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:
Explanation:
A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

F is the force applied and x is the elongation of the spring
k is the spring constant.
negative sign indicates the change in direction from equilibrium position.
In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

Inserting this into Hooke's Law

computing it for x,

This is the model which will tell the length of the spring against change in the mass located in the pan.
Answer: 3.75 joules
Explanation:
Given that:
Mass of acorn = 0.300 kilograms
velocity = 5.oo m/s
Kinetic energy = ?
Since, kinetic energy is the energy possessed by a moving object, its value depends on the mass M and velocity V of the acorn.
Thus, Kinetic energy = 1/2 x mv^2
= 1/2 x 0.300kg x (5.00m/s)^2
= 0.5 x 0.3kg x (5.00m/s)^2
= 0.15 x (5.00m/s)^2
= 3.75 joules
Thus, the kinetic energy of the falling acorn is 3.75 joules
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
Density is defined as [mass] / [volume] .
The only choice listed with those physical dimensions is 'd' .