Answer:
The orbital speed of this second satellite is 5195.16 m/s.
Explanation:
Given that,
Orbital radius of first satellite 
Orbital radius of second satellite 
Mass of first satellite 
Mass of second satellite 
Orbital speed of first satellite = 4800 m/s
We need to calculate the orbital speed of this second satellite
Using formula of orbital speed
From this relation,
Now, 
Put the value into the formula
Hence, The orbital speed of this second satellite is 5195.16 m/s.
Answer:
Saturated zone is area below the water table in which the soil is completely saturated with groundwater.
Explanation:
The saturated zone lies below the ground. It is mainly the lower zone of rock along with the water table where pore spaces are completely filled with water. Even the saturated zone is sometimes separated into 2 subzones: the phreatic zone and the capillary fringe.
The area where pores spaces are not saturated with water is also unsaturated zone. Localized saturated zones can occur within the unsaturated zone. The unsaturated zone lies above the groundwater table.
The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.
The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.
So;
Period(T) = 54 minutes or 3240 seconds
Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz
Learn more: brainly.com/question/14588679
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
