.7 cm. b. 11.4 cm.

Explanation:

a. We know h'/h = d'/d where h' = image height = + 2 cm (since it is a real image), h = object height = + 1.5 cm, d' = image distance from mirror and d = object distance from mirror = 20 cm

So, from h'/h = d'/d

d = h'd/h

= 2 cm × 20 cm/1.5 cm

= 40/1.5 cm

=  26.67 cm

≅ 26.7 cm

The position of the image is 26.7 cm from the mirror

b. Using the mirror formula

1/d + 1/d' = 1/f where d = object distance from mirror = + 20 cm, d' = image distance from mirror = + 26.7 cm (its positive since its a real image) and f = focal length of mirror.

So, 1/d + 1/d' = 1/f

⇒ f = dd'/(d + d')

= 20 cm × 26.7 cm/(20 cm + 26.7 cm)

= 534/46.7

= 11.43 cm

≅ 11.4 cm

The focal length of the mirror is 11.4 cm

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr