Consider the equation 𝜐 = 1/3 z xt 2. The dimensions of the variables υ, x, and t are [L]/[T], [L], and [T], respectively. The numerical factor 3 is di- dimensionless. What must be the dimensions of the variable z, such that both sides of the equation have the same dimensions? Show how you determined your answer.

To solve this problem, we need to ensure that both sides of the equation \(\nu = \frac{1}{3} z \cdot x \cdot t^2\) have the same dimensions. The dimensions of each variable are given as follows:

- \(\nu\) has dimensions \([L]/[T]\), i.e., \([\nu] = [L][T]^{-1}\).
- \(x\) has dimensions \([L]\), i.e., \([x] = [L]\).
- \(t\) has dimensions \([T]\), i.e., \([t] = [T]\).
- The numerical factor \(\frac{1}{3}\) is dimensionless, so it does not affect the dimensions.

Now, let's write out the dimensions of each term on the right-hand side of the equation.

### Step 1: Dimensional Analysis of the Right-Hand Side
The right-hand side of the equation is \(z \cdot x \cdot t^2\).

- The dimension of \(x\) is \([L]\).
- The dimension of \(t^2\) is \([T^2]\).
- Let the dimension of \(z\) be \([z]\).

Thus, the dimension of the right-hand side is:
\[
[z] \cdot [L] \cdot [T^2]
\]
So, the overall dimension of the right-hand side is:
\[
[z][L][T^2]
\]

### Step 2: Equating Dimensions
For the equation to be dimensionally consistent, the dimensions on the left-hand side must equal the dimensions on the right-hand side. Therefore:
\[
[\nu] = [z][L][T^2]
\]
Substitute the dimensions of \(\nu\) from the left-hand side:
\[
[L][T]^{-1} = [z][L][T^2]
\]

### Step 3: Solving for \([z]\)
To solve for \([z]\), we can divide both sides by \([L]\):
\[
[T]^{-1} = [z][T^2]
\]

Now, to isolate \([z]\), divide both sides by \([T^2]\):
\[
[z] = [T]^{-1} \cdot [T]^{-2}
\]
Simplify the exponents:
\[
[z] = [T]^{-3}
\]

### Final Answer
The dimensions of the variable \(z\) must be \([T]^{-3}\), meaning that \(z\) has the dimensions of \(\text{time}^{-3}\).

This ensures that both sides of the equation have the same dimensions.

Would you like any more details or have any questions?

Here are 5 related questions:
1. How would the dimensions of \(z\) change if the equation had a different exponent for \(t\)?
2. Can you determine the dimensions of \(z\) if \(x\) had different dimensions, such as \([L]^2\)?
3. What are some examples of physical quantities that have dimensions of \([T]^{-3}\)?
4. How does dimensional analysis help in verifying the correctness of an equation in physics?
5. What would happen to the dimensions of \(z\) if the equation was \(\nu = \frac{1}{3} z \cdot x^2 \cdot t\)?

**Tip:** Dimensional analysis is a powerful tool in physics for checking the consistency of equations and can often guide you in deriving or confirming formulas.

Understand how to determine the dimensions of variable z in an equation using dimensional analysis. This problem involves equations with variables of different dimensions and requires understanding of dimensional consistency in physics.

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