free fall on the moon is 1.62m/s2 What is the lenght of the pendulum whos period on the moon mtches the period of 1.80m long on Earth?
Period=2PI sqrt (l/g) where g is as given.
squkare both sides. For g=9.8, solve for Period T.
then put that T in the moon example, with g=1.62, and solve for l.
To find the length of the pendulum that matches the period of a 1.80m long pendulum on Earth, you can use the formula for the period of a pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
On the Moon, the acceleration due to gravity is 1.62 m/s^2. We want to find the length of the pendulum on the Moon that matches the period of a 1.80m long pendulum on Earth.
Let's substitute the known values into the formula:
T (Moon) = 2π√(L (Moon) / g (Moon))
T (Earth) = 2π√(L (Earth) / g (Earth))
We know that T (Moon) = T (Earth), L (Earth) = 1.80m, and g (Moon) = 1.62m/s^2.
We can rewrite the formula by substituting the known values:
2π√(L (Moon) / 1.62) = 2π√(1.80 / 9.81)
Now, we can solve for L (Moon):
√(L (Moon) / 1.62) = √(1.80 / 9.81)
Squaring both sides of the equation:
L (Moon) / 1.62 = 1.80 / 9.81
Cross multiplying:
L (Moon) = (1.62 * 1.80) / 9.81
Calculating the value:
L (Moon) = 0.2974 meters (rounded to four decimal places)
Therefore, the length of the pendulum on the Moon that matches the period of a 1.80m long pendulum on Earth is approximately 0.2974 meters.