What is the period of a simple pendulum with a length of 2.25 m on each of the four given planets? Use the acceleration due to gravity on the "surface" of each planet given in the table. Mercury Venus Mars Jupiter Saturn Uranus Neptune 3.70 m/s? | 8.87 m/s² | 3.71 m/s? | 23.1 m/s? | 10.4 m/s? | 8.87 m/s²| 11.0 m/s? Jupiter: Saturn: S Uranus: Neptune: S S

What is the period of a simple pendulum with a length of 2.25 m on each of the four given planets? Use the acceleration due to gravity on the "surface" of each planet given in the table. Mercury Venus Mars Jupiter Saturn Uranus Neptune 3.70 m/s? | 8.87 m/s² | 3.71 m/s? | 23.1 m/s? | 10.4 m/s? | 8.87 m/s²| 11.0 m/s? Jupiter: Saturn: S Uranus: Neptune: S S

Name: What is the period of a simple pendulum with a length of 2.25 m on ea
Uploaded: 2024-03-20T06:58:04.000Z
Channel: Bartleby
Description: What is the period of a simple pendulum with a length of 2.25 m on each of the four given planets? Use the acceleration dueto gravity on the "surface" of each planet given in the table.MercuryVenusMarsJupiterSaturnUranusNeptune3.70 m/s? | 8.87 m/s²

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter15: Oscillations

Section: Chapter Questions

Problem 49P: A pendulum with a period of 2.00000 s in one location (g=9.80m/s2) is moved to a new location where...

See similar textbooks

Similar questions

We do not need the analogy in Equation 16.30 to write expressions for the translational displacement of a pendulum bob along the circular arc s(t), translational speed v(t), and translational acceleration a(t). Show that they are given by s(t) = smax cos (smpt + ) v(t) = vmax sin (smpt + ) a(t) = amax cos(smpt + ) respectively, where smax = max with being the length of the pendulum, vmax = smax smp, and amax = smax smp2.
How do you determine the gravitational acceleration on a planet where the period of a 1.90 m long pendulum is 2.78 s, and express the result in g = in m/s^2?
The period of a simple pendulum that is 1.00 m long on another planet is 1.66s. What is the acceleration due to gravity on this planet if the mass of the pendulum bob is 1.5kg?
A pendulum has a period of 47.9 s. If you were to transfer this pendulum to the surface of Mars (where g is 3.71 meters per second squared), what would the period (in seconds) be?
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 46.0 cm. The explorer finds that the pendulum completes 98.0 full swing cycles in a time of 145 s. What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per secondgPlanet=(?)m/s^2
A 0.013 kg rubber stopper attached to a 0.93 m string is swung in a circle. If the tension in the string is 0.70 N, what is the period of the stopper's rev.?
The theoretical expression for the period of oscillation for a simple pendulum with small oscillations is given by: T = 2pi*sqrt(L/g) where T is the period, L is the length of the pendulum, and g is the value of the Earth’s gravitational field. Period (s) Length (m) 4.49 5.00 6.34 10.00 7.77 15.00 8.97 20.00 10.03 25.00 14.19 50.00 20.08 100.00 Knowing that LENGTH is the indipendent variable and PERIOD is the dependent one please write your equation in y=mx+b form, including the y.
1. In order to swing with a period of exactly 2.0 s, a grandfather clock’s 1.5 kg pendulum must have a length of ______m. 2. Determine the acceleration due to gravity for Planet X. _________________ Given Planet X Information:Mass (kg) = 1.00Length (m) = 0.70Period (s) = 1.3955Gravity (m/s2) = 14.19
The period of a simple pendulum of lengtg 1 meter, on another planet is 1.66 seconds. what is the accelration due to gravity on this planet in m/s^2?
A mountain climber, who has had physics in high school, figures out the gravity at his location in the mountains. He used a 400 cm length of a string and found that with a rock tied at its end, its period as a pendulum was 1.39x10-3 hour.
Consider a place where the gravity is one-fourth the gravity on Earth (g' = g/4), then the frequency of oscillation of a simple pendulum in that place, f', as compared to its frequency on earth is: f’ = 9f f’ = 4f f’ = f/3 f’ = 2f f’ = f/2
Geologists use pendulums to find tiny differences in the earths gravitational field from one position to another because these differences provide information on what is underground. A 1.0108 m simple pendulum is observed to have a period of 2.0005 s. Find the acceleration due to gravity “g”, to 5 significant digits, at that location.