An optical disk drive in a computer can spin a disk at up to 10,000 rpm. If a particular disk is spun at 5050 rpm while it is being read, and then is allowed to come to rest over 0.435 s, what is the magnitude of the average angular acceleration of the disk?

α = 1215.71 rad/s²

Explanation:

The angular acceleration of an object is defined as the time rate of change of angular velocity of he object. The formula for the angular acceleration of an object is given as follows:

α = (ωf - ωi)/Δt

where,

α = angular acceleration of the disk = ?

ωf = Final Angular Velocity = 0 rad/s (Since disk finally stops)

ωi = Initial Angular Velocity = (5050 rpm)(2π rad/rev)(1 min/60 s)

ωi = 528.83 rad/s

Δt = time interval = 0.435 s

Therefore,

α = (0 rad/s - 528.83 rad/s)/(0.435 s)

α = - 1215.71 rad/s²

here, negative sign indicates that the direction of acceleration is opposite to the direction of angular velocity or the angular motion. So, the magnitude of acceleration will be:

α = 1215.71 rad/s²

  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