Three balls with the same radius 21 cm are in water. Ball 1 floats, with
half of it exposed above the water level. Ball 2, with a density 893
kg/m3 is held below the surface by a cord anchored to the bottom of the
container, so that it is fully submerged. Ball 3, of density 1320 kg/m3, is
suspended from a rope so that it is fully submerged. Assume the
density of water is 1000 kg/m3 in this problem.
A. Which is true for Ball 1?
B. What is the tension on the rope holding the second ball, in newtons?
C. What is the tension on the rope holding the third ball in N?

Explanation:

A )

The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³

B )

Tension in the ball will be equal to net force acting on the ball

Net force on the ball = buoyant force - weight .

4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )

= 40.65 x 10⁻⁶ N .

C )Tension in the 3 rd  ball will be equal to net force acting on the ball

Net force on the ball =  weight  - buoyant force

= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000  )

=  121.6  x 10⁻⁶ N .

  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