An astronaut on the Moon attaches a small brass ball to a 1.00- m length of string and makes a simple pendulum. She times 15 complete swings in a time of 75.0 seconds. From this measurement she calculates the acceleration due to gravity on the Moon. What is her result?

Question

Create a captivating image of a female astronaut of Hispanic descent on the Moon. In her space suit, she is performing an experiment: a small brass ball is attached to a precisely 1.00-metre long string, functioning as a simple pendulum. Captured mid-swing, the pendulum is in stark contrast to the pale lunar surface and the black abyss of space behind. The astronaut is carefully timing the swings with a stopwatch in her other hand. Note that there should be no text in this image.

An astronaut on the Moon attaches a small brass ball to a 1.00- m length of string and makes a simple pendulum. She times 15 complete swings in a time of 75.0 seconds. From this measurement she calculates the acceleration due to gravity on the Moon. What is her result?

Answer 1

T =2•π•sqrt(L/g),

T =t/Ná
g = 4• π^2•L•N^2/t^2 =
=4• π^2•1•15^2/75^2 =
=1.58 m/s^2

Answer 2

To calculate the acceleration due to gravity on the Moon, we can use the formula for the period of a simple pendulum. The period, T, is the time it takes for one complete swing.

The formula is given by: T = 2π√(L/g)

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

In this case, the astronaut measured 15 complete swings in a time of 75.0 seconds:

T = 75.0 seconds / 15 swings
T = 5.0 seconds per swing

Now we can rearrange the formula to solve for g:

T = 2π√(L/g)
T^2 = (2π)^2(L/g)
g = (4π^2L) / T^2

Plugging in the known values:
L = 1.00 m
T = 5.0 s

g = (4π^2 * 1.00) / (5.0)^2

Calculating the value:

g = (4 * 3.1416^2 * 1.00) / 25
g ≈ 6.27 m/s^2

So, her calculation for the acceleration due to gravity on the Moon is approximately 6.27 m/s^2.

Answer 3

To calculate the acceleration due to gravity on the Moon, the astronaut needs to use the formula for the period of a simple pendulum, which is given by:

T = 2π * √(L/g)

where:
T is the period of the pendulum (time for one complete swing)
L is the length of the pendulum string
g is the acceleration due to gravity

In this case, the astronaut times 15 complete swings in a time of 75.0 seconds, which means the period (T) is 75.0 seconds divided by 15, or 5.0 seconds.

Substituting the known values into the equation, we get:

5.0 = 2π * √(1.00/g)

Now we need to isolate g to find the acceleration due to gravity on the Moon. Rearranging the equation and solving for g, we have:

√(1.00/g) = 5.0 / (2π)

Squaring both sides of the equation, we get:

1.00/g = (5.0 / (2π))^2

Now we can solve for g:

g = 1.00 / [(5.0 / (2π))^2]

Evaluating this expression gives us the approximate value for the acceleration due to gravity on the Moon as:

g ≈ 1.63 m/s²

Therefore, the astronaut's result for the acceleration due to gravity on the Moon is approximately 1.63 m/s².

Answer 4

An astronaut on the Moon attaches a small brass ball to a 1.00- m length of string and makes a simple pendulum. She times 15 complete swings in a time of 75.0 seconds. From this measurement she calculates the acceleration due to gravity on the Moon. What is her result?

T =2•π•sqrt(L/g),

To calculate the acceleration due to gravity on the Moon, we can use the formula for the period of a simple pendulum. The period, T, is the time it takes for one complete swing.

To calculate the acceleration due to gravity on the Moon, the astronaut needs to use the formula for the period of a simple pendulum, which is given by:

1345 m/s^2