The mass of particles a river can transport is proportional to the sixth power of its speed. Let:

• $m$ represent the mass of particles the river can transport,
• $v$ represent the speed of the river.

The relationship is expressed as:

$m \propto v^6 \quad \text{or equivalently} \quad m = k v^6,$

where $k$ is a proportionality constant.

Step 1: Relating the Masses

For the given problem:

• The usual speed is $v_1$,
• The usual mass transported is $m_1 = k v_1^6$,
• The required mass to be transported is $m_2 = 15 m_1$.

Since $m_2 = k v_2^6$ (where $v_2$ is the required speed), we have:

$15 m_1 = k v_2^6.$

Substituting $m_1 = k v_1^6$ into the equation:

$15 k v_1^6 = k v_2^6.$

Cancel $k$ from both sides:

$15 v_1^6 = v_2^6.$

Step 2: Solving for $v_2$

Take the sixth root of both sides:

$v_2 = v_1 \cdot 15^{1/6}.$

Step 3: Calculating $v_2$

Let $v_1$ be the usual speed of the river. Plugging in $15^{1/6}$:

$15^{1/6} \approx 1.57.$

Thus:

$v_2 = v_1 \cdot 1.57.$

Step 4: Final Answer

If the usual speed $v_1$ is known, multiply it by $1.57$ to find $v_2$. For example, if $v_1 = 5$ miles per hour, then:

$v_2 = 5 \cdot 1.57 = 7.85 \, \text{miles per hour}.$

Let me know the usual speed $v_1$, or we can assume $v_1$ if needed! Would you like additional details? 😊

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Expanding Knowledge

1. What is the physical significance of the sixth power relation in real-world rivers?
2. How does this relationship apply to sediment size in river transport?
3. Can the speed-to-mass relationship be generalized for other powers? Why the sixth?
4. How might real-world factors like turbulence affect the accuracy of this formula?
5. What would the speed be if the mass transported needed to double instead of increasing 15 times?

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Tip:

Always recheck proportionality constants to ensure they cancel correctly in physics-based problems!