Alens forms an image of an object. the object is 16.0 cm from the lens. the image is 12.0 cm from the lens on the same side as the object.

(a) what is the focal length of the lens? is the lens converging or diverging?
(b) if the object is 8.50 mm tall, how tall is the image? is it erect or inverted?
(c) draw a principal-ray diagram.

(a) -48.0 cm, diverging

We can use the lens equation:

where

f is the focal length

p = 16.0 cm is the object distance

q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)

Solving for f, we find the focal length of the lens:

The lens is diverging, since the focal length is negative.

(b) 6.38 mm, erect

We can use the magnification equation:

where

y' is the size of the image

y = 8.50 mm is the size of the object

Substituting p and q that we used in the previous part of the problem, we find y':

and the image is erect, since the sign is positive.

(c)

See attached picture.

the answer is c.)

explanation:

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz