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The climber move 0.19 m/s faster than surfer on the nearby beach.
Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.
Number of seconds in a day, t=24*60*60=86400 sec
The linear speed on the beach is calculated as
V1=[tex]\frac{2πr}{t}[/tex]
Here, t is the time
Plugging the values in the above equation
V1=[tex]\frac{2π*6.4*10^6}{86400}[/tex]=465.421 m/s
Velocity on the mountain is calculated as
V2=[tex]\frac{2π(r+h)}{t}[/tex]
Plugging the values in the above equation
V2=[tex]\frac{2π(6.4*10^6+2600}{86400}[/tex]=465.61 m/s
Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s
The climber is moving 1.00040625 times faster than the man on the beach.
What is the linear velocity?
We know that the linear velocity of can be written as,
[tex]V = \omega \times r[/tex]
Where ω is the angular velocity, and r is the radius.
What is the linear velocity of the climber on the mountain top?
we know the formula for the linear velocity,
[tex]V = \omega \times r[/tex]
substitute the value, where ω is the angular velocity of the earth, r is the radius of the person from the center of the earth,
[tex]V_c = \omega \times (6400+2.6)\\\\V_c = \omega \times (6402.6)[/tex]
What is the linear velocity of the surfer on the beach?
we know the formula for the linear velocity,
[tex]V = \omega \times r[/tex]
substitute the value, where ω is the angular velocity of the earth, r is the radius of the surfer from the center of the earth,
[tex]V_s = \omega \times (6400)[/tex]
How much faster does a climber on top of the mountain move than a surfer at a nearby beach?
We know the linear velocity of the climber and the surfer, taking the ratio of the two,
[tex]\dfrac{V_c}{V_s} = \dfrac{6402.6\omega}{6400 \omega}[/tex]
[tex]\dfrac{V_c}{V_s} = 1.00040625\\\\{V_c} = 1.00040625V_s[/tex]
Hence, the climber is moving 1.00040625 times faster than the man on the beach.
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