if r varies partly as the square of v and partly as the cube of v.find the relationship of r and v
r = k1v^2 + 2v^3
make that k2
If we assume that "r" varies partly as the square of "v" and partly as the cube of "v," we can express this relationship mathematically.
Let's break it down into two parts:
1. Partly as the square of "v":
We can represent this relationship as "r" is directly proportional to the square of "v." Mathematically, it can be written as:
r ∝ v^2
2. Partly as the cube of "v":
We can represent this relationship as "r" is directly proportional to the cube of "v." Mathematically, it can be written as:
r ∝ v^3
To combine both relationships, we need to take both proportions into account. Since "r" contains two different proportionalities, we can express it as:
r ∝ v^2 * v^3
Simplifying the above equation, we get:
r ∝ v^(2 + 3)
r ∝ v^5
Therefore, the relationship between "r" and "v" is that "r" varies directly with the fifth power of "v." Mathematically, it can be written as:
r = k * v^5
where "k" is a constant of proportionality.
To find the relationship between r and v, we need to understand what it means for r to vary partly as the square of v and partly as the cube of v.
Let's denote the relationship as follows:
r = k(v^2)(v^3)
Here, k is a constant of proportionality that relates the magnitudes of r and v. By combining the exponents, we can simplify the equation:
r = k(v^5)
This equation shows that r varies directly with the fifth power of v. Thus, the relationship between r and v is that r is directly proportional to the fifth power of v.