For a simple pendulum: Period = 6.3437 (rounded) seconds
If you mean what units, that would be meters.
To find the area of a square, you square the length of one of its sides. In this case, with a side length of 15 meters, the area would be 15 meters multiplied by 15 meters, which equals 225 square meters. Therefore, the area of a square with a side length of 15 meters is 225 square meters.
To find the area of a rectangle, you multiply the length by the width. In this case, the length is 10 meters and the width is 9 meters. So, the area would be 10 meters x 9 meters = 90 square meters.
Fifteen square meters is a measurement of area, not length. Area is a two-dimensional measurement, while meters are a unit of length. To find out how many meters make up 15 square meters, you would need to know the shape of the area in question. If it is a square, then the length of each side would be the square root of 15, which is approximately 3.87 meters.
With a simple pendulum, provided the angular displacement is less than pi/8 radians (22.5 degrees) it will be a straight line, through the origin, with a slope of 2*pi/sqrt(g) where g is the acceleration due to gravity ( = 9.8 mtres/sec^2, approx). For larger angular displacements the approximations used in the derivation of the formula no longer work and the error is over 1%.
The period of a pendulum is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Since the pendulum's length and mass do not change, its period on the moon would be T = 2π√(L/1.62), assuming the pendulum is the same length. Solving for T gives 2.56 seconds.
Yes. Given a constant for gravity, the period of the pendulum is a function of it's length to the center of mass. In a higher gravity, the period would be shorter for the same length of pendulum.
Assuming a gravitational acceleration of 9.81 m/s^2, a pendulum in Nairobi with a length of approximately 0.25 meters would have a time period of around 1 second. This is calculated using the formula T = 2π√(L/g), where T is the time period, L is the length of the pendulum, and g is the gravitational acceleration.
The length of a pendulum affects its period of oscillation, but to determine the length of a specific pendulum, you would need to measure it. The formula for the period of a pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
The time period of a pendulum is determined by its length and gravitational acceleration. If the length of the second pendulum is one third of the original pendulum, its time period would be shorter since the time period is directly proportional to the square root of the length.
It doesn't matter what unit you use to measure the physical length of the pendulum. As a matter of fact, it doesn't matter what unit you use to measure the duration of its period either. If both are at rest on the same planet, then the penduum with the longer string has the longer period. Period!
A pendulum in Nairobi with a length of approximately 0.25 meters would have a time period of one second. This is because the acceleration due to gravity in Nairobi is approximately 9.81 m/s^2.
The time period of a simple pendulum at the center of the Earth would be constant and not depend on the length of the pendulum. This is because acceleration due to gravity is zero at the center of the Earth, making the time period independent of the length of the pendulum.
Increasing the mass of a pendulum would not change the period of its oscillation. The period of a pendulum only depends on the length of the pendulum and the acceleration due to gravity, but not the mass of the pendulum bob.
For small amplitudes, the period can be calculated as 2 x pi x square root of (L / g). Convert the length to meters, and use 9.8 for gravity. The answer will be in seconds. About 1.4 seconds.
This pendulum, which is 2.24m in length, would have a period of 7.36 seconds on the moon.
Making the length of the pendulum longer. Also, reducing gravitation (that is, using the pendulum on a low-gravity world would also increase the period).