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Estimate the number of steps you would have to take to walk a distance equal to the circumference of the Earth.

Estimate the number of steps you would have to take to walk a distance equal to the circumference of the Earth.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Just wanted to check if my answers are correct :) please let me know if I’ve made any mistakes.
1. Estimate the number of steps you would have to take to walk a distance equal to the circumference of the Earth. The circumference of Earth is approx. 40,075 km. The average step length for an adult is approx. 0.75 meters (2.5 feet). We divide the distance by the step length: 40,075,000 meters / 0.75 meters/step = 53,433,333 steps Answer: 1, 53,433,333 steps is needed if we take 2.5 feet per step. 2. Treat a cell in a human as a sphere of radius 1.0 μm. (a) Determine the volume of a cell - 4.189 x 10^-18 m^3 (b) Estimate the volume of your body - approx. 66 x 10^-3 m^3 (c) Estimate the number of cells in your body - 1.569 x 10^14 cells is the estimated cells in the body.
Transcribed Image Text:1. Estimate the number of steps you would have to take to walk a distance equal to the circumference of the Earth. The circumference of Earth is approx. 40,075 km. The average step length for an adult is approx. 0.75 meters (2.5 feet). We divide the distance by the step length: 40,075,000 meters / 0.75 meters/step = 53,433,333 steps Answer: 1, 53,433,333 steps is needed if we take 2.5 feet per step. 2. Treat a cell in a human as a sphere of radius 1.0 μm. (a) Determine the volume of a cell - 4.189 x 10^-18 m^3 (b) Estimate the volume of your body - approx. 66 x 10^-3 m^3 (c) Estimate the number of cells in your body - 1.569 x 10^14 cells is the estimated cells in the body.
ed States) HW. 1.8 Dimensional Analysis 1. Each of the following equations was given by a student during an examination (a) 1/2mv² = 1/2mv² + √mgh (b) v = v₁ + at² (c) ma = 22 Do a dimensional analysis of each equation and explain why the equation can't be correct. (Hint: Determine the dimensions on each side of the equation and compare them.) a) Dimension of mv^2/2=M[LT^-1]^2=[ML^2T^-2] Similarly dimension of mv0^2/2=[ML^2T^-2] Homework Assignment #1 PH-270-IS1-Spring-2023 Dimension of √(mgh)=[MLT^-2L1^1/2=[M^1/2LT^-1] Therefore, dimensions on each side are not same. b) v=v0+at^2 Dimension of v & VO is [LT^-1] Dimension of at^2 is [LT^-1T^2]=[L] Therefore dimensions on each side are not same. c) dimension of ma is [MLT^-2] Dimension of v^2 is [LT^-1]^2=[[L^2T^-2] HW1.6 Focus
Transcribed Image Text:ed States) HW. 1.8 Dimensional Analysis 1. Each of the following equations was given by a student during an examination (a) 1/2mv² = 1/2mv² + √mgh (b) v = v₁ + at² (c) ma = 22 Do a dimensional analysis of each equation and explain why the equation can't be correct. (Hint: Determine the dimensions on each side of the equation and compare them.) a) Dimension of mv^2/2=M[LT^-1]^2=[ML^2T^-2] Similarly dimension of mv0^2/2=[ML^2T^-2] Homework Assignment #1 PH-270-IS1-Spring-2023 Dimension of √(mgh)=[MLT^-2L1^1/2=[M^1/2LT^-1] Therefore, dimensions on each side are not same. b) v=v0+at^2 Dimension of v & VO is [LT^-1] Dimension of at^2 is [LT^-1T^2]=[L] Therefore dimensions on each side are not same. c) dimension of ma is [MLT^-2] Dimension of v^2 is [LT^-1]^2=[[L^2T^-2] HW1.6 Focus
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