A car goes around a circular track at 30 m/s. If the radius of the curve is 90 m, what is the period of the car's revolution around the track?

18.9s

Explanation:

Using the formula;

ω = v/r

Where;

ω = angular velocity (rad/s)

v = linear velocity (m/s)

r = radius of the circular track (m)

According to the given information, v = 30m/s, r = 90m

ω = v/r

ω = 30/90

ω = 3/9

ω = 0.3333 radians/seconds.

Since ω = 2π/T

Where;

π = 3.142

T = period (s)

ω = angular velocity

0.333 = 2 × 3.142/T

T = 2 × 3.142/0.333

T = 6.284/0.333

T = 18.87s

T = 18.9s

b) phosphorus pentoxide

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