/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A useful and easy-to-remember ap... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A useful and easy-to-remember approximate value for the number of seconds in a year is \(\pi \times 10^{7}\). Determine the percent error in this approximate value. (There are 365.24 days in one year.)

Short Answer

Expert verified
The percent error in the approximate value for the number of seconds in a year is approximately 0.447%.

Step by step solution

01

Calculate the Actual Number of Seconds in a Year

To calculate the actual number of seconds in a year, first break down a year into smaller units of time. There are 365.24 days in a year, 24 hours in each day, 60 minutes in each hour, and 60 seconds in each minute. Multiply these values together to get the actual number of seconds in a year: \(365.24 \times 24 \times 60 \times 60 = 31556921.6\) seconds.
02

Calculate the Approximate Value

The exercise provides an approximate value of \( \pi \times 10^{7} \) seconds, which equals 31415926.5 seconds.
03

Calculate the Percent Error

Percent error is calculated using the formula: \[ Percent \ Error = \frac{|(Actual \ Value-Approximate \ Value)|}{Actual \ Value} \times 100\% \]. So, substituting our values, we have \[Percent \ Error = \frac{|(31556921.6 - 31415926.5)|}{31556921.6} \times 100\% = 0.447 \%\]

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Seconds in a Year
Time is a fundamental concept in our lives, and understanding it in different units offers us a way to comprehend the world around us. When we break time down into seconds for an entire year, it's not often that one appreciates the sheer number of these tiny moments that pass by.

To quantify this, we calculate the actual number of seconds in a year. Given that there are 365.24 days in a year, a reflection of the Earth's orbit around the Sun including the extra leap year time, this calculation involves several time conversion steps. Each day comprises 24 hours, each hour has 60 minutes, and each minute contains 60 seconds. Thus, the chain of multiplication gives us the total seconds in a year: \[\begin{equation}365.24 \times 24 \times 60 \times 60 = 31,556,921.6 \text{ seconds}.\end{equation}\]

Sometimes, you may encounter approximations like the easy-to-remember value of \(\pi \times 10^{7}\), or 31,415,926.5 seconds for simplicity. This is where error calculation becomes relevant to determine the precision of such approximations.
Time Conversion
Understanding time conversion is essential for daily life and scientific computation. Time can be expressed in years, days, hours, minutes, and seconds, each serving various functions depending on the context.

For example, knowing how many seconds there are in a minute (60 seconds), how many minutes in an hour (60 minutes), and hours in a day (24 hours) allows us to convert one unit of time to another. This kind of knowledge is particularly useful in areas like physics and engineering, where precision is key. By mastering time conversion, we not only aid in our daily understanding of time management but also lay the groundwork for more advanced studies involving time's role in scientific theories.
Approximation in Physics
In physics, approximation is a powerful tool used to simplify complex problems into more manageable terms while still retaining essential features of the physical system.

An approximation might involve rounding numbers to make them easier to work with, or estimating values when exact figures are unnecessary or impossible to determine. For instance, in the exercise of finding the seconds in a year, using \(\pi \times 10^{7}\) seconds is a form of approximation that simplifies the actual figure of 31,556,921.6 seconds without greatly diminishing the concept's understanding.

However, approximations come with the understanding that there is a trade-off between simplicity and accuracy, known as the 'percent error'. This is calculated to acknowledge the difference between the true value and the approximation. A low percent error indicates a good approximation, which is often sought after in physics to balance ease of use with fidelity to physical reality.

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

While following a treasure map, you start at an old oak tree. You first walk \(825 \mathrm{~m}\) directly south, then turn and walk \(1.25 \mathrm{~km}\) at \(30.0^{\circ}\) west of north, and finally walk \(1.00 \mathrm{~km}\) at \(32.0^{\circ}\) north of east, where you find the treasure: a biography of Isaac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reasonable, compare it with a graphical solution drawn roughly to scale.

The volume of a solid cylinder is given by \(V=\pi r^{2} h,\) where \(r\) is the radius and \(h\) is the height. You measure the radius and height of a thin cylindrical wire and obtain the results \(r=0.036 \mathrm{~cm}\) and \(h=12.1 \mathrm{~cm}\). What do your measurements give for the volume of the wire in \(\mathrm{mm}^{3} ?\) Use the correct number of significant figures in your answer.

A spelunker is surveying a cave. She follows a passage 180 \(\mathrm{m}\) straight west, then \(210 \mathrm{~m}\) in a direction \(45^{\circ}\) east of south, and then \(280 \mathrm{~m}\) at \(30^{\circ}\) east of north. After a fourth displacement, she finds herself back where she started. Use a scale drawing to determine the magnitude and direction of the fourth displacement. (See also Problem 1.57 for a different approach.)

A rectangular piece of aluminum is \(7.60 \pm 0.01 \mathrm{~cm}\) long and \(1.90 \pm 0.01 \mathrm{~cm}\) wide. (a) Find the area of the rectangle and the uncertainty in the area. (b) Verify that the fractional uncertainty in the area is equal to the sum of the fractional uncertainties in the length and in the width. (This is a general result.)

DATA You are a mechanical engineer working for a manufacturing company. Two forces, \({\boldsymbol{F}}_{1}\) and \({\boldsymbol{F}}_{2}\), act on a component part of a piece of equipment. Your boss asked you to find the magnitude of the larger of these two forces. You can vary the angle between \(\vec{F}_{1}\) and \(\vec{F}_{2}\) from \(0^{\circ}\) to \(90^{\circ}\) while the magnitude of each force stays constant. And, you can measure the magnitude of the resultant force they produce (their vector sum), but you cannot directly measure the magnitude of each separate force. You measure the magnitude of the resultant force for four angles \(\theta\) between the directions of the two forces as follows: $$ \begin{array}{cc} \hline \boldsymbol{\theta} & \text { Resultant force (N) } \\ \hline 0.0^{\circ} & 8.00 \\ 45.0^{\circ} & 7.43 \\ 60.0^{\circ} & 7.00 \\ 90.0^{\circ} & 5.83 \end{array} $$ (a) What is the magnitude of the larger of the two forces? (b) When the equipment is used on the production line, the angle between the two forces is \(30.0^{\circ} .\) What is the magnitude of the resultant force in this case?

See all solutions

Recommended explanations on Physics Textbooks

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.