Class IX Physics Reference Material for SA-I 2014-15
... Total time taken If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity. 3. What does the odometer of an automobile measure? Ans. The odometer of an automobile measures the distance covered by an automobile. 4. What d ...
... Total time taken If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity. 3. What does the odometer of an automobile measure? Ans. The odometer of an automobile measures the distance covered by an automobile. 4. What d ...
Answers to Multiple-Choice Problems Solutions to Problems
... (b) F is maximum when f 5 90°, when v is perpendicular to B. Fmax 5 0 q 0 vB 5 4.32 3 10216 N. F is minimum S S when f 5 0° or 180°, when v is either parallel or antiparallel to B. Fmin 5 0. (c) 0 q 0 is the same for an electron and a proton, so F 5 3.45 3 10216 N, the same as for a proton. Since th ...
... (b) F is maximum when f 5 90°, when v is perpendicular to B. Fmax 5 0 q 0 vB 5 4.32 3 10216 N. F is minimum S S when f 5 0° or 180°, when v is either parallel or antiparallel to B. Fmin 5 0. (c) 0 q 0 is the same for an electron and a proton, so F 5 3.45 3 10216 N, the same as for a proton. Since th ...
Unit 4. Forces - Perry County School District 32
... can be deformed (have their shape changed) if enough force is applied, such as with a sledgehammer or in a collision with another car. Contact Forces Vs Field Forces Forces cause not only deformations (changes in shape), they can also cause motion – if you push or pull on a cart, it may move. When y ...
... can be deformed (have their shape changed) if enough force is applied, such as with a sledgehammer or in a collision with another car. Contact Forces Vs Field Forces Forces cause not only deformations (changes in shape), they can also cause motion – if you push or pull on a cart, it may move. When y ...
Seesaws 9 Balanced Seesaw
... An object’s angular acceleration is equal to the torque exerted on it divided by its rotational mass. The angular acceleration is in the same direction as the torque. angular acceleration = torque/rotational mass torque = rotational mass· angular acceleration ...
... An object’s angular acceleration is equal to the torque exerted on it divided by its rotational mass. The angular acceleration is in the same direction as the torque. angular acceleration = torque/rotational mass torque = rotational mass· angular acceleration ...
lecture1437132938
... (iii) d + b < c + a Adding (i) and (ii) 2d < 2b d < b Adding (ii) and (iii) 2d < 2a d < a Adding (iii) and (i) 2d < 2c d < c Thus the necessary conditions for the link ‘a’to be a crank are that the shortest link is fixed and the sum of the shortest and the longest link is less than the sum ...
... (iii) d + b < c + a Adding (i) and (ii) 2d < 2b d < b Adding (ii) and (iii) 2d < 2a d < a Adding (iii) and (i) 2d < 2c d < c Thus the necessary conditions for the link ‘a’to be a crank are that the shortest link is fixed and the sum of the shortest and the longest link is less than the sum ...
chapter-6-with-changes-thursday-jan-9
... projectile is the same as solving a constant velocity problem. We will be ignoring air resistance. All problems will be treated as if there is no air drag and therefore no horizontal acceleration (no horizontal net force). ...
... projectile is the same as solving a constant velocity problem. We will be ignoring air resistance. All problems will be treated as if there is no air drag and therefore no horizontal acceleration (no horizontal net force). ...
Ch 7 - Keene ISD
... The diameter of an audio compact disk is 12.0 cm. When the disk is spinning at its maximum rate of 540 rpm, what is the speed of a point (a) at a distance 3.0 cm from the center and (b) at the outside edge of the disk, 6.0 cm from the center? Consider two points A and B on the rotating compact disk ...
... The diameter of an audio compact disk is 12.0 cm. When the disk is spinning at its maximum rate of 540 rpm, what is the speed of a point (a) at a distance 3.0 cm from the center and (b) at the outside edge of the disk, 6.0 cm from the center? Consider two points A and B on the rotating compact disk ...
p - Effingham County Schools
... no external forces. They both start at rest, one behind the other. Skater C, the boy, gives skater D, the girl, a push. Find the final velocities of the two in-line skaters ...
... no external forces. They both start at rest, one behind the other. Skater C, the boy, gives skater D, the girl, a push. Find the final velocities of the two in-line skaters ...
Chapter 5: Problems
... assume the box does not slide over the floor of the truck. (a) The truck is traveling at constant velocity up the hill. (b) The truck is traveling at constant velocity down the hill. (c) The truck is at rest at a red light halfway up the hill. (d) The truck is speeding up as it climbs the hill. (e) ...
... assume the box does not slide over the floor of the truck. (a) The truck is traveling at constant velocity up the hill. (b) The truck is traveling at constant velocity down the hill. (c) The truck is at rest at a red light halfway up the hill. (d) The truck is speeding up as it climbs the hill. (e) ...
INTRODUCTORY LECTURES ON FLUID DYNAMICS
... surface. For T > μN, the body will overcome the frictional force and accelerate. A distinguishing characteristic of most fluids in their inability to support tangential stresses between layers without motion occurring; i.e. there is no analogue of limiting friction. Exceptions are certain types of so ...
... surface. For T > μN, the body will overcome the frictional force and accelerate. A distinguishing characteristic of most fluids in their inability to support tangential stresses between layers without motion occurring; i.e. there is no analogue of limiting friction. Exceptions are certain types of so ...
Newton's theorem of revolving orbits
In classical mechanics, Newton's theorem of revolving orbits identifies the type of central force needed to multiply the angular speed of a particle by a factor k without affecting its radial motion (Figures 1 and 2). Newton applied his theorem to understanding the overall rotation of orbits (apsidal precession, Figure 3) that is observed for the Moon and planets. The term ""radial motion"" signifies the motion towards or away from the center of force, whereas the angular motion is perpendicular to the radial motion.Isaac Newton derived this theorem in Propositions 43–45 of Book I of his Philosophiæ Naturalis Principia Mathematica, first published in 1687. In Proposition 43, he showed that the added force must be a central force, one whose magnitude depends only upon the distance r between the particle and a point fixed in space (the center). In Proposition 44, he derived a formula for the force, showing that it was an inverse-cube force, one that varies as the inverse cube of r. In Proposition 45 Newton extended his theorem to arbitrary central forces by assuming that the particle moved in nearly circular orbit.As noted by astrophysicist Subrahmanyan Chandrasekhar in his 1995 commentary on Newton's Principia, this theorem remained largely unknown and undeveloped for over three centuries. Since 1997, the theorem has been studied by Donald Lynden-Bell and collaborators. Its first exact extension came in 2000 with the work of Mahomed and Vawda.