To solve for $v$ in the equation:

$qU = \frac{1}{2}m \cdot v^2$

we can isolate $v$ by following these steps:

1. Multiply both sides by 2 to eliminate the $\frac{1}{2}$ factor: $2 \cdot qU = m \cdot v^2$

2. Divide both sides by $m$ to isolate $v^2$: $v^2 = \frac{2 \cdot qU}{m}$

3. Take the square root of both sides to solve for $v$: $v = \sqrt{\frac{2 \cdot qU}{m}}$

So, the solution for $v$ is:

$v = \sqrt{\frac{2 \cdot qU}{m}}$

Let me know if you'd like further details or have additional questions!

Related Questions

1. What happens to $v$ if $qU$ doubles, assuming $m$ is constant?
2. How would the equation change if $qU$ included potential energy terms?
3. In what types of physical scenarios might this formula apply?
4. How does $m$ affect the value of $v$ in this formula?
5. How does this equation relate to kinetic energy?

Tip: When solving for a variable, isolating it step-by-step helps ensure accuracy and clarity.