P.F. 10-1 | Quizalize

Generate all of these 25 questions Part A: Each correct answer is worth 5. 1. The regular pentagon shown has a side length of 2 cm. The perimeter of the pentagon is (A) 2 cm (B) 4 cm (C) 6 cm (D) 8 cm (E) 10 cm 2 cm 2. The faces of a cube are labelled with 1, 2, 3, 4, 5, and 6 dots. Three of the faces are shown. What is the total number of dots on the other three faces? (A) 6 (B) 8 (C) 10 (D) 12 (E) 15 3. The equation that best represents \a number increased by _ve equals 15" is (A) n 􀀀 5 = 15 (B) n _ 5 = 15 (C) n + 5 = 15 (D) n + 15 = 5 (E) n _ 5 = 15 4. The line graph shows the number of bobbleheads sold at a store each year. The sale of bobbleheads increased the most between (A) 2016 and 2017 (B) 2017 and 2018 (C) 2018 and 2019 (D) 2019 and 2020 (E) 2020 and 2021 Number of 2016 2017 2018 2019 2020 Year Sale of Bobbleheads 2021 Bobbleheads 20 40 60 80 5. Starting at 72, Aryana counts down by 11s: 72; 61; 50; : : : . What is the last number greater than 0 that Aryana will count? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 6. In the diagram, \ABC = 90_. The value of x is (A) 68 (B) 23 (C) 56 (D) 28 (E) 26 Day of the Week 44° x° A B C x° 7. Which of the following values is closest to zero? (A) 􀀀1 (B) 5 4 (C) 12 (D) 􀀀4 5 (E) 0:9 Grade 8 8. A jar contains 267 quarters. One quarter is worth $0.25. How many quarters must be added to the jar so that the total value of the quarters is $100.00? (A) 33 (B) 53 (C) 103 (D) 133 (E) 153 9. A package of 8 greeting cards comes with 10 envelopes. Kirra has 7 cards but no envelopes. What is the smallest number of packages that Kirra needs to buy to have more envelopes than cards? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 10. For the points in the diagram, which statement is true? (A) e > c (B) b < d (C) f > b (D) a < e (E) a > c y x (e, f ) (a, b) (c, d ) Part B: Each correct answer is worth 6. 11. The 26 letters of the English alphabet are listed in an in_nite, repeating loop: ABCDEFGHIJKLMNOPQRSTUVWXY ZABC : : : What is the 258th letter in this sequence? (A) V (B) W (C) X (D) Y (E) Z 12. A public holiday is always celebrated on the third Wednesday of a certain month. In that month, the holiday cannot occur on which of the following days? (A) 16th (B) 22nd (C) 18th (D) 19th (E) 21st 13. A circular spinner is divided into three sections. An arrow is attached to the centre of the spinner. The arrow is spun once. The probability that the arrow stops on the largest section is 50%. The probability it stops on the next largest section is 1 in 3. The probability it stops on the smallest section is (A) 1 4 (B) 2 5 (C) 1 6 (D) 2 7 (E) 3 10 14. A positive number is divisible by both 3 and 4. The tens digit is greater than the ones digit. How many positive two-digit numbers have this property? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 15. A rectangular pool measures 20 m by 8 m. There is a 1 m wide walkway around the outside of the pool, as shown by the shaded region. The area of the walkway is (A) 56 m2 (B) 60 m2 (C) 29 m2 (D) 52 m2 (E) 50 m2 20 m 8 m 1 m Grade 8 16. The results of asking 50 students if they participate in music or sports are shown in the Venn diagram. What percentage of the 50 students do not participate in music and do not participate in sports? (A) 0% (B) 80% (C) 20% (D) 70% (E) 40% Music Sports 15 5 20 17. There are 2 3 as many golf balls in Bin F as in Bin G. If there are a total of 150 golf balls, how many fewer golf balls are in Bin F than in Bin G? (A) 15 (B) 30 (C) 50 (D) 60 (E) 90 18. In the sequence shown, Figure 1 is formed using 7 squares. Each _gure after Figure 1 has 5 more squares than the previous _gure. What _gure has 2022 squares? (A) Figure 400 (B) Figure 402 (C) Figure 404 (D) Figure 406 (E) Figure 408 Figure 1 Figure 2 Figure 3 19. Mateo's 300 km trip from Edmonton to Calgary passed through Red Deer. Mateo started in Edmonton at 7 a.m. and drove until stopping for a 40 minute break in Red Deer. Mateo arrived in Calgary at 11 a.m. Not including the break, what was his average speed for the trip? (A) 83 km/h (B) 94 km/h (C) 90 km/h (D) 95 km/h (E) 64 km/h 20. Equilateral triangle ABC has sides of length 4. The midpoint of BC is D, and the midpoint of AD is E. The value of EC2 is (A) 7 (B) 6 (C) 6:25 (D) 8 (E) 10 Part C: Each correct answer is worth 8. 21. The positive factors of 6 are 1, 2, 3, and 6. There are two perfect squares less than 100 that have exactly _ve positive factors. What is the sum of these two perfect squares? (A) 177 (B) 80 (C) 145 (D) 52 (E) 97 22. In the list p; q; r; s; t; u; v, each letter represents a positive integer. The sum of the values of each group of three consecutive letters in the list is 35. If q + u = 15, then p + q + r + s + t + u + v is (A) 85 (B) 70 (C) 80 (D) 90 (E) 75 Grade 8 23. The net shown is folded to form a cube. An ant walks from face to face on the cube, visiting each face exactly once. For example, ABCFED and ABCEFD are two possible orders of faces the ant visits. If the ant starts at A, how many possible orders are there? (A) 24 (B) 48 (C) 32 (D) 30 (E) 40 A D B C E F 24. The number 385 is an example of a three-digit number for which one of the digits is the sum of the other two digits. How many numbers between 100 and 999 have this property? (A) 144 (B) 126 (C) 108 (D) 234 (E) 64 25. Student A, Student B, and Student C have been hired to help scientists develop a new avour of juice. There are 4200 samples to test. Each sample either contains blueberry or does not. Each student is asked to taste each sample and report whether or not they think it contains blueberry. Student A reports correctly on exactly 90% of the samples containing blueberry and reports correctly on exactly 88% of the samples that do not contain blueberry. The results for all three students are shown below. Student A Student B Student C Percentage correct on samples 90% 98% (2m)% containing blueberry Percentage correct on samples 88% 86% (4m)% not containing blueberry Student B reports 315 more samples as containing blueberry than Student A. For some positive integers m, the total number of samples that the three students report as containing blueberry is equal to a multiple of 5 between 8000 and 9000. The sum of all such values of m is (A) 45 (B) 36 (C) 24 (D) 27 (E) 29

LESSON 4. Cellular Respiration • Define cellular respiration • Identify the stages of clan respiration You have just learned how the energy from the sun is captured, processed, and stored in the form of glucose. Cellular respiration, another important life process, is the means by which cells release the stored energy in glucose to make adenosine triphosphate (ATP). The primary goal of this life process is to convert stored energy into usable form, such as ATP, for the cells to carry out their functions. Cellular respiration involves several chemical reactions. The reactions can be summed up in the following equation: C6 H12 O6 + 602 ----- 6 CO₂ +6H₂O + ATP Glucose oxygen carbon dioxide water energy Aerobic respiration reactions, or cellular respiration that takes place in the presence of oxygen, can be grouped into three stages glycolysis, Krebs cycle, and electron transport chain (ETC). Stage 1: Glycolysis Glycolysis is the process that breaks down one molecule of 6-C glucose into 3-C pyruvates or pyruvic acids. It also releases four molecules of ATP. This process occurs in the cytoplasm of the cell. The following is the step-by-step process of glycolysis. Take note that several enzymes are involved in this process. 1. The first step of glycolysis requires energy. It can only proceed when the two ATP molecules donate energy to the glucose by transferring a phosphate group with the help of an enzyme, producing glucose 6-phosphate 2. Then, a specific enzyme promotes the rearrangement of the atoms, producing the fructose 6-phosphate. 3. The action of the enzyme in step 2 promotes the transfer of a phosphate group from another ATP molecule, forming fructose 1,6-bisphosphate. 4. The resulting fructose 1,6-bisphosphate molecules, with the help of another enzyme, splits into two molecules, each with three carbon backbones. These two sugars are dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. 5. Another important enzyme then rapidly interconverts the molecules of dihydro-xyacetone phosphate and glyceraldehyde 3-phosphate. This produces two molecules of glyceraldehyde 3-phosphate or 3-phosphoglyceraldehyde (PGAL) 6. The succeeding step involves another enzyme-mediated action. The hydrogen (H) from PGAL is transferred to the oxidizing agent, nicotinamide adenine dinucleotide (NAD), which forms NADH. A phosphate (P) is also added from the cytosol of the cell to oxidize the two molecules of PGAL, forming two 1.3-bisphosphoglycerate. 7. A phosphate (P) from 1,3-biphosphoglycerate is transferred to ADP to form ATP. This happens for each of the two 1,3-bisphosphoglycerate. resulting to a yield of two ATP and two 3-phosphoglycerate molecules. 8. A phosphate is transferred from 3-phosphoglycerate molecules from the third carbon to the second carbon, forming 2-phosphoglycerate molecules A hydrogen atom and a hydroxyl ((OH) group is released, which then combines to form water (H2O). The removal of H2O from 2-phosphoglycerate results in the formation of 2- phosphoglycerate molecules. 9. A hydrogen atom and a hydroxyl ((OH) group is released, which then combines to form water (H2O). The removal of H2O from 2-phosphoglycerate results in the formation of two phosphoenolpyruvic acid (PEP) 10. Phosphate (P) from PEP is transferred to ADP (and forms ATP) and the final product, pyruvic acid. This reaction yields two molecules of pyruvic acid and two ATP molecules In summary, a single glucose molecule that undergoes the process of glycolysis produces two molecules of pyruvic acid, four molecules of ATP, two molecules of NADEL and two molecules of H.O. However, only two molecules of ATP are counted as net products since two molecules of ATP are spent throughout the process. Stage II: Krebs Cycle The Krebs cycle, named after its proponent Sir Hans Adolf Krebs, is a cyclical series of enzyme-controlled reactions. This stage of cellular respiration occurs in the matrix of the mitochondria. It is sometimes. called the citric acid cycle (CAC) since it produces citric acid. Citric acid contains three carboxyl (COOH) groups; hence, it is also called the tricarboxylic acid cycle (TCA). This requires the pyruvic acids produced during glycolysis. The main function of this cycle is to produce high-energy-yielding molecules, namely, NADH and flavin adenine dinucleotide (FADH) that will later on be used in the electron transport chain reaction. Figure 6-7. Summary of glycolysis and corresponding products in each reaction presented (See Appendix F on page 285 for an enlarged and complete version of the image.) An initial process is needed for the Krebs cycle to begin. As a pyruvate molecule from glycolysis enters the mitochondrion, it undergoes an important preliminary ate to form acetyl-CoA reaction. Coenzyme-A (COA) combines with pyruvate help of an enzymatic complex. This conversion also produces CO, and NADH. The Krebs cycle is summarized as follows. Take note that several enzymes are involved in this process. 1. The Krebs cycle technically begins when the acetyl-CoA combines with oxaloacetic acid (OAA), a 4-C molecule, to produce citric acid, a 6-C molecule. 2. With the aid of an enzyme, the citric acid now goes through a series of reactions that releases energy. Water molecule is removed from the citric acid and is returned in a different location. The-OH group is repositioned, forming the molecule isocitrate. 3. Isocitrate is then oxidized, forming the a-ketoglutarate, a 5-C molecule. The byproducts of this reaction are NADH and CO, 4 The a-ketoglutarate loses its CO, and a coenzyme-A is added in its place. The decarboxylation occurs with the help of NAD, which then becomes NADH. The resulting molecule is called succinyl-CoA. 5. Succinyl-CoA is converted into succinate. Also in this reaction, a molecule of guanosine triphosphate (GTP) is synthesized. The GTP molecule has similar structure and energy properties to that of ATP and is used by cells the same way. The free phosphate group attacks the succinyl-CoA molecule, which detaches the COA. Then, phosphate is attached to GDP to come up with GTP, similar to the process that occur in ATP synthesis (from ADP to ATP). 6. Two hydrogens are removed from succinate, A molecule of flavin adenine dinucleotide (FAD), a coenzyme similar to NAD, is reduced to FADH, as it takes the hydrogens from the succinate. This reaction produces the fumarate. 7. Fumarate is then converted into malate as the addition of a water molecule is catalyzed. The final reaction is the regeneration of oxaloacetate. The resulting byproduct of this regeneration is NADH Recall that two pyruvate molecules were produced during glycolysis, causing the Krebs cycle to turn twice. Each tuts produces three molecules of NADH, single ATH one FADIH, and the by-product CO, which is exhaled. Stage III: Electron Transport Chain The electron transport chain (ETC) is a series of photon pumps on the inner membrane of the mitochondrion. Electron transport is the last stage of the cellular respiration. In this stage, the energy from NADH and FADH, from the Krebs cycle is transferred to ADP to produce ATP. This process is generally known as oxidative phosphorylation. This energy coupling mechanism in the cell was revealed by the work of Peter stored energy in the form of proton (1) gradient to phosphorylate (add phosphate) ADP and produce ATP. The pumping of hydrogen sons across the inner membrane creates higher concentration ions in the inner membrane than on the outside of the membrane. This chemiosmotic gradient causes the ions to flow back across the membrane where the concentration of ions is lower. ATP synthase lined in the matrix serve as a channel protein, helping the ions to move across the membrane. The chemiosmotic gradient powers the phosphorylation of ADP to ATP, which also occurs in the ATP synthase. After passing through the ETC, the oxygen, being the final hydrogen acceptor, combines with two electrons and two protons, forming a water molecule. Water is a by-product of cellular respiration and is excreted. MINI TEST 6-3 1. Which energy-releasing pathway yields the most ATF in each glucose molecule? 2. Briefly describe the two stages of aerobic respiration that follow glycolysis: (a) Krebs cycle (b) Electron transport chain Anaerobic Respiration Most cells carry out arrobic respiration when oxygen is present. Aerobic respiration is an efficient process that yields a lot of ATP. However, many organisms thrive in mud, marshes, animal gut, canned goods, sewage treatment pond, and deep oceans where oxygen is scarce. Organisms that can live without oxygen are called anaerobes. Cellular respiration that proceeds without the presence of oxygen is called anaerobic respiration. In the event that the oxygen supply becomes low, aerobic cells also perform fermentation and lactic acid fermentation anaerobic pathways. There are two common anaerobic pathways in these cells, alcoholic fermentation and lactic acid fermentation. In alcoholic fermentation, ethyl alcohol and carbon dioxide are produced by some cells using the pyruvate from glycolysis. Each pyruvate molecule is rearranged into acetaldehyde and carbon dioxide, which is eventually released. NADII gives up electrons to acetaldehyde to form ethanol Fermentation is widely used in the industry. Yeast, a fungus used in making bread. can undergo anaerobic respiration. Bakers aux sugar, flour, water, and yeast to form the bread dough. The dough rises due to the carbon dioxide and alcohol released by the yeast cells trapped in air bubbles. Beer and wine manufacturers, we yeast to ferment the sugars in wheat and grape juice, forming alcoholic beverages such as beer and wine. In some cells, glycolysis produces two pyruvates, two NADH molecules, and two ATP molecules. Pyruvate itself becomes the final acceptor of the electrons from the NADH that produces the final product: lactate. Oftentimes, this product is called lactic acid. Human skeletal muscles can carry out fermentation when the blood cannot supply the cells with adequate oxygen during strenuous activities. When lactic acid builds up in the muscles, fatigue, burning sensation, and cramps result. Lactic acid will continue to build up until there is adequate supply of oxygen. Lactic acid is then converted back into pyruvate in the liver. Muscles also restore normal functions. Have you ever wondered why milk or cream turns sour after some time? Bacterial cells that undergo fermentation are responsible in producing lactate that turns the milk sour. These bacteria are used in manufacturing yogurt and sour milk products. Fermentation pathways do not breakdown and utilize the glucose completely. ATP is no longer produced beyond the process of glycolysis. Thus, energy produced is just enough for some single-celled organisms, or the energy can only be used by multicellular organisms for a short period.

Что называют простыми механизмами Умение облегчать себе труд с помощью технологий отличает человека от животного. Тысячи лет назад наши предки научились мастерить простые механизмы, которые способны увеличивать усилие, а также изменять направление прикладываемой силы. Принципы их работы лежат в основе любого орудия труда — от садовой лопаты до подъёмного крана. Простые механизмы — приспособления, служащие для преобразования вектора силы по величине и/или направлению. Подписывайтесь на телеграм-канал Домашней школы Фоксфорда — здесь мы каждый день публикуем полезные посты о лайфхаках обучения, тайм-менеджменте, развитии и поддержке школьников, а ещё делимся бесплатными материалами и шпаргалками. Виды простых механизмов Наклонная плоскость и её разновидности: клин и винт. Рычаг и его разновидности: блок и ворот. Теперь расскажем, как они работают. В этой статье мы рассмотрим действие идеальных механизмов, в работе которых не учитывается сила трения. Работа простых механизмов Наклонная плоскость Подниматься по пологому склону горы легче, чем карабкаться по отвесной скале. Чем меньше наклон — тем легче его преодолеть. Это нехитрое наблюдение помогло людям создать простой механизм — наклонную плоскость. Допустим, нам нужно поднять груз на определённую высоту. Конечно, можно сделать это непосредственно. Поднятие груза на высоту Правда, если груз большой, приложить достаточную силу будет нелегко. Но если поставить его на лёгкую тележку и вкатывать по наклонной плоскости, то понадобится гораздо меньше усилий. Наклонная плоскость Чем меньше угол наклона плоскости, тем больше выигрыш в силе: k = l / h. Чтобы просто поднять груз весом в один килограмм, требуется усилие: F1 = mg = 9,8 H. Теперь посмотрим, какое усилие понадобится, чтобы поднять этот груз на один метр, используя наклонную плоскость длиной десять метров: Использование наклонной плоскости позволило нам выиграть в силе в десять раз. Но путь, который нам пришлось пройти с грузом, также увеличился вдесятеро. Бесплатный доступ к занятиям в Домашней школе Вы получите записи уроков по нескольким предметам, познакомитесь с учителями и попробуете решить домашнее задание Начать бесплатно Клин С помощью наклонной плоскости удобно не только поднимать грузы. Рассмотрим топор: его лезвие — это клин, боковые поверхности которого сходятся под острым углом, образуя наклонные плоскости. Когда мы вонзаем топор в полено, эти плоскости с огромной силой раздвигают волокна древесины и заставляют полено расколоться. Клин При ударе сила P вгоняет топор в дерево, и на его лезвие действуют сдавливающие силы F со стороны полена. Проекция каждой из сил F на плоскость симметрии лезвия (AB) равна Fsinα. Поскольку они действуют с двух сторон, условие равновесия сил таково: P = 2Fsinα. Чем длиннее и острее клин (то есть чем меньше угол), тем меньше может быть P по отношению к 2F. Угол лезвия обычного колуна — около 25°, соответственно, сила Р примерно в пять раз меньше, чем 2F. Иными словами, чтобы расколоть полено, нужно приложить в пять раз меньше усилий, чем требуется, чтобы разорвать его. Люди пользуются топорами уже более 9 тысяч лет. Гвозди, иглы и ножи работают по тому же принципу. Клин придуман не человеком, а самой природой: например, клюв дятла легко вонзается в дерево благодаря оптимальной клиновидной форме. Винт Если свернуть наклонную плоскость в спираль вокруг цилиндра — получится винт. Винт Впервые описание винта встречается в работах древнегреческого учёного Архита Тарентского, жившего в V–IV веках до нашей эры. Знаменитый Архимед в III веке до нашей эры создал с помощью винта устройство для подъёма воды в оросительные каналы. Винты широко используют для крепления деталей, бурения отверстий и даже в качестве движителя сверхпроходимых шнекороторных вездеходов. Резьба винта — это наклонная плоскость длиной l и высотой h, свёрнутая в трубочку. Когда мы наворачиваем гайку на болт, мы перемещаем её по наклонной плоскости. Как и в случае с обычной плоскостью, выигрыш в силе равен отношению h к l, но теперь l рассчитывается по формуле длины окружности: l = πD. ‍Расстояние между витками называют шагом резьбы. Чем оно меньше, тем длиннее плоскость и больше выигрыш в силе. Рычаг Простейший рычаг — это палка, способная вращаться вокруг неподвижной опоры. Принцип рычага используется при работе башенного крана, рычажных весов, кухонных ножниц и даже обычной лопаты. Интересно, что кости в наших конечностях тоже работают как рычаги. Рычаг У любого рычага есть точка опоры (О) и два плеча (длины l1 и l2), к которым в точках A и B прикладываются силы. Вращение рычага зависит от приложенной к нему силы и от длины плеча. Чем больше сила и чем длиннее плечо, тем сильнее вращающее действие. Именно поэтому работать лопатой проще, держа её ближе к концу черенка, а нести груз на согнутой руке легче, чем на вытянутой. На рисунке тело А воздействует на рычаг с большей силой, чем тело B, но плечо l1 короче, чем l2, поэтому тела находятся в равновесии. В таких случаях говорят, что моменты двух сил уравновешены. Момент силы — произведение силы на длину плеча: M = F ⋅ l. Рассчитаем моменты силы для обоих тел: MA = PA ⋅ l1 = mAgl1. MB = PB ⋅ l2 = mBgl2. Тела находятся в равновесии, значит, mAgl1 = mBgl2. Чем больше будет длина плеча l2, тем меньшее усилие понадобится, чтобы уравновесить тело A. Так, при достаточной длине рычага можно поднять даже неподъёмный груз. Действие рычага Чтобы просто поднять тело, нужно преодолеть силу тяжести: F = mg. Чтобы вычислить силу для поднятия тела рычагом, нужно приравнять соответствующие моменты сил: mgl1 = Fx ∙ l2. Следовательно, Если l1 больше l2 в пять раз, то: Увеличивая длину плеча, мы выигрываем в силе, но проигрываем в перемещении. Нам удалось уменьшить силу в пять раз, но чтобы короткое плечо рычага поднялось на 10 сантиметров вверх, придётся опустить длинное на 50 сантиметров. Бесплатное руководство: как перейти на семейное образование Рассказываем, как забрать документы из обычной школы и перейти на домашнее обучение с онлайн‑аттестацией Email Телефон Принимаю условия соглашения и даю согласие на обработку своих персональных данных на условиях политики конфиденциальности Блок Частный случай рычага — блок. Так называют колесо с жёлобом, в который вложен трос. Если ось колеса зафиксировать, к одному концу троса привязать груз, а за другой тянуть, получится простой механизм — неподвижный блок. Неподвижный блок На груз действует сила тяжести F = mg. Чтобы удержать верёвку, требуется приложить такую же силу. Никакого выигрыша в величине силы неподвижный блок не даёт. Зато можно менять её направление — тянуть верёвку в любую сторону. Если прицепить груз к оси колеса, один конец верёвки закрепить, а за другой тянуть, получится подвижный блок, который позволяет выиграть в силе в два раза. Эффект достигается за счёт того, что блок с грузом поднимают как бы сразу две верёвки: за левую тянет человек, а правую натягивает вбитый в потолок гвоздь. Подвижный блок За выигрыш в силе приходится платить проигрышем в перемещении: чтобы поднять груз на нужную высоту h, понадобится выбрать вдвое большую длину и верёвки: l = 2h. Ворот Ворот издревле применяется для поднятия воды из колодца. К барабану, способному вращаться вокруг своей оси, прикреплены верёвка и рукоять. Когда мы вращаем рукоятку — вращается и цилиндр, а верёвка наматывается на него, поднимая или опуская груз. Ворот Ворот действует по тому же принципу, что и рычаг: плечом силы в данном случае становится рукоятка, а плечом груза — радиус барабана. Чем длиннее рукоять относительно радиуса барабана — тем больше выигрыш в силе. На рисунке длина рукояти равна трём радиусам барабана. Значит, он поднимает ведро с силой, в три раза большей, чем сила наших рук. При этом путь, который проходит рукоять ворота, в три раза длиннее куска верёвки, который в это время накручивается на вал. Золотое правило механики Все примеры простых механизмов, которые мы рассмотрели, имеют одно общее свойство, которое называют золотым правилом механики. Во сколько раз мы выигрываем в силе, во столько же раз проигрываем в перемещении. Произведение силы на перемещение в механике называется работой и обозначается буквой А: A = F ⋅ S ⋅ cos α, где α — угол между векторами силы и перемещения. Если направления векторов совпадают, формула работы выглядит проще: A = F ⋅ S. Сэкономить в силе больше, чем проиграть в перемещении — то есть выиграть в работе — не позволяет ни один механизм. Чем меньше силы нужно потратить при подъёме тела по наклонной плоскости, тем длиннее должна быть эта плоскость. Чем меньше сил нужно для воздействия на рычаг — тем длиннее должно быть его плечо. «Дайте мне точку опоры, и я переверну Землю» — заявил Архимед. Теоретически он мог бы поднять груз, равный нашей планете, выбрав рычаг подходящей длины. Масса Земли — примерно 6 000 000 000 000 000 000 000 тонн, в то время как человек в среднем способен поднять груз около 60 килограммов. А значит, плечо силы должно быть больше плеча груза в 100 000 000 000 000 000 000 000 раз. Поэтому, чтобы плечо груза сдвинулось хотя бы на один сантиметр, учёному пришлось бы сдвинуть плечо силы на 1000 000 000 000 000 000 км. Даже со скоростью движения в 1 м/с на это ушло бы тридцать тысяч миллиардов лет. Ответим

Understanding Quantum Theory of Electrons in Atoms The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding. As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels. The energy levels are labeled with an n value, where n = 1, 2, 3, …. Generally speaking, the energy of an electron in an atom is greater for greater values of n. This number, n, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure 9.4.1). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has. This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom: The values nf and ni are the final and initial energy states of the electron. The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital, which is distinct from an orbit, is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located. Another quantum number is l, the angular momentum quantum number. It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, …, n – 1. This means that an orbital with n = 1 can have only one value of l, l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell. In addition, the greater the angular momentum quantum number, the greater is the angular momentum of an electron at this orbital. Orbitals with l = 0 are called s orbitals (or the s subshells). The value l = 1 corresponds to the p orbitals. For a given n, p orbitals constitute a p subshell (e.g., 3p if n = 3). The orbitals with l = 2 are called the d orbitals, followed by the f-, g-, and h-orbitals for l = 3, 4, 5, and there are higher values we will not consider. There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction ψ is zero at this distance for this orbital. Such a value of radius r is called a radial node. The number of radial nodes in an orbital is n – l – 1. Consider the examples in Figure 9.4.2. The orbitals depicted are of the s type, thus l = 0 for all of them. It can be seen from the graphs of the probability densities that there are 1 – 0 – 1 = 0 places where the density is zero (nodes) for 1s (n = 1), 2 – 0 – 1 = 1 node for 2s, and 3 – 0 – 1 = 2 nodes for the 3s orbitals. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found. Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY If an electron has an angular momentum (l ≠ 0), then this vector can point in different directions. In addition, the z component of the angular momentum can have more than one value. This means that if a magnetic field is applied in the z direction, orbitals with different values of the z component of the angular momentum will have different energies resulting from interacting with the field. The magnetic quantum number, called ml, specifies the z component of the angular momentum for a particular orbital. For example, for an s orbital, l = 0, and the only value of ml is zero. For p orbitals, l = 1, and ml can be equal to –1, 0, or +1. Generally speaking, ml can be equal to –l, –(l – 1), …, –1, 0, +1, …, (l – 1), l. The total number of possible orbitals with the same value of l (a subshell) is 2l + 1. Thus, there is one s-orbital for ml = 0, there are three p-orbitals for ml = 1, five d-orbitals for ml = 2, seven f-orbitals for ml = 3, and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure 9.4.3. Figure 9.4.4 illustrates the energy levels for various orbitals. The number before the orbital name (such as 2s, 3p, and so forth) stands for the principal quantum number, n. The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for l ≥ 1, each corresponding to a specific value of ml. In the case of a hydrogen atom or a one-electron ion (such as He+, Li2+, and so on), energies of all the orbitals with the same n are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, n, are called degenerate energy levels. However, in atoms with more than one electron, this degeneracy is eliminated by the electron–electron interactions, and orbitals that belong to different subshells have different energies. Orbitals within the same subshell (for example ns, np, nd, nf, such as 2p, 3s) are still degenerate and have the same energy. While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or ms. The other three quantum numbers, n, l, and ml, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schrödinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schrödinger equation and is not related to the normal spatial coordinates (such as the Cartesian x, y, and z). Electron spin describes an intrinsic electron “rotation” or “spinning.” Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, even though this rotation cannot be observed in terms of the spatial coordinates. The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the α state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number ms=12. The other is called the β state, with the z component of the spin being negative and ms=−12. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having ms=−12 and ms=12 are different if an external magnetic field is applied. Figure 9.4.5 illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the 12 spin quantum number and down (in the negative z direction) for the spin quantum number of −12. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with ms=12 has a slightly lower energy in an external field in the positive z direction, and an electron with ms=−12 has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting. The Pauli Exclusion Principle An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that electrons can share the same orbital (the same set of the quantum numbers n, l, and ml), but only if their spin quantum numbers ms have different values. Since the spin quantum number can only have two values (±12), no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly

Electrostatics The section of CBSE Class 12 Physics electrostatic potential and capacitance notes mainly deals with the in-depth analysis of electromagnetic phenomena when they are not performing any movements. Additionally, it is divided into ten further sub-topics to study the companion processes of reaching the state. These are - 1. Electric charge In this section of Physics ch 2 Class 12 notes, you get to learn about the basic features of electric charge and its expression in Physics. Along with its basics, the sections help to understand the full potential of charge. Different aspects of Charge included in Class 12 Physics Chapter 2 notes are - Definition Type: Positive and Negative Charge Unit and dimensional formula Point Charge Properties of Charge Comparison of Charge and Mass Methods of Charging Electroscope 2. Coulomb's Law Force is created when charges of opposite signs attract each other, and they repulse if the signs are the same. Coulomb's law tries to define this phenomenon through a mathematical formula, explicitly mentioned in Physics Class 12 notes Chapter 2. Moreover, there is key information about the variation of the constant k and its effect on a medium. Coulomb's law's vector form and the principle of superimposition are also explained in ch 2 Physics Class 12 notes. (Image will be uploaded soon) 3. Electric Field As stated in Class 12 Physics Chapter 2 notes, every positively or negatively charged particle has their respective electric fields. It feels a force at the time of interaction which might be attraction or repulsion. As it arises from electric charge, it is crucial to know about its different parts like - Electric field intensity Relation between electric force and electric field Super imposition of electric field Point charge Continuous charge distributions Properties of Electric Field Lines Motion of Charged Particles in an Electric field Learning more about the electric field from electric potential and capacitance notes Class 12 helps a student to get a grasp of upcoming chapters. 4. Electric Potential Energy When energy helps a charge to move from an electric field, it is known as the Electric Potential Energy. This section of electrostatic chapter Class 12 notes requires a student to study the Electron volt (eV), and the potential energy that an n number of charges can hold. 5. Electric Potential This section of Class 12 Physics Chapter 2 notes focuses on in-depth learning of Electric Potential or Voltage. Basically, it defines the potential movement of energy. 6. Relation between Electric Field and Potential Apart from knowing more about the relationship between the two values, Physics Class 12 Chapter 2 notes also discuss equipotential surfaces. 7. Electric Dipole Essentially, 'Dipoles' are two opposite points of charge represented with q and –q, with their distance between each other being 2a. Electric Dipoles are crucial in your study of Physics Class 12 Chapter 2 notes to learn more about electric fields and their potential. Additionally, Class 12 Physics Chapter 2 notes focus on the influence of electric dipoles on a uniform electric field mainly through Force and Torque, Work, and Potential Energy. In the last part of Electrostatics, further focus is on using the formulas to their fullest potential. It includes subsections of Electric Field, Electric Potential Energy, Electric Potential, and Electric Dipole. In the notes for electrostatic potential and capacitance, you will find proper solutions accompanied by clear and crisp diagrams for better understanding. 8. Gauss's Law Apart from just discussing the Gauss's Law, in Physics Class 12 ch 2 notes there is a thorough explanation of its properties and applications. The Gauss' Law states that net electric flux passing through a hypothetical closed surface is equal to the net electric charge present within the same closed surface. Being a broad part of the whole chapter, you may need to spend a little more time on it. Moving forward, it starts discussing the properties of conductors in relation to Gauss's Law. The Class 12 Physics notes Chapter 2 perfectly defines the journey to Gauss' Law from Coulomb's Law. Here is the Gauss's Law present in the Class 12 Physics ch 2 notes, (image will be uploaded soon) 9. Capacitors There is a dedicated section about Capacitors in the Class 12 Physics Chapter 2 notes elucidating its functions and importance as storage of potential electric energy. After explaining the structure of a capacitor, it points out the different types, parallel plate, spherical and cylindrical. The section of Chapter 2 notes of Physics Class 12 is further divided into subheads like: Properties of an ideal battery Grouping of capacitors Simple circuits (Series and Parallel) Dielectric Van de Graaff generator Combination of drops Charge distribution method Wheatstone Bridge-based circuit Extended Wheatstone Bridge Infinite network of capacitors Redistribution of charge between two capacitors Vedantu prepares the Class 12 Physics Chapter 2 notes with help from subject matter experts. In the PDF, you get a comprehensive idea of the topic along with potential answers to the most asked questions. Furthermore, the detailed explanation on each section and subsections are written in a simple language allows a student to ace their exams with wholesome knowledge. These Physics Chapter 2 Class 12 notes are going to be one of the best supplementary study materials besides a student’s textbooks. Visit the Vedantu website or download the app to get your hands on all important notes! Important Questions A charge of 4 × 10–8C is uniformly distributed on the surface of a spherical conductor, having a radius of 15 cm. Determine the electric field just outside this sphere at a point that is 15 cm from the centre of this sphere. Determine the capacitance given that the distance between the two plates has been reduced by half and the parallel plate capacitor holds a capacitance of 20 pF (where 1pF = 10-12 F) having air between the two plates. What will be the total capacitance of a combination where three capacitors, each having a capacitance of 20 pF, are connected in series. A square having a side of 10 cm has a 500 µC charge at its centre. Determine the work done to move a charge of 10 µC between two points that are diagonally opposite each other on the square. At an equatorial point, what will be the electrostatic potential because of an electric dipole? Calculate the work done to move a test charge, q, through a length of 1 cm along the equatorial axis of an electric dipole? Polarisation A capacitor has its plates enclosed in a medium that can be filled by insulating substances. A net dipole moment is then induced by an electric field in the dielectric. This event causes the field in an opposite direction. Equipotential Surface An equipotential surface is a type of surface where the potential always has a constant value. If considered as a point charge, the concentric spheres that are centred at a particular area of this charge are basically equipotential surfaces. Advantages of Vedantu's Revision Notes: A Comprehensive Resource for Effective Learning There are several reasons why one may refer to Vedantu's revision notes for studying a subject like Electrostatic Potential and Capacitance. Here are some key points: Comprehensive Coverage: Vedantu's revision notes provide a comprehensive coverage of the entire topic, ensuring that all important concepts and subtopics are included. Concise and Organized: The notes are designed to be concise, focusing on the key points and core ideas. They are organized in a structured manner, making it easy for students to navigate and revise the content. Simplified Explanation: The revision notes offer simplified explanations of complex concepts, making them more accessible and easier to understand. This helps students grasp the material more effectively. 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