A simple pendulum makes 120 complete oscillations in 2.80 min at a location where g = 9.80 m/s2.
(a) Find the period of the pendulum.
_________s
(b) Find the length of the pendulum.
_________m
A simple pendulum makes 120 complete oscillations in 2.80 min at a location where g =...
A simple pendulum makes 107 complete oscillations in 2.60 min at a location where g = 9.80 m/s2. (a) Find the period of the pendulum. s (b) Find the length of the pendulum. m
A simple pendulum makes 117 complete oscillations in 3.10 min at a location where g = 9.80 m/s2. (a) Find the period of the pendulum. 1.59 s (b) Find the length of the pendulum. m
A simple pendulum makes 129 complete oscillations in 3.60 min at a location where g 9.80 m/'s (a) Find the period of the pendulum. 167 (b) Find the length of the pendulum. 07442x Your response is within accuracy to minimize roundoff error.m 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four digit
A simple pendulum has a length of 52.2 cm and makes 83.8 complete oscillations in 2.00 min. (a) Find the period of the pendulum. (seconds) (b) Find the value of g at the location of the pendulum. (m/s^2)
TUDIEM 14 Geoff counts the number of oscillations of a simple pendulum at a location where the acceleration due to gravity is 9.80 m/s, and finds that it takes 28.0 s for 17 complete cycles. 1) Calculate the length of the pendulum. (Express your answer to three significant figures.) m Submit
A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2s. The length of a seconds pendulum is 0.992 7 m at Tokyo, Japan, and 0.994 2 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations? A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g = 9.80 m/s^2. Find (a) the period of the pendulum and...
The period T of a simple pendulum with small oscillations is calculated from the formula T=2pi sqrt(L/g) where L is the length of the pendulum and g is the acceleration due to gravity. suppose that measured values of L and g have errors and are corrected with new values where L is increased from 4m to 4.5m and g is increased from 9 m/s2 to 9.8 m/s2. Use differentials to estimate the change in the period. Does the period increase...
if a simple pendulum with length 1.50 m makes 73 oscillations in 180 s, what is the acceleration of gravity at this location?
1. How much is the period of 1=1.00 m long pendulum on the Moon (g = 1.600 m/s2) 4.97 sec. 2. On a planet X pendulum of the length 0.500 m makes 50.0 oscillations in 1.00 min. Find the acceleration of gravity on the planet X. g = 13.7 m/s2. 3. Find the period of small oscillations of a meter stick suspended by its end near Earth surface (assume g=9.800 m/s2). Notice that this is not a simple pendulum but...
A simple pendulum on the surface of Earth is found to undergo 11.0 complete small-amplitude oscillations in 8.55 s. Find the pendulum's length. length of the pendulum: .00156 m