/*! 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} Q17Q Is the Young’s modulus for a b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Is the Young’s modulus for a bungee cord smaller or larger than that for an ordinary rope?

Short Answer

Expert verified

The Young’s modulus for a bungee cord is smaller than that of an ordinary rope.

Step by step solution

01

Understanding Young’s modulus

Young’s modulus is one of the essential properties of an element that estimates the stiffness and identifies the ductility or brittleness of the element.

02

Evaluating the change in Young’s modulus with ordinary rope and bungee cord

The relation of Young’s modulus is given by:

\(E = \left( {\frac{F}{A}} \right)\left( {\frac{L}{{\Delta L}}} \right)\)

Here, Fis the force, Ais the area, Lis the length of the cord, and\(\Delta L\)is the change in the length of the cord.

In the above relation, it is observed that Young’s modulus relies on the change in the length of the particular chord. This indicates that the value of Young’s modulus reduces with an increase in change in length.

The variation in the length of a bungee cord is much larger than an ordinary rope. So, Young’s modulus for a bungee cord will be smaller.

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!

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

(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door’s weight (Fig. 9–69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

(II) Calculate\({F_{\rm{A}}}\)and\({F_{\rm{B}}}\)for the uniform cantilever shown in Fig. 9–9 whose mass is 1200 kg.

(I) Calculate the mass m needed in order to suspend the leg shown in Fig. 9–47. Assume the leg (with cast) has a mass of 15.0 kg, and its CG is 35.0 cm from the hip joint; the cord holding the sling is 78.0 cm from the hip joint.


A uniform meter stick with a mass of 180 g is supported horizontally by two vertical strings, one at the 0-cm mark and the other at the 90-cm mark (Fig. 9–82). What is the tension in the string (a) at 0 cm? (b) at 90 cm?


(III) Four bricks are to be stacked at the edge of a table, each brick overhanging the one below it, so that the top brick extends as far as possible beyond the edge of the table. (a) To achieve this, show that successive bricks must extend no more than (starting at the top) \(\frac{{\bf{1}}}{{\bf{2}}}{\bf{,}}\frac{{\bf{1}}}{{\bf{4}}}{\bf{,}}\frac{{\bf{1}}}{{\bf{6}}}\) and \(\frac{{\bf{1}}}{{\bf{8}}}\)of their length beyond the one below (Fig. 9–75a). (b) Is the top brick completely beyond the base? (c) Determine a general formula for the maximum total distance spanned by n bricks if they are to remain stable. (d) A builder wants to construct a corbeled arch (Fig. 9–75b) based on the principle of stability discussed in (a) and (c) above. What minimum number of bricks, each 0.30 m long and uniform, is needed if the arch is to span 1.0 m?

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.