Problem 8--circular motion and projectile motion: tarzan (mass 80 kg) swings at the end of an 8.0-m long vine. when directly under the vines support, he releases the vine and files across a swamp. when he releases the vine, he is 5.0 m above the swamp and 10.0 m horizontally from the other side. determine the force the vine exerts on him at the instant before he lets go (the vine is straight down when he lets go).

1000 N

Explanation:

Let v be the his horizontal speed when he releases the vine and t be the time it takes for him to fly across the swamp and land.

Assume

When he releases the vine, there's only gravity acting on him downward. There's also no vertical velocity at this point.

Therefore, for him to travel 5 m vertically:

Horizontally we have

For Tarzan to achieve this horizontal speed, the centripetal acceleration and force exerting on the vine at the instant before he let go must be

  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