The free-fall acceleration on mars is 3.7 m/s2.

(a) what length of pendulum has a period of 1.2 s on earth? cm

(b) what length of pendulum would have a 1.2-s period on mars? cm an object is suspended from a spring with force constant 10 n/m.

(c) find the mass suspended from this spring that would result in a period of 1.2 s on earth. kg

(d) find the mass suspended from this spring that would result in a period of 1.2 s on mars.

(a) The length of the pendulum on Earth is 36.8cm

(b) The length of the pendulum on Mars is 13.5cm

(c) Mass suspended from the spring on Earth is 0.37kg

(d) Mass suspended from the spring on Mars is 0.36kg

Explanation:

Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2

( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm

(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m

(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg

(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )