You throw a bowling ball of mass M and radius 1m has an initial speed v0 down the lane. It has an initial angular velocity of 3v0/R just after its release. The coefficient of kinetic friction is 0.08. It begins to roll without slipping at t=2s. What is the speed of the ball just as it begins rolling without slipping?

Explanation:

Given

Initial angular velocity

Initial speed

Coefficient of kinetic friction

As No external torque is applied therefore angular momentum is conserved

For rolling without slipping

acceleration provided by surface

Speed of ball after

using

v=final velocity

u=initial velocity

thus

  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

im guessing the west of station b