Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.2 m and is released with a speed of 12 m/s .

229.11 N

Explanation:

Mass of the shot (m) = 7.0 kg

Distance moved by the shot (S) = 2.2 m

Initial velocity of the shot (u) = 0 m/s (assume)

Final velocity of the shot (v) = 12 m/s

First we will determine the acceleration of the shot using the following equation of motion.

So, the acceleration of the shot is 32.73 m/s².

Now, using Newton's second law, the force acting on the shot is equal to the product of mass and acceleration. So,

Force = Mass × Acceleration

Therefore, the average force exerted by the shot-putter is 229.11 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