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Ch 03: Motion in Two or Three Dimensions
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 03: Motion in Two or Three DimensionsProblem 23a
Chapter 3, Problem 23a

The earth has a radius of 6380 km and turns around once on its axis in 24 h. What is the radial acceleration of an object at the earth's equator? Give your answer in m/s2 and as a fraction of g.

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First, convert the radius of the Earth from kilometers to meters. Since 1 km = 1000 m, the radius in meters is 6380 km * 1000 m/km = 6380000 m.
Next, calculate the angular velocity (ω) of the Earth. The Earth completes one full rotation in 24 hours, which is equivalent to 24 * 3600 seconds. The angular velocity is given by ω = 2π / T, where T is the period of rotation. So, ω = 2π / (24 * 3600) s.
The radial (centripetal) acceleration (a_r) at the Earth's equator can be calculated using the formula a_r = ω² * r, where r is the radius of the Earth. Substitute the values of ω and r to find a_r.
To express the radial acceleration as a fraction of g (acceleration due to gravity), use the value g = 9.81 m/s². Calculate the fraction by dividing the radial acceleration by g: fraction = a_r / g.
Ensure the units are consistent and check the calculations for accuracy. The final result will give you the radial acceleration in m/s² and as a fraction of g.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Acceleration

Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is calculated using the formula a_c = v^2/r, where v is the tangential velocity and r is the radius of the circular path. This concept is crucial for determining the radial acceleration of an object at the Earth's equator.
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Tangential Velocity

Tangential velocity refers to the linear speed of an object moving along a circular path. It is given by the formula v = 2πr/T, where r is the radius of the circle and T is the period of rotation. For the Earth, this period is 24 hours, and understanding this concept helps in calculating the centripetal acceleration at the equator.
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Gravitational Acceleration (g)

Gravitational acceleration, denoted as g, is the acceleration due to Earth's gravity, approximately 9.81 m/s². When calculating the radial acceleration as a fraction of g, it provides a comparative measure of how the centripetal acceleration at the equator relates to the standard gravitational pull experienced by objects on Earth's surface.
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Related Practice
Textbook Question

The earth has a radius of 6380 km and turns around once on its axis in 24 h. If arad at the equator is greater than g, objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

Textbook Question

A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. What is the radial acceleration of the blade tip expressed as a multiple of g?

Textbook Question

A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. What is the linear speed of the blade tip, in m/s?

Textbook Question

A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. How high is the balloon when the rock is thrown?

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Textbook Question

A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. At the instant the rock hits the ground, how far is it from the basket?

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Textbook Question

A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m/s at an angle of 33.0° above the horizontal. Ignore air resistance. Calculate Draw x-t, y-t, vx–t, and vy–t graphs for the motion.

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