Old naval ships fired 14 kg cannon balls from a 250 kg cannon. it was very important to stop the recoil of the cannon, since otherwise the heavy cannon would go careening across the deck of the ship. in one design, a large spring with spring constant 2.0×10^4 n/m was placed behind the cannon. the other end of the spring braced against a post that was firmly anchored to the ship's frame.

what was the speed (in m/s) of the cannon ball if the spring compressed 56 cm when the cannon was fired?

Speed of canon ball = V = 89.443 m/s

Explanation:

the spring potential energy will be converted into kinectic energy

M= mass of cannon

m= mass of cannon balls

k= 2.0*10^4 N/m

compression x= 0.56 m

0.5kx^2 = 0.5Mv^2

2 x 10^4 x 0.56^2 = 250 x v^2

Speed of canon = v = 5.008 m/s

Now,

we have, by momentum consevation

m x 0 + M x 0 = mV - Mv

0 = 14 x V - 250 x 5.008

Speed of canon ball = V = 89.443 m/s

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂