Consult concept simulation 6.1 in preparation for this problem. a golf club strikes a 0.039-kg golf ball in order to launch it from the tee. for simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 7810 n, and is in contact with the ball for a distance of 0.013 m. with what speed does the ball leave the club?

72.15723 m/s

Explanation:

F = Force = 7810

s = Displacement = 0.013 m

m = Mass of club = 0.039

v = Velocity

Here work done will be the change in kinetic energy

The speed at which the ball leaves the club is 72.15723 m/s

it is the second option.

  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