Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 64

The Earth rotates on its own axis once per day \((24.0 \mathrm{h})\) What is the tangential speed of the summit of Mt. Kilimanjaro (elevation 5895 m above sea level), which is located approximately on the equator, due to the rotation of the Earth? The equatorial radius of Earth is \(6378 \mathrm{km}\).

Short Answer

Expert verified
Answer: The tangential speed of the summit of Mt. Kilimanjaro due to Earth's rotation is approximately \(464.28 \, m/s\).

Step by step solution

01

Determine the total radius of the circular path

To find the total radius, add the height of Mt. Kilimanjaro to the equatorial radius of Earth. Convert both values to meters to ensure we have consistent units. Radius of Earth = \(6378 km\) Height of Mt. Kilimanjaro = \(5895 m\) Total radius \(r\): \(r = 6378 \times 10^3 m + 5895 m = 6,378,000 m + 5895 m = 6,383,895 m\)
02

Calculate the circumference of the circular path

Use the circumference formula of a circle, \(C = 2 \pi r\), to calculate the distance Mt. Kilimanjaro's summit moves during one complete rotation of Earth. \(C = 2 \pi \times 6,383,895 m = 2 \pi \times 6,383,895 m \approx 40,107,514 m\)
03

Calculate the tangential speed

To find the tangential speed, divide the circumference by the time taken for one complete rotation, which is 24 hours or 86400 seconds. Tangential speed \(v_t\): \(v_t = \frac{C}{T} = \frac{40,107,514 m}{86,400 s} \approx 464.28 \, m/s\) The tangential speed of the summit of Mt. Kilimanjaro due to Earth's rotation is approximately \(464.28 \, m/s\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Grace is playing with her dolls and decides to give them a ride on a merry-go- round. She places one of them on an old record player turntable and sets the angular speed at 33.3 rpm. (a) What is their angular speed in rad/s? (b) If the doll is \(13 \mathrm{cm}\) from the center of the spinning turntable platform, how fast (in \(\mathrm{m} / \mathrm{s}\) ) is the doll moving?
The time to sunset can be estimated by holding out your arm, holding your fingers horizontally in front of your eyes, and counting the number of fingers that fit between the horizon and the setting Sun. (a) What is the angular speed, in radians per second, of the Sun's apparent circular motion around the Earth? (b) Estimate the angle subtended by one finger held at arm's length. (c) How long in minutes does it take the Sun to "move" through this same angle?
Verify that all three expressions for radial acceleration $\left(v \omega, v^{2} / r, \text { and } \omega^{2} r\right)$ have the correct dimensions for an acceleration.
A child pushes a merry-go-round from rest to a final angular speed of 0.50 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 2.0 revolutions. What is the angular acceleration of the merry- go-round?
A spacecraft is in orbit around Jupiter. The radius of the orbit is 3.0 times the radius of Jupiter (which is \(\left.R_{1}=71500 \mathrm{km}\right) .\) The gravitational field at the surface of Jupiter is 23 N/kg. What is the period of the spacecraft's orbit? [Hint: You don't need to look up any more data about Jupiter to solve the problem.]
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Space Physics

Read Explanation

Translational Dynamics

Read Explanation

Thermodynamics

Read Explanation

Wave Optics

Read Explanation

Electrostatics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free