Aman located at a point a on one bank of a river that is 1000 ft wide observed a woman jogging on the opposite bank. when the jogger was first spotted, the angle between the river bank and the man's line of sight was 25°. one minute later, the angle was 35°. how fast was the woman running if she maintained a constant speed? (round your answer to the nearest integer.)

Refer the given figure

Here A is the position of observer and B1 is the first position of lady and B2 is the position after one minute.

The distance traveled by lady = B1B2 = AC1 - AC2

We have AC1 = 1000 = 2144.51 m

               AC2 = 1000 = 1428.15 m

The distance traveled by lady = B1B2 = AC1 - AC2 = 2144.51 - 1428.15 = 716.36 m

So velocity of lady = distance / time = 716.36/60 = 11.94 m/s ≈ 12 m/s

  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