Two people are playing tug-of-war. Due to their choice of footwear, theircoefficient of static friction is different. Participant 1 has a mass of 60 kg, acoefficient of static friction of 2.0, and can pull with a maximum force of1000 N. Participant 2 has a mass of 80 kg and a coefficient of staticfriction of 1.2, and can pull with a maximum force of 1200 N. Who wins?

Participant 1 wins

Explanation:

Coefficient of static, μ = F/N, where F is the frictional force and N is the normal force.

The force exerted by each participant is the sum of the frictional force acting on each of them and the maximum force with which each participant pulls on the rope.

Frictional force, F =  μ * N

Normal force, N = mass * acceleration due to gravity, g

For Participant 1; μ = 2.0, mass = 60 kg, g = 9.8 m/s²

Frictional force = 2.0 * 60 * 9.8 = 1176 N

Total force = (1176 + 1000) = 2176 N

For Participant 2; μ = 1.2, mass = 80 kg, g = 9.8 m/s²

Frictional force = 1.2 * 80 * 9.8 = 940.8N

Total force = (940.8 + 1200) N = 2140.8 N

From the values obtained above, Participant 1 exerts more force than Participant 2, therefore, Participant 1 wins

  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

real, smaller and upside-down.   so the answer should be b.