Suppose two skiers (A and B) are racing. Assume a frictionless surface! They start from the top of a mountain at the same time, and glide down to the flat area below. Just before the finish line there is a ditch. The skiers can either go down into the ditch or take a flat bridge over the ditch. Both the bridge and the ditch are covered with frictionless snow. Skier A decides to go down into the ditch. Skier B decides to go over the bridge. . Which skier gets to the finish line first, or do they arrive at the same time? a. Skier A (ditch) arrives first
b. Skier B (bridge) arrives first
c. The skiers arrive at the same time
d. Neither skier arrives at the finish line

b. Skier B (bridge) arrives first

Explanation:

This is because, skier B continues along the bring with the same velocity he started with before moving over the bridge and since the bridge is frictionless, he losses no kinetic energy and his speed is constant.

Whereas, skier A losses kinetic energy as he goes into the ditch. This is due to his change in potential energy. He thus emerges from the ditch with lesser kinetic energy than skier B and thus a slower speed.

Therefore, skier B arrives first since he moves at a constant speed.

  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