Essential University Physics (3rd Edition)

3rd Edition

ISBN: 9780134202709

Author: Richard Wolfson

Publisher: PEARSON

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

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Textbook Question

Chapter 22, Problem 24E

The classical picture of the hydrogen atom has the electron or-biting 0.0529 nm from the proton. What’s the electric potential associated with the proton’s electric field at this distance?

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Chapter 22, Problem 23E

Chapter 22, Problem 25E

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Please help me on the following question (Please refer to image below)An Olympic lifter (m = 103kg) is holding a lift with a mass of 350 kg. The barexerts a purely vertical force that is equally distributed between both hands. Each arm has amass of 9 kg, are 0.8m long and form a 40° angle with the horizontal. The CoM for each armis 0.5 m from hand. Assuming the lifter is facing us in the diagram below, his right deltoidinserts 14cm from the shoulder at an angle of 13° counter-clockwise from the humerus.A) You are interested in calculating the force in the right deltoid. Draw a free body diagramof the right arm including the external forces, joint reaction forces, a coordinate system andstate your assumptions.B) Find the force exerted by the right deltoidC) Find the shoulder joint contact force. Report your answer using the magnitude and directionof the shoulder force vector.

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Question 6: Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of Cl₂ gas is 8.70 L at 895 torr and 24°C. (a) How many grams of Cl₂ are in the sample? ⚫ Atomic mass of CI = 35.453 g/mol • Molar mass of Cl₂ = 2 x 35.453 = 70.906 g/mol Solution: Use the Ideal Gas Law: Step 1: Convert Given Values • Pressure: P = 895 torr → atm PV= = nRT 1 P = 895 × = 1.1789 atm 760 • Temperature: Convert to Kelvin: T24273.15 = 297.15 K • Gas constant: R = 0.0821 L atm/mol. K Volume: V = 8.70 L Step 2: Solve for n . PV n = RT n = (1.1789)(8.70) (0.0821)(297.15) 10.25 n = = 0.420 mol 24.405 Step 3: Calculate Mass of Cl₂ Final Answer: 29.78 g of Cl₂. mass nx M mass= (0.420)(70.906) mass= 29.78 g

Chapter 22 Solutions

Essential University Physics (3rd Edition)

Ch. 22.1 - What would happen to the potential difference Vab...Ch. 22.1 - (1) A proton (charge e), (2) an alpha particle...Ch. 22.1 - The figure shows three straight paths AB of the...Ch. 22.2 - You measure a potential difference of 50 V between...Ch. 22.2 - The figure shows three paths from infinity to a...Ch. 22.3 - The figure shows cross sections through two...Ch. 22 - Why can a bird perch on a high-voltage power line...Ch. 22 - One proton is accelerated from rest by a uniform...Ch. 22 - Would a free electron move toward higher or lower...Ch. 22 - The electric Field at the center of a uniformly...

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