What is the frequency of a pendulum of length 1.50 m at a location where 1 point the acceleration due to gravity is 9.79 m/s^2?
1 answer:
You might be interested in
a) For the motion of car with uniform velocity we have , , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0
Substituting
The constant velocity of car a = 2.33 m/s
b) We have
s = 520 m, t = 223 seconds, u =0 m/s
Substituting
Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209
Answer:
50.2 m
Explanation:
We can solve the problem by using the following SUVAT equation for the vertical position of the rock:
where
h is the initial height (the depth of the canyon), taking the bottom of the canyon as reference position
u = 0 is the initial velocity of the rock
t is the time
is the acceleration of gravity
When the rock reaches the bottom, t = 3.2 s and y = 0. Substituting these numbers and solving for h, we find the depth of the canyon: