Consider the expression 2π times square root of ℓ/g, where ℓ is length and g is gravitational acceleration in units of length divided by time squared.Evaluate its units.

this is the period of oscillation of the simple pendulum ==>[T] = seconds

To evaluate the units of the given expression, let's break it down into individual components:

- 2π: This is a dimensionless constant.
- Square root of ℓ: The given value of ℓ represents length, and taking the square root of a length does not affect the units. Therefore, the square root of ℓ is in units of length.
- g: The given value of g represents gravitational acceleration, which has units of length divided by time squared.

Now, let's multiply these components together:

2π * square root of ℓ * g

The units can be determined as follows:

- 2π: This is dimensionless.
- Square root of ℓ: As mentioned above, it is in units of length.
- g: It is in units of length divided by time squared.

Combining these units, we have:

2π * (length) * (length/time^2)

Simplifying further:

2π * length^2 / time^2

Therefore, the units of the given expression are length squared divided by time squared.

To evaluate the units of the given expression, we need to consider the units of each term separately and then combine them. Let's break it down step by step:

1. The term 2π is a dimensionless constant, meaning it does not have any units associated with it.

2. The square root of ℓ represents a length. Since we don't have any specific units for length given, we will denote it as [L].

3. The term g represents gravitational acceleration and is given in units of length divided by time squared. Let's denote it as [L/T^2].

Now, let's evaluate the units of the entire expression by combining the individual units:

2π times square root of ℓ/g = (2π) × (√ℓ/g)

= 2π[L] × (√ℓ/[L/T^2])

= 2π√(ℓ/L) × T

Thus, the final units of the expression are T, representing time.

Therefore, the expression 2π times square root of ℓ/g has units of time (T).