Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 49

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. Glenda drops a coin from ear level down a wishing well. The coin falls a distance of \(7.00 \mathrm{m}\) before it strikes the water. If the speed of sound is \(343 \mathrm{m} / \mathrm{s}\), how long after Glenda releases the coin will she hear a splash?

Short Answer

Expert verified
Answer: It takes Glenda 1.21 seconds to hear the splash after she drops the coin.

Step by step solution

01

Calculate the time it takes for the coin to fall

Since we know the distance the coin falls (\(7.00 \mathrm{m}\)) and the acceleration due to gravity (\(9.80 \mathrm{m} / \mathrm{s}^{2}\)), we can use the free-fall equation to find the time it takes for the coin to fall: $$h = \frac{1}{2}gt^2$$ We know h, and g, and we want to find t. To do this, rearrange the equation to solve for t: $$t = \sqrt{\frac{2h}{g}}$$ Now plug in the values for h and g: $$t = \sqrt{\frac{2(7.00 \mathrm{m})}{9.80 \mathrm{m} / \mathrm{s}^{2}}}$$
02

Calculate the time it takes for the sound to travel back to Glenda

To find the time it takes for the sound to travel back to Glenda, we can use the speed of sound (\(343 \mathrm{m} / \mathrm{s}\)). The time it takes for the sound to travel is given by the equation: $$t_s = \frac{d}{v}$$ where t_s is the time it takes for the sound to travel, d is the distance it travels, and v is the speed it travels. Plug in the values for d (same as h) and v: $$t_s = \frac{7.00 \mathrm{m}}{343 \mathrm{m} / \mathrm{s}}$$
03

Find the total time

To find the total time, add the time it takes for the coin to fall (t) and the time it takes for the sound to travel back (t_s): $$t_{total} = t + t_s$$ First, calculate the values for t and t_s obtained in the previous steps: $$t = \sqrt{\frac{2(7.00 \mathrm{m})}{9.80 \mathrm{m} / \mathrm{s}^{2}}} = 1.19 \mathrm{s}$$ $$t_s = \frac{7.00 \mathrm{m}}{343 \mathrm{m} / \mathrm{s}} = 0.020 \mathrm{s}$$ Now, add these values: $$t_{total} = 1.19 \mathrm{s} + 0.020 \mathrm{s} = 1.21 \mathrm{s}$$ Thus, Glenda will hear the splash 1.21 seconds after she releases the coin.

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

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A penny is dropped from the observation deck of the Empire State building (369 m above ground). With what velocity does it strike the ground?
A squirrel is trying to locate some nuts he buried for the winter. He moves \(4.0 \mathrm{m}\) to the right of a stone and digs unsuccessfully. Then he moves \(1.0 \mathrm{m}\) to the left of his hole, changes his mind, and moves \(6.5 \mathrm{m}\) to the right of that position and digs a second hole. No luck. Then he moves \(8.3 \mathrm{m}\) to the left and digs again. He finds a nut at last. What is the squirrel's total displacement from its starting point?
A chipmunk, trying to cross a road, first moves \(80 \mathrm{cm}\) to the right, then \(30 \mathrm{cm}\) to the left, then \(90 \mathrm{cm}\) to the right, and finally \(310 \mathrm{cm}\) to the left. (a) What is the chipmunk's total displacement? (b) If the elapsed time was \(18 \mathrm{s},\) what was the chipmunk's average speed? (c) What was its average velocity?
To pass a physical fitness test, Massimo must run \(1000 \mathrm{m}\) at an average rate of \(4.0 \mathrm{m} / \mathrm{s} .\) He runs the first $900 \mathrm{m}\( in \)250 \mathrm{s} .$ Is it possible for Massimo to pass the test? If so, how fast must he run the last 100 m to pass the test? Explain.
In a cathode ray tube in an old TV, electrons are accelerated from rest with a constant acceleration of magnitude $7.03 \times 10^{13} \mathrm{m} / \mathrm{s}^{2}\( during the first \)2.0 \mathrm{cm}$ of the tube's length; then they move at essentially constant velocity another \(45 \mathrm{cm}\) before hitting the screen. (a) Find the speed of the electrons when they hit the screen. (b) How long does it take them to travel the length of the tube?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Torque and Rotational Motion

Read Explanation

Scientific Method Physics

Read Explanation

Physics of Motion

Read Explanation

Nuclear Physics

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