The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. when steam is shut off, the friction of the bearings and the air brings the wheel to rest in 1.0 $h$. what is the magnitude of the constant angular acceleration of the wheel in $rev/min^2$? do not enter the units.

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t      ⇒     α = (ωfin - ωin) / t

⇒     α = (0 rev/min - 113 rev/min) / (60 min)

⇒     α = - 1.883 rev/min²

f=ma newton

f=100x2.5=250n

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg