The part of a river that would have animals with muscular bodies and adaptations that let survive in turbulent water is in the transition zone, the mid-transition zone to be precise.
Water at the source zone possesses a lot of potential energy and as it flows from the upper reaches the potential energy is turned into kinetic energy when the course of the river begins to gradually level out and this translates into increase in velocity. By the time river water reaches the middle of the transition zone, most of the potential energy would have been turned into kinetic energy and thus water velocity would be quite high here.
Animals living here would develop muscles because of constantly fighting against the strong current to avoid being swept downstream.
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description.
From the definition we know that the speed of a body can be described as a function of gravity and height



Then applying the kinematic equation of displacement, the height can be written as

Re-arrange to find t,



Thus the calculation of the displacement would be subject to



Therefore the required distance must be 0.547m