The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:

Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation.

is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.

Our final equation is:

Colatitude is:

The answer is:
Question:
The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.
Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.
Answer:
The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³
Explanation:
Here, we have mass density of cloud = 2.0×10⁻²¹ g/cm^3
Density = Mass/Volume
Volume = Mass/Density = If the mass is 40 kg and the body is made up of 70% by mass of water, we have
28 kg water = 28000 g
Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.
Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.
The radius of the cylinder is equal to half the diameter:

The volume of the cylinder is given by:

where h is the heigth of the cylinder. Converting into meters,

And the density of the material will be given by the ratio between the mass and the volume:

Answer:
Cannot be determined from the given information
Explanation:
Given the following data;
Velocity = 24 m/s
Period = 3 seconds
To find the amplitude of the wave;
Mathematically, the amplitude of a wave is given by the formula;
x = Asin(ωt + ϕ)
Where;
x is displacement of the wave measured in meters.
A is the amplitude.
ω is the angular frequency measured in rad/s.
t is the time period measured in seconds.
ϕ is the phase angle.
Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.
However, the given parameters can be used to calculate the frequency and wavelength of the wave.