On an alien planet, a simple pendulum of length 0.300 m has oscillation frequency 0.709 Hz. Find the period of the pendulum. Find the acceleration due to gravity on this planet's surface. What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass 0.450 kg attached to a spring of force constant 9.75 N/m?

1.41s

5.95m/s

0.2746m

Explanation:

The time period

T = 1/f

= 1/0.709s

= 1.41 seconds

We have

T = 2π√l/g

T² = 4π²l/g

g = 4π²l/T²

g = 4x3.14²x0.300/1.41²

g = 5.95m/s² this is the acceleration due to gravity.

Then the time period of the glide

T2 = 2π√m/k

Length of pendulum = l

Time period T

T2 = 2π√l/g

Then T1 = T2

2π√m/k = 2π√l/g

M/k = l/g

L = g.m/k

L = 5.95x0.450/9.75

L = 0.2746

This must be the length of the simple pendulum

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