ZPH 123 Problem set Forces 1. Two 25.0-N weights are suspended at opposite ends of a rope that passes over a light‚ frictionless pulley. The pulley is attached to a chain that goes to the ceiling. a. What is the tension in the rope? b. What is the tension in the chain? 2. A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with constant speed of 3.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.20. a. What horizontal force must be applied by
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Multiple Choice Question 6.31 A 1000-kg car moving at 10 m/s brakes to a stop in 5 s. The average braking force is 3000 N 5000 N 2000 N ***(answer) 1000 N 4000 N Force = mass x acceleration. Acceleration = velocity/time = -10/5 = -2 m/s/s. (- sign means a deceleration from velocity of 10 to 0) Force = 1000 x -2 = -2000 Newtons (i.e. 2000N in opposite direction to motion) Multiple Choice Question 6.11 When you jump from an elevated position you usually bend your knees upon reaching
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[pic] AP® Physics C 2002 Free response Questions The materials included in these files are intended for use by AP teachers for course and exam preparation in the classroom; permission for any other use must be sought from the Advanced Placement Program®. Teachers may reproduce them‚ in whole or in part‚ in limited quantities‚ for face-to-face teaching purposes but may not mass distribute the materials‚ electronically or otherwise. These materials and any copies made of them
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Page 12 1 a) Between t = 30 and t = 45 mins b) 7.5 mins c) i) distance travelled = area under graph between t = 0 and t = 12½ mins ii) average speed = total distance travelled total time for journey = total area under graph 60 mins 2 a) ∆v = 32 m/s a =10 m/s² t = ∆v = 32 = 3.2 s a 10 b) 3 a) OP – constant acceleration PQ – constant acceleration (greater than OP) QR – constant speed RS – constant deceleration b) O and S c) 6 m/s d) 70 s e) Total distance travelled = area under
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1 Problem 1-4. An Analysis of the transactions made by Acme Consulting for the month of July 1. Explain each transaction. 1) $20‚000 is the amount invested by the owner/s into the business 2) $7‚000 is the cost of the equipment purchased for the business. $5‚000 has been paid. $2‚000 is the amount still owed for the purchase. 3) $1‚000 is the amount expended to purchase inventories and supplies for the business. 4) $4‚500 is the amount paid for
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Pendulum Problems ACTIVITY 1: Copy and paste the example problem and the steps‚ so that the steps are in the correct order into a new Word document and upload it to Moodle. Example Problem 1: A hypnotist swings her watch from 20.0cm chain in front of a subject’s eyes. What is the period of the swing of the watch. Thus‚ we see that the pendulum used by the hypnotist has a period of 0.898s. | Before we can use this formula‚ however‚ we must ensure all our variables are in the correct units.
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Conservative force - Any force which conserves mechanical energy‚ as opposed to a nonconservative force. See statement of conservation of mechanical energy. Conservative System - A system in which energy is conserved. Energy - The ability to do work. Kinetic Energy - The energy of motion. Nonconservative Force - Any force which does not conserve mechanical energy‚ as opposed to a conservative force. Path independence - Property of conservative forces which states that the work done
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PHYS 101-2 STUDY GUIDE 1. The study of stationary or resting charges. ____________________________ 2. ________________________ is a connection made to Earth. 3. ________________________ is the process by which electrons are added or removed from a body. 4. List the three modes of electrification. __________________________________________________________________________________________________________________________________________________________________________________________
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Completion Problems (Elasticity) 1. A steel wire of length 4.7m and cross-sectional area 3×10-5m2 stretches by the same amount as a copper wire of length 3.5m and cross-sectional area of 4×10–5m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper? For Steel l1= 4.7 m‚ a1 = 3.0 x 10-5 m2 Y1= Fl1/a1∆l Y1= (F x 4.7m)/ (3.0 x 10-5 x ∆l) For Copper L2=3.5 m‚ a2 = 4.0 x 10-5 m2 Y2= (F x 3.5m)/ (4.0 x 10-5 x ∆l) Dividing Steel by Copper Y1/ Y2
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ANSWERS TO GUIDE QUESTIONS 1. What happens to the distribution of magnetic flux lines when the iron ring was placed in between the U-magnets? Magnetic Flux‚ by definition‚ is the measure of the quantity or the strength of the magnetic force in a given area‚ whether a closed or an open area. When the iron ring is placed in between the U-magnets‚ there has been a region by which has been enclosed‚ following the area of the iron ring. And according to Gauss’s Law for magnetism‚ the total magnetic
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