Answer: D. 5cm
Explanation:
Given the following :
Focal length (f) = - 6.0 cm
Height of object = 15.0cm
Distance of object from mirror (u) = 12.0cm
Height of image produced by the mirror =?
Firstly, we calculate the distance of the image from the mirror.
Using the mirror formula
1/f = 1/u + 1/v
1/v = 1/f - 1/u
1/v = 1/-6 - 1/12
1/v = - 1/6 - 1/12
1/v = (- 2 - 1) / 12
1/v = - 3 / 12
v = 12 / - 3
v = - 4
Using the relation :
(Image height / object height) = (- image distance / object distance)
Image height / 15 = - (-4) / 12
Image height / 15 = 4 / 12
Image height = (15 × 4) / 12
Image height = 60 / 12
Image height = 5cm
Answer:
36.125 J
Explanation:
The formula for kinetic energy is KE = .5(m)(v²).
Using the given information, mass = 1 g and v = 8.50. Plug that information into the equation. KE = .5(1)(8.50²) = 36.125 J.
Is always virtual (meaning that the light rays do not actually come from the image), upright, and of the same shape and size as the object it is reflecting. A virtual image is a copy of an object formed at the location from which the light rays appear to come.
Answer:
23.8 m
Explanation:
The distance travelled by the zebra can be calculated by using the equation:
where
u is the initial velocity
t is the time
a is the acceleration
For this zebra,
u = 0 since it starts from rest
is the acceleration
Substituting t = 5 s, we find the distance travelled by the zebra:
