When It begins to drop because that when gravity will have its strongest pull on the object.
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance

Where, d = distance
A = amplitude
Put the value into the formula


Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Answer:
Explanation:
The time period of geosynchronous satellite must be equal to T .
The radius of its orbit will be ( R+ h )
orbital velocity V₀ = 
Time period T = 2π( R + h ) / V₀
= 2π( R + h ) x 
= R +h
h =
- R.
the more pressure put on the string, the more frequency and higher pitch.