A 50.0 kg child is riding in a 120.0 kg bumper car at an amusement park. The child drives the bumper car at 4.5 m/s head-on into the wall of the ride and rebounds at the speed. What is the magnitude of the change in momentum of the child in the bumper car?

Explanation:

Total mass of child and car = 50 + 120 = 170 kg .

initial velocity u = 4.5 m/s

initial momentum = mass x velocity = 170 x 4.5 = 765 kg m/s

final velocity = - 4.5 m/s [ - ve is due to reversal of direction ]

final momentum = - 765 kg m/s

change in momentum = final momentum - initial momentum

= - 765 - ( 765 )

= - 1530 kg m/s .

  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

0.863333

86%

hope this