Arotating wheel has a constant angular acceleration. it has an angular velocity of 5.0 rad/s at time t = 0 s, and 3.0 s later has an angular velocity of 9.0 rad/s. what is the angular displacement of the wheel during the 3.0-s interval?

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 5 rad/s

     Final velocity, v = 9 rad/s    

     Time, t = 3 s

     Substituting

                      v = u + at  

                      9 = 5 + a x 3

                      a = 1.33 rad/s²

     Angular acceleration is 1.33 rad/s²

Now we have equation of motion v² = u² + 2as

     Initial velocity, u = 5 rad/s

     Final velocity, v = 9 rad/s    

     Angular acceleration = 1.33 rad/s²

Substituting  

v² = u² + 2as

9² = 5² + 2 x 1.33 x s

s = 21 radians

Angular displacement during 3 seconds is 21 rad

  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