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Get Full Access to Physics - 4 Edition - Chapter 29 - Problem 78
Get Full Access to Physics - 4 Edition - Chapter 29 - Problem 78

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# Predict/Explain An uncharged capacitor is charged by ISBN: 9780321611116 152

## Solution for problem 78 Chapter 29

Physics | 4th Edition

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Problem 78

Predict/Explain An uncharged capacitor is charged by moving some electrons from one plate of the capacitor to the other plate. (a)Is the mass of the charged capacitor greater than, less than, or the same as the mass of the uncharged capacitor? (b) Choose the best explanation from among the following: I. The charged capacitor has more mass because it is storing energy within it, just like a compressed spring. II. The charged capacitor has less mass because some of its mass now appears as the energy of the electric eld between its plates. III. The capacitor has the same mass whether it is charged or not because charging it only involves moving electrons from one plate to the other without changing the total number of electrons.

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ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16) Pedigrees  Foundational Information To Understand Inheritance o The genotype of the parents o How genes interact o Mendel’s Principles o Information on genotype and allele frequency  Pedigree Structure and Information o Inbreeding  When two genetically related animals breed.  Inbred Pedigree  When a common ancestor is on the top and bottom of a pedigree.  Crossing in an inbred line gives you hybrid vigor. Ex. AAbbCC x aaBBcc F1Generation: AaBbCc o Line Breeding  Picking someone famous to breed to gives a pedigree more identity. ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16)  Other Views of Pedigrees (Mendelian Flow Diagram) o Additional Symbols You Might See  Curly Calf Syndrome In Cattle and Pedigrees Animal Bovine (Cattle) Trait Curly Calf Syndrome (Arthrogryposis Multiplex Congenita or AM) Phenotype(s) Can give rise to a healthy born calf or a still-born calf Involved Chromosomes N/A Involved Genes N/A Biology/Physiology Neurologic The spine is bent and twisted in affected calves. Mutation(s) (Genotype) Homozygous Recessive AMF=Normal AMC=Carrier Can It Be Tested For Through Yes A DNA Test Cost To Perform DNA Test \$15-\$25 ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16) o Transforming GAR Grid Maker Pedigree Into Mendelian Flow Diagram ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16) Inheritance and Probability  Foundational Information To Understand Inheritance o The genotype of the parents o How genes interact o Mendel’s Principles o Information on genotype and allele frequency  Mendelian Inheritance Reflects Rules of Probability o You can relate Mendel’s laws to that of the probability of flipping a coin or rolling a dice.  Probability is ranked on a scale from zero to one.  An event is ranked at zero when an event has no chance of occurring.  An event is ranked at one, when an event is certain to take place.  The Probability of Tossing A Coin o One has a ½ or a 50% chance of flipping heads or tails on a coin.  The Probability of Rolling a Dice o One has 1/6 or a 16% chance in rolling a specific number (ex. 6) on a six- sided dice. o One has 5/6 or an 83% chance of rolling any other number than the specific number (for ex. 6) on a six-sided dice.  Probability and Segregation of Gametes o Just like a coin toss, where each toss is an independent event, the distribution of alleles is random, where each gamete from a heterozygous parent has 50% chance of inheriting the dominant form of the allele and a 50% chance of inheriting the recessive allele.  Rules of Multiplication o How do I use the rules of multiplication  Calculate the probability of each independent event, then multiply the individual probabilities together in order to obtain the final probability that these events will occur together.  Ex. A heterozygous polled cow (Pp) mates with a heterozygous pulled bull (Pp). What is the probability that they with produce a horned calf (pp)  From the cow whose alleles are Pp, she has 50% chance of passing the small p down her offspring. The bull whose alleles are also Pp, he too has a 50% ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16) chance of passing down the small p. So since each parent has a 50 % or ½ chance of passing down the small, you multiply ½ times ½ and you get ¼. Thus, the calf has a 25% or ¼ chance of being horned. You can double check your work with a Punnett square. o When do I use the rules of multiplication  When you want to know the probability that two or more independent events will occur in a specific combination.  Curly Calf Syndrome In Cattle and Probability o What is the probability that GAR Grid Maker with be free from Curly Calf Syndrome GAR Grid Maker Possible Genotypic Ratio: 1:1 (2 FF:FC) Grid Maker Has A 50% Chance of Either Being Curly Calf Syndrome Free or a 50% chance of Being a Carrier of Curly Calf Syndrome ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16)  Rules of Multiplication And Di-Hybrid Crosses o Allows us to predict the probability of th1 F generation without the need to construct a 16-part Punnett square.  Ex. When mating two heterozygous parents with the same genotypes of BbPp what is the probability of producing offspring with the gamete BP  Since each parent has a 50% of producing a big B and big P, then you multiply ½ and ½ and you get ¼. Meaning the offspring will have a ¼ or 4/16 chance in inheriting the BP genotype.  Ex. When mating two heterozygous parents with the same genotypes of BbPp what is the probability of producing offspring with the gamete BBPP  Since the dam has a 50% chance of producing a big B and big P, then you multiply ½ and ½ and you get ¼. Since the sire also has a 50% chance of producing a big B and big P, then you multiply ½ and ½ and you get ¼. When you multiply ¼ and ¼ you get 1/16, meaning the offspring will have a 1/16 chance of having the BBPP gamete.  Rule Of Addition o Use the rules of addition when the probability of an event can occur in two of more different ways.  Ex. Probability of a heterozygote.  There is 50% chance that the dominant allele could come from the sperm, and a 50% chance the recessive allele can come from the egg. Therefore, multiplying ½ and ½ gets you ¼. OR  There is 50% chance that the dominant allele could come from the egg, and a 50% chance the recessive allele can come from the sperm. Therefore, multiplying ½ and ½ gets you ¼.  So the probability of a heterozygote is the addition of ½ from the first event possibility and ½ from the second event possibility. ½ + ½= ¼  Combining Rules of Multiplication and Addition to Solve Complex Problems in Simple Traits o First use the rule of multiplication to calculate the probability for each of the genotypes and then use the rule of addition to calculate the probabilities of the recessive traits. ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16)  Ex. Determine the probability of producing an offspring with the genotypes of bbppCc, Bbppcc, bbPpc, bbPPcc, and bbppcc. Then, find the probability of two recessive phenotypes for at least two of three resulting from a tri- hybrid cross in cattle that are BpPpCc (dam) and bbPpcc (sire).  An example on how to find the probability of one of the above genotypes. What is the probability of producing an offspring with the genotype of bbppCc with parents that have the following genotypes BpPpCc (dam) and bbPpcc (sire)  The probability of producing bb is a ½ (50%) chance from the dam and 1 (100%) chance from the sire. So by multiplying ½ and 1 you get ½.  The probability of producing pp is a ½ (50%) chance from the dam and a ½ (50%) chance from the sire. So by multiplying ½ and ½ you get ¼.  The probability of producing Cc is a ½ (50%) chance from the dam and 1 (100%) chance from the sire. So by multiplying ½ and 1 you get ¼. Thus, the probability of producing an offspring with the genotype bbppCc is ½ x ¼ x ½ which equals 1/16. Therefore there is a 1/16 chance in that the offspring will have the bbppCc genotype.  After calculating the probability of the remaining genotypes (bbppCc, Bbppcc, bbPpc, bbPPcc, and bbppcc), find the probability of two recessive phenotypes for at least two of three resulting from a tri-hybrid cross in cattle that are BpPpCc (dam) and bbPpcc (sire).  From the calculation above the probability of producing an offspring with the genotype bbppCc is ½ x ¼ x ½ which equals 1/16.  The probability of an offspring having the genotype Bbppcc is ½ x ¼ x ½ which equals 1/16. ANEQ 328 Foundations In Animal Genetics Week 10 Notes (3/22/16-3/24/16)  The probability of an offspring having the genotype bbPpcc is ½ x ½ x ½ which equals 2/16.  The probability of an offspring having the genotype bbPPcc is ½ x ¼ x ½ which equals 1/16.  The probability of an offspring having the genotype bbppcc is ½ x ¼ x ½ which equals 1/16. Therefore the probability of an offspring possessing two recessive traits is 1/16 + 1/16 + 2/16 + 1/16 + 1/16 which equals 6/16.

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##### ISBN: 9780321611116

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