The period of a simple pendulum is given by:

where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:

(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is

And its period is the reciprocal of its frequency:

So now we can use eq.(1) to find the gravitational acceleration of the planet:
The horizontal force : f = k*N
k- coefficient of friction
k = f /N
N = m * g = 45 kg * 9.81 m/s² = 441.45 N
k = 25 N : 441.45 N = 0.057
Answer C) 0.057
Answer:
(A) 3.1 m/s
(B) 2.0 s
Explanation:
At the minimum speed, the force of gravity equals the centripetal force.
mg = m v² / r
v = √(gr)
v = √(9.8 m/s² × 1.0 m)
v = 3.1 m/s
The time is the circumference divided by the speed.
t = (2π × 1.0 m) / (3.1 m/s)
t = 2.0 s
Answer:
t should be 3.57 second
Explanation:
Formula used is v = u+at
In which v is final velocity, u is initial velocity, a is acceleration and t is time.
Substitute each of the info given into the formula and calculate.
49 = 24 + (7)t
t = 3.57s