A 136 kg defensive lineman is about to tackle a 100 kg quarterback. Right before impact, the lineman is moving north at 6.0 m/s and the the quarterback is moving north at 1.0 m/s. Determine the speed and direction of the two players immediately after the tackle.

Explanation:

We shall apply law of conservation of momentum .

Momentum before impact = momentum after impact

Since they are moving towards the same direction , their momentum will add up .

Total momentum before impact

= 136 x 6 + 100 x 1 = 916 kg m/s .

Total mass of lineman and quarterback = 236 kg , if their velocity after impact be v

their momentum = 236 v

According to law of conservation of momentum

236 v = 916

v = 3.88 m /s

Both of them will move with velocity of 3.88 towards the north .

  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