A Micro –Hydro turbine generator rotor is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity of 837 rpm. The radius of the rotor is 0.62 m and its rotational acceleration is 5.9 rad/s2 . What is the rotor’s angular velocity (in rad/s) after 3.2 s?

Explanation:

Given:

wo = 610 rpm

To rad/s,

= 610 rpm ×2pi rad/60 s

= 63.88 rad/s

wf = 837 rpm

= 87.65 rad/s

ao = 5.9 rad/s^2

Using equations of circular motion,

wf = wo + aot

87.65 = 63.88 + 5.9×t

t = 4.025 s

Using the same equation,

At t = 3.2s (first 3.2 s of the motion),

wf = 63.88 + 5.9 × 3.2

= 82.76 rad/s

= 82.8 rad/s

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures