The period of the pendulum is directly proportional to the square root of the length of the pendulum
Explanation:
The period of a simple pendulum is given by the equation

where
T is the period
L is the length of the pendulum
g is the acceleration of gravity
From the equation, we see that when the length of the pendulum increases, the period of the pendulum increases as the square root of L,
. This means that
The period of the pendulum is directly proportional to the square root of the length of the pendulum
From the equation, we also notice that the period of a pendulum does not depend on its mass.
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Answer:
The answer is 4200 J.
Explanation:
The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :
W = F×D
F = 42N
D = 100m
W = 42 × 100
= 4200 J
Temperature. The other three dont have anything to do with determining climate
Answer:

Explanation:
The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:






