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i = +3.0 m/s; ∆t = 2.0 s; a = +1.2 m/s2; F f = 0 N Required: F a; ∆d; W (a) The force exerted by the string on the cart: Analysis: F net = ma Solution: F net =ma F a!F f =ma F a!0=(0.50 kg)(1.2 m/s2) =0.6kgim/s2 =0.6 N F a =0.60 N Statement: The string exerts a force of 0.60 N on the cart. (b) The displacement of the cart: Analysis: ∆d = v i∆t + 1 2 a∆t2 Solution: First, find v 2. a



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____ 1. A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface. What work is done by the 100-N force? a. 405 J b. 500 J c. 900 J A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface.



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Title: Microsoft Word - Phys 11U Ch5 Section5s3.doc Author: Eileen Jung Created Date: 1/5/2011 11:25:28 AM

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Chapter 5 Work Energy Power and Society

11. True Understanding 12. The formula for momentum is #! p = m! v. Thus, the units are [kg]! m s " # $ % & ' or "kg!m/s$%. The units for force multiplied by time are "#kg!m/s2 $%![s] "or #kg!m/s$%. The units are equal. 13. Answers may vary. Sample answer: The stretchiness of the cloth lengthens the time interval of the collision, reducing the average force exerted on it by the watermelon. 14

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Section 5.3 Types of Energy (b) Total energy of ball at

i = +3.0 m/s; ∆t = 2.0 s; a = +1.2 m/s2; F f = 0 N Required: F a; ∆d; W (a) The force exerted by the string on the cart: Analysis: F net = ma Solution: F net =ma F a!F f =ma F a!0=(0.50 kg)(1.2 m/s2) =0.6kgim/s2 =0.6 N F a =0.60 N Statement: The string exerts a force of 0.60 N on the cart. (b) The displacement of the cart: Analysis: ∆d = v i∆t + 1 2 a∆t2 Solution: First, find v 2. a

panchbhaya.weebly.com uploads 1 3 7 0 13701351 phys11_sm_05_1 pdf

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Answers to Conceptual Integrated Science End-of-Chapter Questions Chapter 1: About Science Answers to Chapter 1 Review Questions 1 The era of modern science in the 16th century was launched when Galileo Galilei revived the

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Section 5.3 Types of Energy (b) Total energy of ball at

i = +3.0 m/s; ∆t = 2.0 s; a = +1.2 m/s2; F f = 0 N Required: F a; ∆d; W (a) The force exerted by the string on the cart: Analysis: F net = ma Solution: F net =ma F a!F f =ma F a!0=(0.50 kg)(1.2 m/s2) =0.6kgim/s2 =0.6 N F a =0.60 N Statement: The string exerts a force of 0.60 N on the cart. (b) The displacement of the cart: Analysis: ∆d = v i∆t + 1 2 a∆t2 Solution: First, find v 2. a

panchbhaya.weebly.com uploads 1 3 7 0 13701351 phys11_sm_05_1 pdf

Nelson Chemistry 12 3.3 Answers fullexams.com

____ 1. A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface. What work is done by the 100-N force? a. 405 J b. 500 J c. 900 J A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface.

panchbhaya.weebly.com uploads 1 3 7 0 13701351 phys11_sm_05_1 pdf

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____ 1. A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface. What work is done by the 100-N force? a. 405 J b. 500 J c. 900 J A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface.

Panchbhaya.weebly.com uploads 1 3 7 0 13701351 phys11_sm_05_1 pdf - Section 5.3 Types of Energy (b) Total energy of ball at

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