A 1.2 kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and bounces off with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 milliseconds, what is the magnitude of the average force on the object by the wall

Average force will be equal to 8400 N

Explanation:

We have give mass of the object m = 1.2 kg

Initial velocity of the object u = - 8 m/sec

And after collision to the wall final velocity of the object v = 6 m/sec

Change in momentum is equal to

Time is given t = 2 milliseconds = 0.002 sec

We know that force is equal to rate of change of momentum

So

So average force will be equal to 8400 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