A painted tooth on a spinning gear has angular acceleration α=(20−t)rad/s2, where t is in s. Its initial angular velocity, at t = 0 s, is 320 rpm.

I think the question will be to find the angular speed at any time t

Since we are given the initial conditions of the angular speed

Explanation:

Given that,

α=(20−t)rad/s2

Give the initial condition

At t=0 w=320rpm

So modeling a differential equation

α=dw/dt

Therefore

dw/dt=(20−t)

Applying variable separation

dw=(20-t)dt

Integrate Both sides

∫dw=∫(20-t)dt

w = 20t-t²/2 + C

C is constant of integration

Using initial conditions

At t=0 w=320rpm

320=0-0+C

C=320rpm

Therefore the equation becomes

W=20t - t²/2 + 320

This is the model angular speed at any time t

d is the answer i came up with

  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