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

d) the number of neutrons plus the number of protons

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this