Ahot-air balloon is descending at a rate of 1.9 m/s when a passenger drops a camera. if the camera is 47 m above the ground when it is dropped, how much time does it take for the camera to reach the ground? if the camera is 47 m above the ground when it is dropped, what is its velocity just before it lands? let upward be the positive direction for this problem.

The camera lands in t=2.91s with a velocity:

Explanation:

The initial velocity of the camera is the same as the hot-air ballon:

Kinematics equation:

when the camera lands, y=0:

We solve this equation to find t:

t1=-3.29s       This solution have not sense in our physical point of view

t2=2.91s    

So, the camera lands in t=2.91s

We replace this value in v(t):

  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

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