Ahelicopter, initially hovering 40 feet above the ground, begins to gain altitude at a rate of 21 feet per second. which of the following functions represents the helicopter’s altitude above the ground y, in feet, t seconds after the helicopter begins to gain altitude?

y = 40 + 21t

Explanation:

Apply the equation of distance covered.

d = vt + C

Where d is the distance covered

v = velocity , t = time

C = constant = initial distance covered

For the case above

d = y

y(t) = vt + C

But y(0) =40 = C

C = 40ft

velocity v = 21 ft/s

Therefore, the equation of the altitude is given by;

y(t) = 21t +40

y = 40 + 21t

  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