A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have when he reaches the top?

8352,54 J

Explanation:

PE= mgh

m=90kg

h=9.47

g= 9.8 m/s2

PE= 90*9.47*9.8

PE= 8252,54 J

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.