At the top of a hill that is 4.37 meters high, a 70 kg skate boarder has 3000 1 point joules of potential energy. What is his speed at the bottom of the hill where
his potential energy is zero?

9.26m/s

Explanation:

Given parameters:

Height of the hill  = 4.37m

Mass of skate boarder = 70kg

Potential energy  = 3000J

Unknown:

Speed at the bottom of the hill  = ?

Solution:

At the bottom of the hill, the total potential energy is the same as the kinetic energy.

          Potential energy  = Kinetic energy

                  mgh  =  mv²  

                       2mgh  = mv²

                           v² = 2gh

                           v = √2gh

 g is the acceleration due to gravity

h is the height

                     v  = √2 x 9.8 x 4.37  = 9.26m/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