Physics, 09.07.2019 04:30 purplefive85

A2600-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 nearby beach? the earth's radius is 6400 km.

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Answer from: hadwell34

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,

$V = \omega \times r$

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,

$V = \omega \times r$

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,

$V_c = \omega \times (6400+2.6)\\\\V_c = \omega \times (6402.6)$

What is the linear velocity of the surfer on the beach?

we know the formula for the linear velocity,

$V = \omega \times r$

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,

$V_s = \omega \times (6400)$

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,

$\dfrac{V_c}{V_s} = \dfrac{6402.6\omega}{6400 \omega}$

$\dfrac{V_c}{V_s} = 1.00040625\\\\{V_c} = 1.00040625V_s$

Hence, the climber is moving 1.00040625 times faster than the man on the beach.

Learn more about Velocity:

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Answer from: murtaghliam1

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= $\frac{2πr}{t}$

Here, t is the time

Plugging the values in the above equation

V1= $\frac{2π*6.4*10^6}{86400}$ =465.421 m/s

Velocity on the mountain is calculated as

V2= $\frac{2π(r+h)}{t}$

Plugging the values in the above equation

V2= $\frac{2π(6.4*10^6+2600}{86400}$ =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

Answer from: Quest

explanation:

could i give you a exaplo that could you?

Answer from: Quest

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

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A2600-m-high mountain is located on the equator. how much faster does a climber on top of the mounta...