Static and Kinetic Friction Introduction The amount of friction force between two surfaces in contact depends on the type of the surfaces in contact and the amount of compression between the surfaces. Static friction is the force that is acting against your force before the object begins to move. If you exert a small push on the box‚ the box will not move because static friction is directly opposite to the push. If you apply a greater force than the static friction force‚ the friction increases
Premium Friction Force
THE MOLECULAR BASIS OF INHERITANCE I. History A. Discovery of “transformation” – a change in genotype and phenotype due to the uptake of external DNA by a cell 1. Griffith 1920s did experiments with Streptococcus pneumoniae (p294 fig16.2) a. took two strains of S. pneumoniae‚ one virulent‚ one not b. heat killed virulent strain‚ then mixed them with the living nonvirulent strain c. living nonvirulent strain became virulent d. nonvirulent strain took on virulent strain’s DNA became virulent
Premium DNA DNA replication
mechanisms that regulate cellular senescence‚ organismal ageing‚ and species-specific lifespan depend on a synergy of pathways that are multifactorial and extremely complex‚ though not yet completely understood. Recently‚ the development of new molecular techniques has elucidated‚ at least in part‚ the primary pathways involved in ageing. In parallel with the search to uncover the factors that control ageing is the endeavor to discover methods of extending lifespan‚ in hopes of living both youthfully
Premium Senescence Gerontology
1 UNIT 9 CHEMICAL KINETICS 1. (c) mole 1–1 sec–1 The rate law for a reaction A + B →products‚ is rate = k [A]1[B]2. 7. Then‚ which one of the following statements is false ? (a) If [B] is held constant while [A] is doubled‚ the reaction will proceed twice as fast. (b) If [A] is held constant while [B] is reduced to one quarter‚ the rate will be halved (c) If [A] and [B] are both doubled‚ the reaction will proceed 8 times as fast. 2. Fro a first order reaction‚ a straight
Premium Chemical kinetics Rate equation Chemical reaction
permeable membrane. During diffusion‚ molecules that are small enough to pass through a membrane’s pores or molecules that can dissolve in the lipid section of a membrane move from an area of higher concentration to an area of lower concentration. The kinetic energy that all molecules possess is the motivating force in diffusion. Facilitated diffusion occurs when molecules are too large to pass through a membrane or are lipid insoluble. In this process‚ carrier protein molecules located in the membrane
Free Diffusion Molecular diffusion Solution
2. A 40 kg crate rests on a horizontal floor‚ and a 75 kg person is standing on the crate. Determine the magnitude of the normal force that (a) the floor exerts on the crate and (b) the crate exerts on the person. [(a) 1.13 x 103 N‚ (b) 735 N] 3. A worker stands still on a roof sloped at an angle of 45° above the horizontal. He is prevented from slipping by a static frictional force of 450 N. Find the mass of the worker. [85 kg] 4. A 4.0-kg bucket of water is raised from a well by a rope
Premium Friction Mass Force
Learning Goals: • Predict the kinetic and potential energy of objects. • Examine how kinetic and potential energy interact with each other. In the space provided‚ define the following words: Kinetic energy-is the energy of motion. An object that has motion - whether it is vertical or horizontal motion Potential energy-is the energy of an object or a system due to the position of the body or the arrangement of the particles of the system Open Internet Explorer. From the FMS
Premium Kinetic energy Energy Potential energy
and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest. a) When the 3 kg weight is moving with a speed of 2.2 m/s‚ what is the kinetic energy of the entire system? KETOT = KEwheel+KEweight = (1/2)(I)(w2)+(1/2)(m*v2) =(0.5* v2)(m+1/2M) =0.5*(2.2^2)*(3+(.5*15)) J b) If the system started from rest‚ how far has the weight fallen? H = KETOT/MG = 0.5*(2.2^2)*(3+(.5*15))/(3*9.8) m
Premium Mass Classical mechanics Energy
Kinetics of Hydrogen Peroxide February 22‚ 2007 Chem. 1130 TA: Ms. Babcock Room 1830 Chemistry Annex PURPOSE OF THE EXPERIMENT Kinetics of Hydrogen Peroxide The major purpose of this experiment is to determine the rate law constant for the reaction of hydrogen peroxide and potassium iodide. In this experiment‚ the goal will be to try to measure the rate law constant at low acidity‚ since at low acidity‚ anything less than 1.0 x 10-3M‚ the effect of the hydrogen ion is negligible. To calculate
Premium Rate equation Reaction rate Chemical kinetics
Mark Scheme (Results) Summer 2012 International GCSE Chemistry (4CH0) Paper 1C Science Double Award (4SC0) Paper 1C Edexcel Level 1/Level 2 Certificate Chemistry (KCH0) Paper 1C Science (Double Award) (KSC0) Paper 1C Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson‚ the world’s leading learning company. We provide a wide range of qualifications including academic‚ vocational‚ occupational and specific programmes for employers. For further information‚
Premium Atom Chlorine Atomic number