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| Question | Answer | |||||||
|---|---|---|---|---|---|---|---|---|
| The displacement from A to B is the vector | AB (arrow overhead), while the displacement from B to A is the vector | BA (arrow overhead), thus AB* | -1. | |||||
| If the displacement from A to B is f, while the displacement from B to C is g, the displacement from A to C is | f+g. | |||||||
| A vector quantity has both | magnitude and direction. | |||||||
| A scalar quantity only has | magnitude. | |||||||
| As distance isn't affected by d[] | direction, direction is a | scalar. | ||||||
| Distance in a particular direction is called | displacement, hence displacement is a | vector. | ||||||
| The displacement from O to P is | OP. The distance from O to P is the [] of the vector OP | magnitude--. | ||||||
| Speed is not affected by the [] of movement | direction--, speed is a | scalar. | ||||||
| Speed in a particular direction is called | velocity, thus velocity is a | vector. | ||||||
| The speed of an object is its velocity's | magnitude. | |||||||
| For an object moving with constant velocity: velocity = | displacement/time, thus displacement = | velocity x time. | ||||||
| In kinematics, one studies an object's | motion without worrying about why it's moving in a particular way. In the graphs, we don't take into account the object's m[] | mass or any forces that may act on the object; generally things are simpified further by only considering motion in a straight line, and taking any acceleration to be | constant. | |||||
| In terms of value, distance can only be either | positive or zero. | |||||||
| Generally, with graphs, the independent variable tend to be shown on the [] axis | horizontal, while on the vertical axis, what tends to be shown is the | dependent variable. | ||||||
| speed = | distance/time; in terms of value, time is only ever taken to be | 0 or positive, thus, in terms of value, speed can only be | positive or zero. | |||||
| In terms of value, displacement along a straight line may be | positive, negative or zero. | |||||||
| Displacement is measured from | the starting point. | |||||||
| acceleration = | change in speed/time taken | |||||||
| A negative acceleration is sometimes called | retardation | |||||||
| In terms of value, velocity along a straight line may be | positive, negative or zero. | |||||||
| The area under a velocity/time graph represents | displacement. Area below the horizontal axis counts as | negative displacement. If there is some area above the horizontal axis (A), and some area below (B), then the final displacement equals | A+B (if B is taken to be negative). | |||||
| The area under an acceleration/time graph represents a [] in v[] | change in velocity. | |||||||
| A particle comes to rest when it's final velocity equals | 0. | |||||||
| The pulling force that a planet exerts on every object is called | gravity. On Earth, if we ignore the effect of air resistance, the acceleration of a falling object due to gravity is | 9.8 m/s². {On another planet, the acceleration due to that planet's gravity may different to ours' depending on the mass of the planet.} | ||||||
| Throughout this unit, the model used for calculations assumes: air resistance is small enough to be | ignore, and the acceleration due to gravity is the same for every object regardless of its weight and that it is c[] | constant--. | ||||||
| Projecting an object downwards is much like dropping it, except its initial velocity isn't | 0. | |||||||
| When an object is projected upwards, it is easier to chose [] as the positive direction | upwards--; the acceleration due to gravity is then given a [] sign | negative--. | ||||||
| If asked to find the greatest height reached by an object - find the point where the final velocity (v) equals | 0. | |||||||
| If an object is projected upwards with speed x m/s from a height of h metres above the ground, to calculated the time taken for the object to reach the ground (i.e. after it's reached its top height), in the suvat equations: s=[], u=[] and a=[] | s=-h, u=x, a=-9.8. | |||||||
| Force is measured in | Newtons. | |||||||
| Weight is a type of f[] | force. | |||||||
| The weight of an object with mass of m kg is | 9.8m N: this is the force that is exerted on it by the | Earth. | ||||||
| Tension is a type of f[] | force. | |||||||
| Newton's first law: if an object is stationary, then it will be [] unless it is acted upon by a [] | stationary--force--; if an object is moving, unless it's acted upon by a force, it will continue to move at the same [] in a [] line | speed--straight--. | ||||||
| A common way of simplifying problems in mechanics is to represent objects as a | dot (/particle); forces and their directions are represented with labelled | arrows. | ||||||
| If there is only one force acting on an object, then the object would | move, in the direction of the | force. If there is a force (A) acting on an object, but the object doesn't move, then there is another force (B) with the [] size and in the [] direction | same--opposite--; if the other force (B) was smaller than A, then the object would [] in the [] of A | move--direction--(and vice-versa) ; if B wasn't opposite A, then the object would move in the direction of [] at an [] | A--angle--. | |||
| If an object is on a suface, then the surface exerts an [] force on the object | upwards--, called the | normal reaction force (so called as it and the surface are | perpendicular). | |||||
| If an object is released from a height: it will accelerate | downwards because, acting on it is only one | force - its | weight (i.e. in the model). Newton's first law tells us that the apple must move; but it can't move with constant velocity as this only happens when there is no | force, so the object must | accelerate. | |||
| An object's horizontal component of velocity is the force that determines its [] velocity | horizontal--; an object's vertical component of velocity is the force that determines its [] velocity | vertical--. | ||||||
| If a projected object has constant horizontal component of velocity of h: its path of movement when its constant vertical component of velocity (v) is 0 is a | straight line; when v is a constant amount greater than 0 downwards, the path of movement is a [] line tending [] | straight--downwards--, and if v (going downwards) increases, the path of movement is a straight line sloping downwards, but at a [] angle | steeper--. If throughout, no [] is [] horizontally, then the ball will, horizontally, continue to have the same speed | force--acting--. | ||||
| If the downward component of velocity is constant, then the object will move in | a straight line, however, projected objects have curved paths as the downward component of the velocity [] at a [] rate | increases--constant-- (remember Newton's first law - the force acting is the weight);if no force is acting horizontally, the horizontal componenent of velocity remains | constant, but the vertical component [] with [] | increases with time at a [] rate due to [] | constant--gravity--. | |||
| In the standard model for mechanics, a string which suspends a weight is taken to be light and | inextensible; 'light' means it has | no weight and 'inextensible' means that it | won't stretch. A string may be used to apply, to an object, some | force; the force is the [] in the string and is an example of a [] force | tension--pulling--. | |||
| Let there be an object, weight - w N, suspended by a string, with tension T N. By Newton's 1st law, if an object remains stationary, there is no | net force: the downward weight is balanced by the [] [] | upward tension (i.e. T = | w). | |||||
| In the standard model for mechanics, a pully is taken to be | light and frictionless. Let there be two weights (w1 N and w2 N), one either side of a pulley; let the string have tension T N one either side (the same tension on both sides). If there's equal weight on both sides, then the objects will be | stationary. If the weights are unequal, then the heavier object goes [] and the lighter one goes [] | down--up--; through that knowledge, one can deduce information about the value of T, often using | inequalities. | ||||
| [#A43#] (1,2 ss) A and B, attached by strings, hang in a vertical line below a fixed point C: find the tension in the upper (T1 N) and lower (T2 N) string, assuming they are light and inextensible. A has weight 15 N and T2 acts [] on A | upward; B has weight 10 N, T2 acts [] on B and T1 acts [] on B | downwards--upwards--. From Newton's first Law, there must be no net force acting on A (as it is stationary), thus T2 = | 15; additionally, there must be no net force acting on B, thus T1 = [] + [] = [] | 10+15=25. | ||||
| If A N pulls downwards on a stationary pulley with tension T N on either side, T = | A/2 (A = 2T). | |||||||
| When an object is acted upon by force a resistance force may increase to match the pulling force, to stop the object | moving, but at some point, the resistance force can't | increase further; increasing the other force beyond that point causes the object to | move. | |||||
| An object is pulled witha horizontal force of P N and the corresponding resistance force is R N. The value of R when P = A N and the object remains stationary is | A; the value of R when P = A N and the object moves at constant speed is | A. The value of R when P = A N and the object gains speed is R where R is (using inequalities) | 0<R<A. The value of R when P = A N and the object slows down is R where R is (using inequalities) | R>A. | ||||
| With Newton's second law, doubling the net force produces double | the acceleration, as the mass remains | constant. | ||||||
| Applying the same force to an object of twice the mass produces [] of the acceleration | half--. Note: 2 * 1/2 = 1: both sides have been multiplied by the same amount. | |||||||
| The acceleration due to gravity on Earth is denoted by | g [#A44#]. | |||||||
| The weight of an object is the [] that gives it an acceleration of [] | force--g--. Using F = ma with a = g, the weight of an object of mass m grams is | mg N. | ||||||
| On a forces diagram, write acceleration as [] arrows | double--. | |||||||
| Let there be a jumper (J) of mass m kg, thus weight mg N. If J is attached to a string, there is tension in the string (T N). mg is [], while, as the string stretches, T [] | constant--increases--. The net downward force (F) is | (mg - T)N (=ma). While mg-T > 0, J continues to [], but at a [] rate (in comparison to when there was no tension) | accelerate--reduced--. When mg-T=0: J moves [], but there is no [] | downwards--acceleration--(J moves with constant velocity). When mg-T<0, J [] in the direction that's [] | accelerates--upwards--; the upward acceleration increases with | T, thus if mg-T was > 0 previously, J has to first [], then move upwards | stop--; the upwards acceleration is greatest at the [] point, when J is momentarily [] | lowest point--stationary--. |
| When an object moves over a rough surface the [] force must be taken into account | resistance (to be given)--. | |||||||
| Newton's 3rd law tells us that every action has an | equal and opposite reaction; it applies whether or not the object | moves. If action and reaction are equal but opposite, they don't [] each other [] | cancel--out--; that is because the action and reaction forces act on [] objects | different. | ||||
| A book of mass m kg lies on the floor of a lift, thelift accelerates upwards at a rate of a m/s². What force does the book exert on the floor of the lift? The book accelerates upwards at | a m/s². Vertically, the forces acting on the book are its [] and the [] force R N from the [] | weight--upward--floor--. Find R, using ma = | R - mg (upwards is the positive direction here). As R is the force that the floor exerts on the book, the force that the book exerts on the floor is | R. Note: the weight of the book is not a force that the book is exerting on the floor, but a force is exerted on the book by the | planet; if the lift accelerated downwards, adapt by taking the positive direct to be | downwards. | ||
| Two objects of mass X kg and Y kg (pulled by a force of F N) are connected by a light inextensible string on a smooth horizontal surface. The equal and opposite force that each object exerts on the other is the [] in the [] | tension (T N)--string--. If we want to find the acceleration of the system and the tension in the string, we can consider each object seperately: in the X kg object: the equation for the force is | T = Xa, while for the Y kg object: the equation for force is | F-T = Ya. The equations should be treated as | simulaeneous equations, and used to find the value of [] and [] | a (the acceleration) and T (the tension) - thus is the answer. | |||
| If a string is on either side of a pulley and on each side has tension T N, then the total force that the string exerts on the pulley is | 2T N. | |||||||
| A force has both magnitude and direction, thus is a [] quantity | vector--. Multiple forces combine to act as a [] force called the [] force | single--resultant--. The resultant force is found by [] the forces as [] | adding--vectors--. | Any force can be represented by two perpendicular components (i.e. i and j). | ||||
| F = ma is a formula for motion in | a straight line. The good news is that the formula still works in two or even three dimensions: to make this work, F and a are given as | vectors and m is a | scalar. You may also use the constant acceleration formulae. | |||||
| If two forces (A and B, B<A) act on the same line on an object, then the resultant force is | (A-B)N; the particle will behave as if the only force acting on it is the | resultant force, so it will [] in that direction | accelerate--. | |||||
| If there are two forces that do not act along the same line, then they can be taken to be two adjacent sides of a | parallelogram; the resultant force, R N, can be represented by the parallelogram's | diagonal. The value of R can be found using the | cosine rule, and the angle that the resultant makes with the horizontal can be found using the | sine rule. Note: if the two primary forces are perpendicular, the resultant is represented by the diagonal of a | rectangle. | |||
| If you have a triangle with hypothenuse 'h', opposite side 'o', adjacent side 'a' and angle θ between p and o: o = | hsinθ, a = | hcosθ. Thus if you have a resultant force, you can find its c[] | components, that is, the two [] forces that it's a resultant of | perpendicular--. | ||||
| When more than one force act on an object, it is often useful to resolve them in a particular direction. This means that you find the [] of the forces in that [] | components--direction-- and | add them together. Components in the opposite direction are taken to be | negative. Adding components in this way gives what we call the [] [] in the given direction | algebraic sum--. You can write R (-->) to show that you intend to resolve the forces in the direction of | the arrow. | |||
| We often use [] to represent a resultant force | R--. Its horizontal and vertical components may be written, respectively, | RH and RV (H,V ss) - these can be found by [] horizontally and vertically, respectively | resolving--. | |||||
| RV and RH (V,H ss) may be shown as the sides of a | right-angled triangle or as the | resultant force. | ||||||
| A particle is in equilibrium if the resultant force acting on it is | zero. To solve problems on the equilibrium of a particle, use the fact that the reolved forces, horizontal and vertically, or in any two perpendicular directions, amount to | zero. Through this, one can find the value of unknowns in the equilibrium. | ||||||
| If a particle of mass m is suspended in eqilibrium by a light inextensible string, then the tension must balance the | weight (i.e. T=mg N). If the particle is them pulled to one side by a horizontal force of F N, the particle remains in | equilibrium: you can find the angle that the string makes with the vertical and the new tension in the string by creating a [] with the forces | triangle--(i.e. a triangle of forces). The hypothenuse would be [] N, and the two legs would be the [] and the [] force | T--weight--horizontal force; θ is the angle between | T N and the weight; using Pythagoras' theorem and trigonometry, T N and θ may be found. | |||
| In mechanics, a device that we often use to deliver a pushing force is a | rod. The pushing force is called the [] in the rod | thrust. In the model, the rod is light and rigid - it has not weight and it doesn't | compress. | |||||
| While the tension in a string acts [] []lly from each end | inwards equally--, the thrust in a rod acts [] []lly from each end | outwards equally--[#A45#]. It is also possible for a rod to be under [], which makes it act like a string | tension--. You can [] with a rod as well as push | pull--, however a string cannot be under | thrust (you can't push something with a piece of string)! | |||
| A particle pushed by a rod (inclined at x° to the horizontal) with a force of FN; the particle has mass m, thus weight | mg N [#A49#]; the particle moves horizontally with constant speed. Since the head moves in a straight line with constant speed, we know from Newton's 1st law that it must be in | equilibrium. What is the force exerted on the particle by the floor? One approach to finding the force is to use the [] of forces | triangle of forces: the three sides of the triangle are | FN, mgN and RN (the sought force) [#A50#]; using the sine rule and the cosine rule, R and θ can be found. An alternative approach would be to find the two [] []s of the required force | perpendicular components-- by [] []lly and []lly | resolving horizontally and vertically; you could the find the resultant force. | ||
| If a particle is on a smooth horizontal plane, the two forces acting on the particle are | its weight (W N) and the normal reaction from the plane (R N). If the plane, is inclined at an angle of θ: the angle between the particle's weight and the plane is | 90-θ, so the angle between W and R is θ. In [#A46#] the component of weight normal to the plane is | Wcosθ, thus R = | Wcosθ; the component of weight acting down the plane is | Wsinθ. If θ (the inclination of the plane) increases, cosθ | reduces, making the component of weight onto the plane, balanced by R, | smaller; additionally, if θ (the inclination of the plane) increases, sinθ [], making the component of weight acting down the plane [] | increases--greater--. |
| Braking force on a car opposes its | motion, thus in a diagram, would be represented with an arrow that's facing in a direction [] to the direction of movement | opposite--. | ||||||
| If asked to find the acceleration of an item in a system or the acceleration of a system, and you don't know, for example, the tension, but can find two equations containing the T, then treat them as | simultaeneous equations. | |||||||
| If a particle is on a smooth inclined plane and a light rigid rod pushes the particle with a horizontal thrust, it exerts a force of F N (in terms of direction) | horizontally and vertically. Note: if the plane is inclined at θ, then the amount of degrees between the rod and the plane is | θ; the component of the thrust acting on the plane is [] N, and the component of thrust acting along the plane is [] N | Fsinθ--Fcosθ--. | |||||
| Generally, an object on a rough surface, with force acting on it, may not move, as | friction is acting on it; when the force increases, | friction increases until it reaches its | maximum value. When the pushing force is equal to the maximum value for the friction, the object will be in | limiting equilibrium; the box is then 'on the point of | slipping'. | |||
| Pressing an object harder onto the surface means that a [] force will be needed to overcome friction | larger--; increasing the downward force, thus the normal reaction force and the [] value of friction will be [] | maximum--increased--; (let 'fm [m ss] be the maximum value of the friction and R the normal reaction force): as R and fm vary, fm/R remains | constant, thus they are in direct [] to each other | proportion--. | ||||
| Let Fm (m ss) be the maximum value of friction and R the normal reaction force: the value of Fm/R is called the | coefficient of friction; it is usually represented with | μ. The connection between Fm, R and μ may be written as | Fm = μR. In general, if an object remains stationary, then the force of friction, F, is given by F | =< μR. | ||||
| The value of μ (the coefficient of friction) depends on the []s in contact | surfaces--; a combination of [] sufaces has a relatively high value of μ, possibly even greater than 1 | rough--; a combination of smoother surfaces has a relatively [] value of μ | low--, possibly as low as 0.04. | |||||
| If an object is on an inclined, and is prevented from sliding down by a pushing force of P N acting up the plane. In the scenario, friction acts [] the plane | up--. The minimum value of P occurs when friction acts up the plane at its | maximum value (i.e. when the object is overcome the friction and would go down if there wasn't the | pushing force). | |||||
| What happens to the force due to friction once an object is moving? For mechanics 1, we use a simplified model in which: the force of friction remains | constant, regardless of | speed; the size of the force of friction is the same as the [] force of static friction | maximum--(i.e. F = μR). Note: as friction is a force, F = ma, may be used with it. | |||||
| Sometimes in questions involving dynamic friction: some elements (such as mass) are not told, in which case, just represent them using | algebra. Q) A object slides along a horizontal floor. The coefficient of friction is 0.6. What is the acceleration of the object? A) Let 'g' represent the constant of gravitational acceleration. Resolving vertically, R = | mg. Horizontally the only force is the | friction (F). F equals [] and [], thus [] = [] | ma and μR--ma = μR; you can substitute μ for 0.6 and [] for [] | R--mg--. The result is | 0.6mg = ma, hence acceleration equals | 0.6g = 0.6 x 9.8 = 5.88. | |
| If a string is being pulled, the pulling force and the tension are | equal. | |||||||
| The difference between the two forces becomes more significant as the angle made with the horizontal is | increased. | |||||||
| If the acceleration is defined in an equation, where the other side is a constant (e.g. a = gsin40-μgcos40), then the acceleration is | constant, thus you can use the suvat equations. | |||||||
| If there are two objects (A and B) moving in opposite directions at the same speed, where B has greater mass than A (A is heading rightward, while B is heading leftward): if they collide, A [] direction | switches--, while B moves [] at a [] speed | left--reduced--. However, A would be able to affect B enough to make it switch direction if it approached it with great enough | speed. | |||||
| The [] and [] of an object are used to define its momentum | mass--velocity--. | |||||||
| Momementum is a useful quantity to help us work out what happens in a c[] | collision. | |||||||
| If an object has mass m kg and velocity v m/s in some direction, then its momentum is given by momentum = | mv Ns. Note: product of the unit of mass (kg) and the units of velocity is kgms^-1, and through F = ma, we see 1 N = | 1 kgms^-2, thus 1 Ns = | kgms^-1. | |||||
| Does momentum have direction? | Yes. Does momentum have size? | Yes. Thus, it's a [] quantity | vector--. | |||||
| The direction of an object's momentum is the same as the direction of its | velocity. Note: with momentum, choose one direction of velocity to be positive; momentum in the opposite direction will be taken as negative. | |||||||
| If object A has momentum X Ns and object B has momentum Y Ns, then their total momentum is | (X+Y) Ns. | |||||||
| If two objects collide and there are no external forces: total momentum before collision = | total momentum after collision: this is known as the | 'principle of conservation of momentum'. Note: the principle can work in the presence of external forces provided that the resultant external force on each object is | 0. | |||||
| If two object coalesce: they [] [] and move as [] | stick together--one--. | |||||||
| Questions about momentum using vectors are treated in almost exactly the same way as one-dimensional question. The differences are: each velocity is given as a | (usually a column) vector; the speed of an object is the velocity vector's | magnitude. | ||||||
| The change in momentum is called | impulse and is measured in | Ns. Impulse is a [] quantity | vector--. | |||||
| The direction of the impulse on an object is the same as the direction of the | force that produced it. | |||||||
| When two objects collide, they exert an impulse on each other of [] magnitude and in [] directions | equal--opposite--. | |||||||
| Where F is a force, t is time, u is initial velocity and v is final velocity, the formula for impulse is | Ft = mv - mu. | |||||||
| The magnitude of an impulse is its | modulus. | |||||||
| The impulse exerted on object A by object B is the change in momentum in object | A. | |||||||
| A light horizontal beam will be in [] if the forces acting upon it act along the same line | equilibrium--; if the forces don't, a [] effect may be created | turning-- (and the beam would no longer be in equilibrium). The greater the distance [] the forces, the greater the turning effect | distance between--. | |||||
| When considering the equilibrium of a large object, we need to take into account, the [] effect of all forces | turning--. | |||||||
| The turning effect of a force about some point depends on: the force's [] and [] | size and direction--, and the [] distance from the point to the line of action of the [] | perpendicular--force--. | ||||||
| Where d m is the perpendicular distance to the line of action of the force, and the force is F N, then the moment of the force about P is | Fd Nm [#A47#]. | |||||||
| If there are multiple moments about a point (P), they can be combine to make a single [] moment | resultant--. For example: if you take clockwise as the positive direction, and there is a clockwise moment of A and an anticlockwise moment of B: the total moment of forces about P is | (A-B) Nm clockwise. | ||||||
| With moments: in place of a light rod, one may have a uniform beam of given mass. When taking moments, the weight of a uniform bean acts at its | midpoint. | |||||||
| For a 'larger object' (i.e. a beam) to be in equilibrium, the [] force acting on it must be zero | resultant-- and the resultant [] about any point must be zero | moment--. This is useful to, for example, find the force that supports exert on a beam. | ||||||
| Taking moment about A can be shown in writing as | M(A) and is likely to be followed by a | colon (e.g. M(A): X = Y). | ||||||
| Any object which is projected from a point is a | projectile. The only force acting during the flight is | the weight, meaning that the only acceleration which acts on a projectile is v[] | vertical. | |||||
| If a projectile hits the ground at the same height it was projected from: its path is s[] | symmetrical. | |||||||
| A projectile moves in two directions simultaeneously: we resolve the velocity [] and [] | vertically and horizontally. | |||||||
| (x,y ss) With projectiles: vertically: initially velocity = [], velocity at time t later = [], acceleration = [], displacement = [] | uy--vy--g (=9.8m/s²)--sy--; horizontally: displacement = [], initial velocity = [], acceleration = [] | sx--ux--0--; as the horizontal acceleration is 0, the velocity at time t later can be written (instead of vx) | ux. | |||||
| An object is projected at v m/s at a direction of θ above the horizontal, resolving horizontally - ux = | vcosθ, resolving vertically - uy = | vsinθ. | ||||||
| You can use the suvat equations with projectiles if the type (vertical or horizontal) specific quanities (e.g sx [x ss]) are all | of the same type. | |||||||
| If any number of objects are propelled (straight or at a decline) or dropped (from the same height), they will fall at an [] rate | equal-- (because the acceleration affecting them is the same); the horizontal speed doesn't affect the rate at which they fall. | |||||||
| If velocity is constant, displacement = | velocity x time. | |||||||
| If the horizontal velocity of an object is x and the vertical velocity is y: as a vector, the velocity is | (x, y) [in a column vector], the speed is thus | √(x²+y²). | ||||||
| The time of flight for a projectile is determined by its [] motion | vertical--. | |||||||
| The time of flight for a projectile to return to the level of projection is t = | 2usinθ/g (u is the initial velocity, θ is the angle of projection and g is the acceleration due to gravity). | |||||||
| The horizontal distance that a projectile travels before it lands is its | range; range = [...] x [...] | horizontal component of velocity x time of flight. | ||||||
| The 'time of flight' is the time a particle takes to move from its point of projection to the point where it [] the [] [] | strikes the horizontal plane--. | |||||||
| If you know the initial velocity of a projectile, then you can easily find an expression for both the horizontal and vertical [] at time t | displacement--: sx [x ss] and sy [y ss] respectively. You can then eliminate [] to produce a single equation connecting sx and sy | t--; this equation is called a | trajectory equation. However, sx is usually just called [] and sy is usually called [] | x--y--. | ||||
| [#A42#] With constant acceleration: where u is the initial veclocity in m/s, v is the final velocity in m/s after t seconds, a is the acceleration in ms^-2 and s is the displacement of the particle in m after t seconds: a = | v-u/t, thus v= | u+at, thus v² = | u² + 2as; s = [] = [] | (u+v/2)t = ut + at²/2. Note: the decorum units of measurement are given, but they are obviously adaptable, and also work where u, v, a and s are | vectors. | |||
| A vector has no vertical displacement when the j component equals | 0. | |||||||
| A support of a beam will exert a [] on the beam | force, and an object on the beam will exerts its [] on it | weight--. | ||||||
| A vector quantity has both | magnitude and direction. | |||||||
| A scalar has only | magnitude. | |||||||
| Distance is not affected by d[] | direction, thus it's a | scalar. | ||||||
| Newton's second law: | F = ma. 'F' is the [] [] acting on the object | net force-- (when there is more than one force, they combine to give the [] overall force | net--); 'm' is the object's []; 'a' is the object's | mass--acceleration-- - the acceleration and the net force have the same | direction. The force is measured in [], the mass is measured in [] and the acceleration is measured in [...] | Newtons (N)--kilograms (kg)--metres per second per second (ms^-2). F newtons is the force required to give a mass of m kg an | acceleration of a m/s². |
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Toluo
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