Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 70

\( 1\) parsec \(=\ldots \ldots \ldots \ldots \ldots\) (a) \(10^{-15} \mathrm{~m}\) (b) \(1.496 \times 10^{11} \mathrm{~m}\) (c) \(1.496 \times 10^{15} \mathrm{~m}\) (d) \(3.08 \times 10^{16} \mathrm{~m}\)

Short Answer

Expert verified
1 parsec \( \approx 3.08 \times 10^{16} \, \mathrm{m} \)

Step by step solution

01

Recall the definition of a parsec

A parsec (pc) is a unit of distance used in astronomy. It is derived from the term "parallax of one arcsecond." 1 parsec is the distance at which an object has an annual parallax angle of one arcsecond. To convert parsecs to meters, we need to know its relationship with other units in the metric system.
02

Conversion of astronomical units to meters

One parsec can be defined as approximately 3.26 light-years, and one light-year is the distance light travels in one year in a vacuum. We also know that the speed of light in a vacuum (c) is about \(c = 3.0 \times 10^8 \, \mathrm{m\cdot s^{-1}}\). Now, let's calculate the distance covered by light in one year (1 light-year): 1 year = 365 days \(\times\) 24 hours/day \(\times\) 60 minutes/hour \(\times\) 60 seconds/minute = 31,536,000 seconds 1 light-year = \( \mathrm{distance} = \mathrm{speed} \times \mathrm{time} = c \times \mathrm{1 \, year} = (3.0 \times 10^8 \, \mathrm{m\cdot s^{-1}}) \times (31,536,000 \, s) \)
03

Calculation of distance in meters for 1 parsec

Now, let's now calculate the distance of 1 parsec in meters, using that 1 parsec \(\approx\) 3.26 light-years: \( 1 \, \mathrm{parsec} = 3.26 \, \mathrm{light-years} = 3.26 \times (3.0 \times 10^8 \, \mathrm{m\cdot s^{-1}}) \times (31,536,000 \, s) \)
04

Compare the calculated value with the given options

After calculating the distance of 1 parsec in meters, we now compare that value with the given options (a, b, c, and d) to determine the correct conversion. The calculation for 1 parsec in meters yields: \( 1 \, \mathrm{parsec} \approx 3.08 \times 10^{16} \, \mathrm{m} \) Comparing this result with the given options, we can see that the correct answer is: (d) \( 3.08 \times 10^{16} \, \mathrm{m} \)

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

Write the unit of angular acceleration in the SI system. (a) \(\mathrm{N} \cdot \mathrm{Kg}\) (b) \(\mathrm{rad} /(\mathrm{sec})^{2}\) (c) \(\mathrm{m} / \mathrm{sec}\) (d) \(\mathrm{N} / \mathrm{kg}\)

Which wave length of light is used in an optical microscope? (a) radio wave (b) \(\mathrm{X}\) - ray (c) infrared (d) visible

Dimensional formula for torque is (a) \(\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-3}\) (b) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\)

The radius of circle is \(1.26 \mathrm{~cm}\). According to the concept of significant figures area of it can be represented as - (a) \(4.9850 \mathrm{~cm}^{2}\) (b) \(4.985 \mathrm{~cm}^{2}\) (c) \(4.98 \mathrm{~cm}^{2}\) (d) \(9.98 \mathrm{~cm}^{2}\)

\( 1 \mathrm{Mev}=\ldots \ldots \ldots \ldots \ldots\) (a) \(10^{7}\) (b) \(10^{4}\) (c) \(10^{5}\) (d) \(10^{6}\)
See all solutions

Recommended explanations on English Textbooks

Language Analysis

Read Explanation

Syntax

Read Explanation

History of English Language

Read Explanation

English Grammar Summary

Read Explanation

Textual Analysis

Read Explanation

Free Response Essay

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