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:
Answer:
The answer is A and C.
Explanation:
Only two factors are relevant when dealing with the gravitational force between two objects - their mass and their distance apart from one another. Gravity's force is proportional to the product of the masses of the two objects and is inversely proportional to the square of the distance between them.
Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6

- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6
*(15s)²
Solving:
x=½*6
*(15s)²
x=½*6
*225 s²
x= 675 m
<u><em>
The object will travel 675 m during that time.</em></u>
<span>At the top of the waterfall, the water has potential energy. Once it goes over</span>