Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

- StudyBlue
- Arizona
- Arizona State University - Tempe
- Physics
- Physics 131
- Adams/menendez
- [SOLUTIONS] Mastering Physics HW1.pdf

Anonymous

Advertisement

HW1 Due: 11:59pm on Monday, August 31, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] Beginning with HW1, there ia a 3% penalty for each incorrect answer. So if you have 5 incorrect answers on a 10 point problem, and then your sixth answer is correct, your credit for that problem will be 8.5 points instead of 10. GBA Problem 21.1 Excess electrons are placed on a small lead sphere with a mass of 7.95 so that its net charge is 3.00×10 ?9 . Part A Find the number of excess electrons on the sphere. Use 1.60×10 ?19 for the magnitude of the charge on an electron. ANSWER: 1.88×10 10 Correct Part B How many excess electrons are there per lead atom? The atomic number of lead is 82, and its molar mass is 207 . Use 6.02×10 23 for Avagadro's number. ANSWER: 8.11×10 ?13 Correct Charged Aluminum Spheres Two small aluminum spheres, each of mass 0.0250 kilograms, are separated by 80.0 centimeters. Part A How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 grams per mole, and its atomic number is 13.) Hint A.1 The definition of mole and atomic number In one mole ( ) of any material, there are approximately atoms present. The number of atoms per mole is called Avogadro's number. The atomic number of an element is the number of protons (and therefore also the number of electrons) in an atom of that element. Hint A.2 How many electrons per mole of aluminum? How many electrons are there in a mole of aluminum? Express your answer numerically. ANSWER: 7.83×10 24 Correct Hint A.3 How many electrons per kilogram of aluminum? How many electrons are in a kilogram of aluminum? Express your answer numerically. ANSWER: 2.90×10 26 Correct Express your answer numerically. ANSWER: 7.25×10 24 Correct Part B How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude (roughly one ton)? Assume that the spheres may be treated as point charges. Hint B.1How to approach the problem Use Coulomb's law to find the charge needed to produce the given force. Then use the charge of an electron to determine the number of electrons necessary to produce the calculated charge. Hint B.2Find the relationship between the charges of the spheres Assume that after some electrons have been removed from it, the first sphere ends up with a net charge of . What would be the charge on the other sphere, , after these extra electrons are added to it? Express your answer in terms of and any necessary constants. ANSWER: = Answer not displayed Express your answer numerically. [ Print ] ANSWER: 5.27×10 15 Correct Part C What fraction of all the electrons in one of the spheres does this represent? Express your answer numerically. ANSWER: 7.26×10 ?10 Correct Placing Charges Conceptual Question Below are free-body diagrams for three electric charges that lie in the same plane. Their relative positions are unknown. Part A Along which of the lines (A to H) in the figure should charge 2 be placed so that the free-body diagrams of charge 1 and charge 2 are consistent? Hint A.1 How to approach the problem Newton?s 3rd law states that the forces exerted by a pair of objects on each other are always equal in magnitude and opposite in direction. Identifying the forces that correspond to 3rd-law pairs in the free-body diagrams will enable you to place the particles in their proper relative position. Hint A.2 Placing charge 2 The two forces acting on charge 2 correspond to the forces exerted on it by charge 1 and charge 3. This means that one of these forces must pair with a force on charge 1 of equal magnitude and opposite direction and the other must pair with a force on charge 3 of equal magnitude and opposite direction. Also note that charge 2 should be repelled by charge 1, since both are negative. Therefore, the vector that represents the force of charge 1 on charge 2 must point away from charge 1. This information is all you need to place charge 2 in its correct position. ANSWER: Correct Part B Along which of the lines (A to H) in the figure should charge 3 be placed so that the free-body diagrams of charge 1, charge 2, and charge 3 are consistent? ANSWER: D Correct Part C Along which of the lines (A to H) in the figure should charge 2 be placed so that the free-body diagrams of charge 1 and charge 2 are consistent? ANSWER: H Correct Part D Along which lines (A to H) in the figure should charge 3 be placed so that the free-body diagrams of charge 1, charge 2, and charge 3 are consistent? ANSWER: F Correct Coulomb's Law Tutorial Learning Goal: To understand how to calculate forces between charged particles, particularly the dependence on the sign of the charges and the distance between them. Coulomb's law describes the force that two charged particles exert on each other (by Newton's third law, those two forces must be equal and opposite). The force exerted by particle 2 (with charge ) on particle 1 (with charge ) is proportional to the charge of each particle and inversely proportional to the square of the distance between them: , where and is the unit vector pointing from particle 2 to particle 1. The force vector will be parallel or antiparallel to the direction of , parallel if the product and antiparallel if ; the force is attractive if the charges are of opposite sign and repulsive if the charges are of the same sign. Part A Consider two positively charged particles, one of charge (particle 0) fixed at the origin, and another of charge (particle 1) fixed on the y-axis at . What is the net force on particle 0 due to particle 1? Express your answer (a vector) using any or all of , , , , , , and . ANSWER: = Correct Part B Now add a third, negatively charged, particle, whose charge is (particle 2). Particle 2 fixed on the y-axis at position . What is the new net force on particle 0, from particle 1 and particle 2? Express your answer (a vector) using any or all of , , , , , , , , and . ANSWER: = Correct Part C Particle 0 experiences a repulsion from particle 1 and an attraction toward particle 2. For certain values of and , the repulsion and attraction should balance each other, resulting in no net force. For what ratio is there no net force on particle 0? Express your answer in terms of any or all of the following variables: , , , . ANSWER: = Correct Part D Now add a fourth charged particle, particle 3, with positive charge , fixed in the yz-plane at . What is the net force on particle 0 due solely to this charge? Hint D.1 Find the magnitude of force from particle 3 Hint not displayed Hint D.2 Vector components Hint not displayed Express your answer (a vector) using , , , , , , and . Include only the force caused by particle 3. ANSWER: = Correct For Hanging Charged Spheres, the force-vector drawing program is rather sensitive. Start by drawing the gravity force vector as large as possible given the specified starting location. Then, to draw the other force vectors accurately, move the gravity vector so as to create a vector triangle indicating that the three vectors actually sum to zero. Finally move all three vectors so that each one starts from the specified location. GBA Hanging Charged Spheres Two small spheres with mass = 15.8 are hung by silk threads of length = 1.28 from a common point . When the spheres are given nonequal charges, so that , each thread hangs at an angle = 23.0 from the vertical. Part A Draw a diagram showing the forces on the left sphere. Treat the spheres as point charges. Draw the force vectors with their tails at the left sphere. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: View Correct Part B Draw a diagram showing the forces on the right sphere. Treat the spheres as point charges. Draw the force vectors with their tails at the right sphere. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: View Correct Part C Find the magnitude of . Give your answer in . ANSWER: = 2.21 Correct Part D Find the magnitude of . Give your answer in . ANSWER: = 3.31 Correct Problem 21.78 Part A Suppose all the electrons in a quantity of carbon atoms with a mass of 18.0 were located at the North Pole of the earth and all the protons at the South Pole. What would be the total force of attraction exerted on each group of charges by the other? The atomic number of carbon is 6, and the atomic mass of carbon is 12.0 . Use 8.85×10 ?12 for the permittivity of free space, 1.60×10 ?19 for the magnitude of the charge on an electron, 6.02×10 23 for Avagadro's number, and 6.38×10 6 for radius of the earth. ANSWER: 4.16×10 7 Correct N Part B What would be the magnitude of the force exerted by the charges in part (a) on a third charge that is equal to the charge at the South Pole, and located at a point on the surface of the earth at the equator? Use 8.85×10 ?12 for the permittivity of free space, 1.60×10 ?19 for the magnitude of the charge on an electron, 6.02×10 23 for Avagadro's number, and 6.38×10 6 for radius of the earth. ANSWER: 1.18×10 8 Correct N Part C What would be the direction of this force? ANSWER: south-to-north direction north-to-south direction Correct Score Summary: Your score on this assignment is 97.2%. You received 48.62 out of a possible total of 50 points. clockwork MasteringPhysics

Advertisement

Want to see the other 5 page(s) in [SOLUTIONS] Mastering Physics HW1.pdf?
JOIN TODAY FOR FREE!

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly!Â I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU
StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2015 StudyBlue Inc. All rights reserved.

© 2015 StudyBlue Inc. All rights reserved.