A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a
1 answer:
The climber's speed is 0.19 m/s greater than that of the surfer on the beach.
Both individuals are on Earth and share a consistent angular velocity, but their linear speeds differ.
Calculating the seconds in a day gives us t=24*60*60=86400 sec
The linear speed on the beach is computed as
V1=
Where t is the duration
Substituting values into the equation leads to
V1==465.421 m/s
The mountain's linear speed is determined as
V2=
Inserting values into the equation results in
V2==465.61 m/s
Thus, the climber exceeding the surfer's speed is calculated as 465.61-465.421=0.19 m/s
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Answer:
All three pendulums will have the same angular frequencies.
Explanation:
For a simple pendulum, the time period using the approximation is expressed as:
The angular frequency is defined as
Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.