The acceleration due to gravity on the moon is 1.62 m/s2. What is the length of a pendulum whose period on the moon matches the period of a 4.50-m-long pendulum on earth

T^2 = 4(pi)^2 * (L/g).

L2 = (g2/g1)L1 = (1.62/9.8)4.5=0.74m.

To find the length of the pendulum on the moon that matches the period of a 4.50-m-long pendulum on Earth, we can use the formula for the period of a pendulum:

T = 2π√(L / g)

Where:
T = period of the pendulum
L = length of the pendulum
g = acceleration due to gravity

Given:
Acceleration due to gravity on the moon (g_moon) = 1.62 m/s^2
Length of the pendulum on Earth (L_earth) = 4.50 m
Period of the pendulum on Earth (T_earth) = unknown

We need to find the length of the pendulum on the moon (L_moon) that matches the period of the 4.50-m-long pendulum on Earth.

First, let's find the period of the pendulum on Earth using the formula:

T_earth = 2π√(L_earth / g_earth)

Since the acceleration due to gravity on Earth is approximately 9.8 m/s^2, we can use this value:

T_earth = 2π√(4.50 m / 9.8 m/s^2)

Now, we can calculate T_earth:

T_earth = 2π√(0.459 m)

Next, let's find the period of the pendulum on the moon:

T_moon = 2π√(L_moon / g_moon)

Since we want T_moon to be the same as T_earth, we can set T_moon equal to T_earth:

T_moon = T_earth

2π√(L_moon / g_moon) = 2π√(0.459 m)

Now, we can solve for L_moon:

√(L_moon / 1.62 m/s^2) = √(0.459 m)

Squaring both sides:

L_moon / 1.62 m/s^2 = 0.459 m

Now, solve for L_moon:

L_moon = 0.459 m * 1.62 m/s^2

L_moon = 0.74358 m

Therefore, the length of a pendulum on the moon that matches the period of a 4.50-m-long pendulum on Earth is approximately 0.74358 meters.

To find the length of a pendulum on the moon that matches the period of a 4.50-m-long pendulum on Earth, you can use the formula for the period of a simple pendulum:

T = 2π√(L/g)

Where:
T = Period of the pendulum
L = Length of the pendulum
g = Acceleration due to gravity

We know that the acceleration due to gravity on the Moon is 1.62 m/s², and we need to find the length of the pendulum on the Moon that matches the period of a 4.50-m-long pendulum on Earth.

First, let's find the period of the 4.50-m-long pendulum on Earth. We can use the same formula, with the appropriate values:

T_earth = 2π√(L_earth/g_earth)

Plugging in the values:
T_earth = 2π√(4.50/9.81) [Note: Earth's acceleration due to gravity is approximately 9.81 m/s²]

Calculating:
T_earth ≈ 2π√(0.458)

Next, we need to find the length of the pendulum on the Moon that will match this period. Again, we use the formula for the period of a pendulum, but now with the values for the Moon:

T_moon = 2π√(L_moon/g_moon)

Plugging in the values:
T_moon = 2π√(L_moon/1.62) [Note: Moon's acceleration due to gravity is 1.62 m/s²]

Since we want the T_moon to be equal to T_earth, we can equate them:

2π√(0.458) = 2π√(L_moon/1.62)

By canceling out the common terms, we get:

√(0.458) = √(L_moon/1.62)

Squaring both sides, we get:

0.458 = L_moon/1.62

Now, solving for L_moon:

L_moon = 0.458 × 1.62

L_moon ≈ 0.742 m

Therefore, the length of the pendulum on the Moon that matches the period of a 4.50-m-long pendulum on Earth is approximately 0.742 meters.