A24 cm candle is placed 6 m in front of a thin diverging lens with a focal length magnitude of 3 m. find the image's height and classify it as real or virtual, upright or inverted, enlarged or reduced.

Image is virtual and formed on the same side as the object. 2 m from the lens.

The size of the image is 7.97 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance =  6 m

v = Image distance

f = Focal length = -3 m (concave lens)

= Object height = 24 cm

Lens Equation

Image is virtual and formed on the same side as the object. 2 m from the lens.

Magnification

The size of the image is 7.97 cm

Image is upright as the magnification is positive and smaller than the object.

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

  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