A grinding wheel, initially at rest, is ro-tated with constant angular acceleration of2.52 rad/s2for 7.03 s. The wheel is thenbrought to rest with uniform deceleration in11.9 rev. Find the angular acceleration required tobring the wheel to rest. Note that an in-crease in angular velocity is consistent with apositive angular acceleration. Answer in units of rad/s2.

Starting from rest, the wheel accelerates to an angular velocity of

ω = (2.52 rad/s²) (7.03 s) ≈ 17.7 rad/s

then undergoes a new acceleration α until it comes to a rest. It does so in 11.9 revolutions, or with an angular displacement of (11.9 rev) • (2π rad/rev) = 23.8π rad. So α satisfies

0² - ω² = 2 α (23.8π rad)

α = - ω² / (47.6π rad)

α ≈ -2.10 rad/s²

neither accurate nor precise.

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by