Find the dimensions of Planck's constant h from the equatioin `E=hv` where E is the energy and v is the frequency.

Explanation

To find the dimensions of Planck's constant \( h \) from the equation \( E = h \nu \), where \( E \) is the energy and \( \nu \) is the frequency, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the dimensions of energy \( E \)**:
- The dimension of energy is given by the formula:
\[ [E] = [M][L^2][T^{-2}] \]
- Therefore, the dimensions of energy \( E \) are:
\[ [E] = M^1 L^2 T^{-2} \]

2. **Identify the dimensions of frequency \( \nu \)**:
- Frequency \( \nu \) is defined as the reciprocal of the time period \( T \):
\[ \nu = \frac{1}{T} \]
- Thus, the dimension of frequency is:
\[ [\nu] = [T^{-1}] \]

3. **Rearrange the equation \( E = h \nu \) to find \( h \)**:
- From the equation \( E = h \nu \), we can express \( h \) as:
\[ h = \frac{E}{\nu} \]

4. **Substitute the dimensions into the equation**:
- Now substituting the dimensions of \( E \) and \( \nu \):
\[ [h] = \frac{[E]}{[\nu]} = \frac{M^1 L^2 T^{-2}}{T^{-1}} \]

5. **Simplify the dimensions**:
- When we divide \( M^1 L^2 T^{-2} \) by \( T^{-1} \), we subtract the exponents of \( T \):
\[ [h] = M^1 L^2 T^{-2} \cdot T^{1} = M^1 L^2 T^{-1} \]

6. **Final result**:
- Therefore, the dimensions of Planck's constant \( h \) are:
\[ [h] = M^1 L^2 T^{-1} \]

### Final Answer:
The dimensions of Planck's constant \( h \) are \( M^1 L^2 T^{-1} \).