College Physics
College Physics
OER 2016 Edition
ISBN: 9781947172173
Author: OpenStax
Publisher: OpenStax College
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Chapter 9, Problem 28PE
To determine

The direction and the magnitude of the force exerted by the kneecap on the upper leg bone.

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4.) The diagram shows the electric field lines of a positively charged conducting sphere of radius R and charge Q. A B Points A and B are located on the same field line. A proton is placed at A and released from rest. The magnitude of the work done by the electric field in moving the proton from A to B is 1.7×10-16 J. Point A is at a distance of 5.0×10-2m from the centre of the sphere. Point B is at a distance of 1.0×10-1 m from the centre of the sphere. (a) Explain why the electric potential decreases from A to B. [2] (b) Draw, on the axes, the variation of electric potential V with distance r from the centre of the sphere. R [2] (c(i)) Calculate the electric potential difference between points A and B. [1] (c(ii)) Determine the charge Q of the sphere. [2] (d) The concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common terminology to describe different types of fields. [1]
3.) The graph shows how current I varies with potential difference V across a component X. 904 80- 70- 60- 50- I/MA 40- 30- 20- 10- 0+ 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 VIV Component X and a cell of negligible internal resistance are placed in a circuit. A variable resistor R is connected in series with component X. The ammeter reads 20mA. 4.0V 4.0V Component X and the cell are now placed in a potential divider circuit. (a) Outline why component X is considered non-ohmic. [1] (b(i)) Determine the resistance of the variable resistor. [3] (b(ii)) Calculate the power dissipated in the circuit. [1] (c(i)) State the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P. [1] (c(ii)) Describe, by reference to your answer for (c)(i), the advantage of the potential divider arrangement over the arrangement in (b).
1.) Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A. The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10-s N. (a) State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P. [2] (b) Calculate the current in wire Q. [2] (c) Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown. wire P wire R wire Q 0.05 m 0.05 m The net magnetic force on wire Q is now zero. (c.i) State the direction of the current in R, relative to the current in P.[1] (c.ii) Deduce the current in R. [2]

Chapter 9 Solutions

College Physics

Ch. 9 - Why are the forces exerted on the outside world by...Ch. 9 - Explain why the forces in our joints are several...Ch. 9 - Why are the forces exerted on the outside world by...Ch. 9 - Explain why the forces in our joints are several...Ch. 9 - Certain of dinosaurs were bipedal (walked on two...Ch. 9 - Swimmers and athletes during competition need to...Ch. 9 - If the maximum force the biceps muscle can exert...Ch. 9 - Suppose the biceps muscle was attached through...Ch. 9 - Explain one of the reasons why pregnant women...Ch. 9 - (a) When opening a door, you push on it...Ch. 9 - When tightening a bolt, you push perpendicularly...Ch. 9 - Two children push on opposite sides of a door...Ch. 9 - Use the second condition for equilibrium (net =0 )...Ch. 9 - Repeat the seesaw problem in Example 9.1 with the...Ch. 9 - Prob. 6PECh. 9 - Two children of mass 20.0 kg and 30.0 kg sit...Ch. 9 - Prob. 8PECh. 9 - A person carries a plank of wood 2.00 m long with...Ch. 9 - Prob. 10PECh. 9 - Prob. 11PECh. 9 - Prob. 12PECh. 9 - Prob. 13PECh. 9 - Prob. 14PECh. 9 - Prob. 15PECh. 9 - Prob. 16PECh. 9 - To get up on the roof, a person (mass 70.0 kg)...Ch. 9 - Prob. 18PECh. 9 - Prob. 19PECh. 9 - Suppose you needed to raise a 250-kg mower a...Ch. 9 - Prob. 21PECh. 9 - Prob. 22PECh. 9 - Prob. 23PECh. 9 - Prob. 24PECh. 9 - Prob. 25PECh. 9 - Prob. 26PECh. 9 - Prob. 27PECh. 9 - Prob. 28PECh. 9 - Prob. 29PECh. 9 - Prob. 30PECh. 9 - Prob. 31PECh. 9 - Prob. 32PECh. 9 - Prob. 33PECh. 9 - Prob. 34PECh. 9 - Prob. 35PECh. 9 - Integrated Concepts Suppose we replace the 4.0-kg...Ch. 9 - Prob. 37PECh. 9 - You have just planted a sturdy 2-m-tall palm tree...Ch. 9 - Unreasonable Results Suppose two children are...Ch. 9 - Construct Your Own Problem Consider a method for...Ch. 9 - Prob. 1TPCh. 9 - Prob. 2TPCh. 9 - Prob. 3TPCh. 9 - Prob. 4TPCh. 9 - Prob. 5TPCh. 9 - Prob. 6TPCh. 9 - Prob. 7TPCh. 9 - Prob. 8TPCh. 9 - Prob. 9TPCh. 9 - Prob. 10TPCh. 9 - Prob. 11TP
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