The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain river normally flows at a speed of 2 miles per hour. What must its speed be in order to transport particles that are 15 times as massive as usual? Round your answer to the nearest hundredth.
m = ks^6
m/s^6 = k a constant
If m is replaced with 15m, then you want a new s such that
15m/s^6 = m/2^6
s^6 = 15*2^6
s = 2*15^(1/6) = 3.14 mi/hr
We are given that the mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. Let's denote the mass of the particles that the river can currently transport as M, and the speed of the river as S.
According to the given information, we have the following relationship: M = k * S^6, where k is the proportionality constant.
In this case, we can find k by using the current mass and speed of the river. Since the river normally flows at a speed of 2 miles per hour, and the mass of the particles it can transport is the usual mass, we have: M = k * 2^6.
Now, we are given that we want to transport particles that are 15 times as massive as usual. Therefore, we need to find the new speed, which we will call S'. The new mass of the particles will be 15 times the usual mass, or 15M.
Plugging in the values, we have: 15M = k * (S')^6.
We can now equate the two equations we have for the masses: M = 15M. k * 2^6 = k * (S')^6.
Simplifying the equation by canceling out k, we get: 2^6 = (S')^6.
Taking the sixth root of both sides of the equation, we have: (S') = (2^6)^(1/6).
Evaluating this expression, we have: S' = 2^(6/6).
Simplifying further, we have: S' = 2^1.
Therefore, the speed of the river must be 2^1 = 2 miles per hour in order to transport particles that are 15 times as massive as usual.
To solve this problem, we need to use the relationship between mass and speed of the river, which states that the mass is proportional to the sixth power of the speed.
Let's first represent the mass of the particles as M, and the speed of the river as v. We can write the relationship as:
M ∝ v^6
Now, we are given that the speed of the river is 2 miles per hour when the particles have their usual (standard) mass. Let's call this usual mass M_standard. We can write this as:
M_standard ∝ 2^6
Next, we are asked to find the speed of the river when the particles are 15 times as massive as usual. Let's call this mass M_new. We can write this relationship as:
M_new ∝ v^6
Since the mass is 15 times as massive, we can write this as:
M_new = 15 * M_standard
Substituting M_standard = 2^6, we have:
15 * 2^6 ∝ v^6
To solve for v, we can simplify the equation as:
v^6 = (15 * 2^6)^(1/6)
Now, calculate (15 * 2^6)^(1/6) using a calculator:
(15 * 2^6)^(1/6) ≈ 3.527
So, the speed of the river should be approximately 3.53 miles per hour in order to transport particles that are 15 times as massive as usual.