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

Image distance = 3.33cm; image height = 1.33cm

Explanation:

The question is incomplete, according to the data given in the question, we can get the distance of image from the mirror and the image height.

First we must know that the focal length (f) of a converging mirror is negative and the image distance from the mirror (v) is negative since it produce virtual images only.

Using the mirror formula to get the image distance;

1/f = 1/u+1/v where;

f is the focal length

u is the object distance

v is the image distance

Given u = 10cm, f = -5cm(for converging mirror)

v = ?

1/v = 1/f-1/u

1/v = -1/5 - 1/10

1/v = -2-1/10

1/v = -3/10

-3v = 10

v = 10/-3

v = -3.33cm

This means the image is 3.33cm behind the mirror.

To get the image height, we will use the magnification formula

Magnification = image height(H1)/object height (H2) = Image distance (v)/Object distance (u)

Since M = H1/H2 = v/u

Therefore, H1/H2 = v/u... (1)

Given H1= 4cm(object 4cm tall)

H2 = ? v = 3.33cm u = 10cm

Substituting in eqn 1 to get H2, we have;

H2/4=3.33/10

10H2 = 3.33×4

10H2= 13.32

H2 = 13.32/10

H2 = 1.33cm

The image height will be 1.33cm

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

a displacement distance is size of displ