A race car accelerates from rest to a velocity of +90 m/s over a distance of 423m. Determine the acceleration of the race car.

9.57m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity  = 90m/s

Distance  = 423m

Unknown:

Acceleration of the race car  = ?

Solution:

To solve this problem, we should apply one of the appropriate motion equations;

      V²  = U²   + 2as

Where V is the final velocity

           U is the initial velocity

           a is the acceleration

           s is the distance

  90²  = 0²  + 2 x a x 423

  8100 = 846a

      a  = 9.57m/s²

  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

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