Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

An 600 mH inductor and a 3 W resistor are connected in series with a 1200-V dc power suppl, Quizzes of Physics

A mathematical problem where a piece of rope is cut into pieces and used to form a circle and a square. The function representing the area of the circle as a function of the length of one side of the square is derived. The solution involves using the formulas for circumference of a circle and perimeter of a square, and solving for the radius of the circle. The area of the circle is then expressed as a function of the length of one side of the square. step-by-step instructions and equations to arrive at the solution.

Typology: Quizzes

2017/2018

Available from 05/05/2022

roneil-algara
roneil-algara 🇵🇭

27 documents

1 / 1

Toggle sidebar

Related documents


Partial preview of the text

Download An 600 mH inductor and a 3 W resistor are connected in series with a 1200-V dc power suppl and more Quizzes Physics in PDF only on Docsity! A piece of rope that is 18 feet long is cut into pieces. One piece is used to form a circle and the other used to form a square. Write a function f representing the area of the circle as a function of the length of one side of the square s. Hint : If C is the diameter of the circle and P is the perimeter of the square, then C + P = 18. Answer f(s) = (9-2s)2/π , the area of the circle as a function of the length of one side of the square. Explanation Let Length of the rope= 18 ft C = circumference of circle P = perimeter of square s = side of the square r = radius of the circle Required f(s) = ? , Area of circle as a function of the length of one side of the square Solution C = 2πr , circumference of a circle P = 4s , perimeter of a square C+P = 18 ; substitute the formulas 2πr + 4s = 18 ; get the expression for radius,r ; 2πr = 18-4s r = (18-4s)/2π ; simplifying by dividing by 2 r = (9-2s)/π >>equation 1 Solve for the area of circle as a function of the length of one side of the square, f(s) ; f(s) = πr2 ; substitute equation 1 to the r = π((9-2s)/π))2 = π (9-2s)2/π2 = π (81-36s+4s2)/π2 f(s) = (81-36s+4s2)/π f(s) = (9-2s)2/π , the area of the circle as a function of the length of one side of the square.
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved