An electric fan is turned off, and its angular velocity decreases uniformly from 550 rev/min to 180 rev/min in a time interval of length 4.30 s.
a.) find the angular acceleration in revolutions per second per second.
= -1.43 rev/s^2
b.) find the number of revolutions made by the fan blades during the time that they are slowing down in part a.
= rev
c.) how many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part a?
= s

(a)

(b) 4.17 rev

(c) 6.40 sec

Explanation:

We have given the angular velocity of fan decreases from 550 rev/min to 180 rev/min

So initial angular speed

Final angular speed

Time is given as t = 4.3 sec

(a) We know that angular acceleration is given by

(b) We know the relation \Theta =\omega _0t+\frac{1}{2}\alpha t^2=9.166\times 4.3-\frac{1}{2}\times 1.43\times 4.3^2=26.1935rad=\frac{26.1935}{2\pi }=4.17rev

(c) Now final angular velocity

So time

  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