Measurements PDF

MEASUREMENTS · Measurement is the process of comparing something with a standard. ↑ The International System of Units or SI is a system of units , , widely used around the world. - It consists of seven...

MEASUREMENTS · Measurement is the process of comparing something with a standard. ↑ The International System of Units or SI is a system of units , , widely used around the world. - It consists of seven base units which all other units of , measurement can be derived from. List of prefixes : eXa- E 10ts Base Quantity Unit Symbol 1075 peta- Length meter M tera- giga- i 1072 109 106 Mass kilogram kG mega- Kilo- M 103 Time second S hecto- H 102 deka- da 107 10-1 Electrical current ampere A deci- Centi- a 10-2 K Milli- 10-3 i Temperature Kelvin substance amount mole Not micro- nano- 1 O 12 10 - PICO- Luminous intensity candela od fento- P 10-15 atto- A 10-18 · Accuracy is how close a value is to the accepted value. Precision is how close two or more measurements. ⑳ ⑧ o - 0 ⑧ ⑧ ⑧ ⑳ Low accuracy Low accuracy High accuracy Low precision High precision Low precision high accuraa (systematic error) (Random errors Percent error is computed when comparing an experimental result to · a value determined by theory or to an accepted known value. (ltheoretical value)) value - experimental % Err = x 100 theoretical value Percent difference is computed when comparing two values that are both · determined by experimentation. C ( I value 1-value 21 % Diff value value 2 100 1 = X · The scientific notation is used to shorten extremely large or small numbers. It is expressed as : - number of times to N X 1 On more the decimal (positive to the right Mantissa negative to the left) (Number between 1 to 10) Example 1 Express 0 0000072 in scientific notation.. 7 2 10 6/ - 0 0000072. =. x nu 6 times Example 2 Express 568 762 in scientific notation.. 568 762. = 5. 68762x102 & 2 times - To add or subtract using scientific notation Write each quantity. with the same exponent n. add the mantissas and copy the , exponent. (7. 4 x 103) + (2. 1x 103) (4 31 x104) + (3 9x183).. = 9 5x103. = (4 31 x104) + (0 39x104).. = 4 70X104v. (2 22 x 10 2) (4 10 x 10 3) - - -.. (2 22 x 10 2) (0 41x10- 2) - = -.. (2 63 x 10 2) - =. - To multiply using scientific notation multiply the Mantissa , and add the exponents To divide divide and subtract them. , instead. (8 104) x(5 102) 0 x 0 x - X107 (-5) -.. = (8 0x5 0) x 104.. +2 = 40 0x100. = 2. 3x1072/ Significant figures represent the margin of · error in a measurement. Guidelines to SigFigs. 1 Any non-zero digits are 2 2. Zeros between non-zeros significant. are significant. 845 3 Sf =.. 606 3 S. =. f 1 234. 4S. =. f 40 501. 5 Sf =.. 3. Zeros to the left of the. 4 Trailing decimal zeros are first non-zero digit are significant. not significant.. 0600 3 = 5 S. f.. 006 0 = 1 Sf.. 0 600. = 3 S.. f 0 082. = 2 S.. f Sigfigs in Calculations.In 1 addition and subtraction round the answer to the , fewest number of decimal places · 89 332. 2 097. -8 232 -> 90 4 /. = 1 98.. In 2 multiplication and division , round the answer to the fewest number of sigfigs. 2 8. -804 = 0 0611388789. 45032 - 13 *. > - 0 0611 /. · Dimensional analysis is based on the relationship between different Units that express the same physical quantity. By definition 1inch is equal to. It can also be written in.54cm We can write this as 2. reverse since they share the the ratio of : same value : in - - -54 We can use this ratio to Using this concept , to solve convert a unit to another : dimensional analysis problems : desired unit 12 00 in.. mien unit unit ven = 30 48 cm ,. => desired unit v This lets us cancel the units and calculate using the desired unit.  SCALARS AND VECTORS Scalars · are quantities that only have a magnitude and answer the question , "How much ?" - Represented using either an Italic letter (A) or a letter in the absolute value symbol (1A · Vectors have both magnitude and direction. It adds the question , "Which way ? " and is represented by both a letter and an arrow : > 7B The arrow's length represents the magnitude - AB. The tail is the point of origin of the vector. The arrowhead represents direction A. Each Vector can be expressed with a cardinal direction and angle , if visualized with a cartesian plane. or N -NE ↑ WofNE of N · angle from y W NOW-No * S of E & E to X E of S WofS L - SE S Cardinal and Ordinal Directions We can even express vectors in different ways in different perspectives : 900 There three ways we can express are & 900 400= 500 the vector : 18 m/s2 > 500 1) 8 M/s2 400 relative to the + X-axis 7 400f , 400 18 m/52 400 N of E 1800 #0%3600 , 1)8 M/s 2 , 500E of N - 2700 · Vector addition is the process of combining two or more vectors and finding their resultant vector or the shortest path from the tail of the first vector to the , final vector.  Can be done graphically or analytically. - Graphical Method 1) Sketch the vectors , connecting them "head/tip to tail". *. 2 Sketch the resultant vector E from the first Vector's R # E = + tail to the final vector's head. B ⑱. 3 Solve for the length of E using a ruler, and its * angle using a protractor. When using a ruler use the scale factor to convert length to actual ,. 1 cm in a ruler = 1 km in the word problem). magnitude (e g. Example Ana walked 9 00m East from her house and walked another 5 00m.. North to reach her school What's her overall displacement ?. - # - * 0 = 00 0 = 00 5m 3m 5m - 10. 1 line = Im 1 line = Im 10 = 29 10 3 3. D D 0 = 00 am 1 line = Im 0 = 00 am 1 line = Im Scale factor Solve for magnitude of R using connect the given vectors. ruler and its angle using protractor. The rule also applies when connecting multiple vectors. Analytical Method In this method , we can simply work with individual vectors : & & ↑ vector be separated into two components : can Ty * the X-component and y-component. & By getting the sum of the vectors' X-and - #x components we can get the X-and y-components , of the resultant vector. We can then find two things : To find the magnitude , Use the To find the direction and angle. Use 4ythagorean Theorem : the inverse tangent : a2 + b2 = ch tan - ( The flow of this method are as follows : 1. Find the X-andy-components of each individual vector (if their magnitude and direction/angle are already given). 2. Add up all the X-components to get the X-component of the resultant vector. Do the same with the y-components..solve 3 for the magnitude and direction/angle of the resultant vector. Example Ana walked 9 00m East from her house and walked another 5 00m.. North to reach her school What's her overall displacement ?. We can use this method to verify the graphical one : - # * MAGNITUDE DIRECTION 0 = 00 b2 = c = 5m a2 + +an 3m 10 (9m) 2 (5m)2 2. = 1 line = Im + = #4m) (5m) =( 2 + = c O = tan - 129. 10 D 3 #06M2 C = 0 = 00 am 1 line = Im Scale factor c = 10 3 M. 8 = 29 10N of E. Example Find the components of a vector with 100 ON. as its magnitude Pointing. 30. 00 North of East. * We can find the components using sine and cosine : CAH T SOH sing = cost =d To solve forx (adi) : To solve fory (opp) : #x - COS 30 08 - sin 30 00 =. 100 ON. 100 ON.. 100 ON. (cos 30 0%. = Ex 100 ON (sin 30 0% =.. My #x = 86 6N. #y = 50 ON. Example Rose went to the grocery store and walked 12 Om 48 0 "West of North. ,. to the end of an aisle She then turned right and walked 8 50m (East).. down the end aisle Finally she made another turn and walked 9 50m ,. ,.. 2.00 North of East Determine the overall displacement of Rose. 2.  We can create a table of values to organize them and easily find the components : VECTORS X- COMPONENT Y COMPONENT - Quadrant# - - # = 12 Om 48 00 WofN. ,. #x = - A sin O #n = A cos O (sin 48 00 (cos 48 0 % - 12 Om 12 Om - = - =.... & - = 8 91m. = 8 03m. B = 8 50m , E. -x = 8 50m. By = Om [ V -- - > - = 9 50m 22 00N of E... &x = C COSE Cy = J sin O = 9 50 COS 22 0.. = 9 50m Sin 22 00.. 1 = 8 81m. = 3 56M. = Now we can solve for R : * = + Ex x x x + y = #y + By + Ty = (8 91m) + 8 50m + 8 81M... = 8 03m + On + 3 56m.. = 8 40m. = 11 GM. Since both Ex and Ey are positive , I is located in Quadrant I. MAGNITUDE DIRECTION 1 - - - - - - 7 O = tan-1 ( EN E = 18. 40m + A = 54 10 relative to. + x-axis ° [ = 14 3M. = 54 1. N of E = 35 90E of N ~. Therefore the final answers, are : 14 3 m 54 10 relative to ·. ,. + x-axis ° · 14 3 n. , 54 1. N of E · 14 3 m.. 35 90 E of N. Example If Bambi's overall displacement after running is 458m 49 00.. East of South determine the Path D taken by Bambi if she , walked the following paths : · Path A : 108m South · Path B : 360m , 25 00 East of North. Path C 275m 31 00 South of East · : ,. we can use the table method again : VECTORS X COMPONENT - Y COMPONENT - # = 108m South, #x = Om #y = - 108m · B = 360. M. 25 00 E of N. x Esint = By = BCSE = 360 m(sin 25 0%).. = 360. m (COS 25 0%. & = 152M = 326m v - = 275m , 31 00 S of E. *= cos Ty = - [sinG = 275m (COS 31 0 %. = = 275m(sin 31 0%. ↑Te 236m 142M = = - [ I = ??? Tx = ??? Ty = ??? & = 458m 49 00E of S ,. #x Esin O = Ey = - RcosO = 458m(sin 49 00 458m (coS 49 0% = >.. 1 = 346M = - 300. M ↳ Tx - To find and Dy , we can subtract the components of the starting vectors to the resultant vector : [x Ex (*x + Bx = - + [x) Ty Ey (Ay + By () = - +  # = 346m - (0m + 152m + 236m) Ty = - 300. m - [ - 108m + 326m + ( - 142m)] = 346m - 388m = - 300. M - 76M = - 42 Om. - = 376m Both components are negative , so i is in Quadrant #. We can now find the magnitude and direction : MAGNITUDE DIRECTION [ > OI 0 tan())) I ↓ = =42 0m)2 / I / 376m)2. - = 83 60S of W. = 378m = 6 400W of S. k --- v = 2640 relative to X-axis Therefore the final answers are : , 378m · 83 60S of W.. 378m · 6 400W of S,. · 378m 2640 relative to X-axis. A unit vector is a vector with a magnitude of. · 1 - It's only purpose is to describe direction in space a. This also allows us to express vectors in unit vector notation : - Unit vectors are represented by the letter with a caret or nat (1) on top AxY" & # = + AyY + Azk" It thi a - - - wouldoklike The value Ax points to the + X aXiS - The value Ay points to the + y -axis The value Az points to the + 2 - axis - This makes computing for the vector sum easier as we simply add like , components : # = AxY + AyY + Azt & = (Ax + Bx)4 + (Ay + By)y + (Az + Bz)π B = BxY + ByY + B2π = Rx4 + RyY + R2Y And also easily lets us find the magnitude : & = Rxx + RyY + R2Y (Fl ((Rx)2 + (Ry)2 + (Rz)2 = - This also lets us do arithmetic with it easily : Example Given T (6 004 + 3 005 1 001) m and E (4 004 =.. -. =. - 5 00y. + 8. 00)M find the magnitude of displacement 25-E.  Substitute vectors... 2T - E = 2(6 004 +. 3 005. - 1. 00M) - (4 004. - 5 005. + 8. 00M) = (12 004 +. 6. 005 - 2. 00K) - (4 004. - 5. 005 + 800M). = (12 00 4 00)4 + (6 00 + 5. -... 00)5 + 72 00 8. -. 00)4 = 8 004 + 11 005 10 00m.. -. Then get the magnitude : 1 2T El - = /(8 001 2. + (11 00)2. + 7- 10 00) 2. = 2205 00m. = 16 9m. · Vectors can be multiplied in two ways. - The dot product only yields a scalar value , and is denoted by F B.. (or scalar product) a b We can see how it works visually : MagnitudeMagnie Perpendicular line 1AI (IBICOSO) ↑ ! · V & cosine 10 > y -- > Line Projection * IBI cos O Given Vectors and B... we get the component of # that lies alongside T... Then multiply like usual. It's like we're aligning vector I to Vector A to easily multiply them both. This process is also commutative meaning the reverse is true : , co ~ IBI (lAlcost) ·perpendicular V & I > > (x1 10 * - IAI - Given Vectors and B... we get the component of * that lies alongside B... Then multiply like usual. Either way this gives us /AllBIcost as an equation for the dot product. This also , has multiple properties : 7 1 COSO = + COS 900 8 = COSO = - < 20 - If 0 < 0 < 900 and the The dot product of two If 90001800 the dot , dof product is positive. Perpendicular vectors is 0. product is negative. The dot product of two vectors can also be expressed as the sum of the products of their respective components. This is the result of the multiplication of vectors in unit vector notation. # B. = (AxY + AyY + AzY)(BxY + By4 + Bzπ) = (AxBx)4 4. + (AxBy)4 j - + (AxBz)4 k +. (AyBx) 5 4 · + (AyBy) 5 Y · + (AyB2) Y R + · (AzBx)( : 4 + (AzBy)Yoj + (AzB2) Yo Multiplying two unit vectors give us : Thus , through substitution : => U = (AxBx)(1) + ( x By))0) + (x Bz)(0) +.Y 4 = yy. = EY. = (1)(1) cos00 #y Bx) (0) + (AyBy)(1) + B2)(0) + = 1 ( zBx)(0) + ( 2By)(0) + (AzBz)(1) ↳ x ↑j = Y.k = yk. = (1)(1) COS900 = AxBx + AyBy + AzB2 = G This gives us two ways to find the dot product : 1AIIBIcos@ AxBx + AyBy + AzBc Method 1 Method 2 Example Find. B of the given two vectors. B 5 00 - =. A = 4 00. METHOD 1 To get O between the vectors : 1300 55 = 77 00. Then substitute : [ #B = 1AIIBIcoSO = (4 00) (5 00) cos 77 00... = 4 50. METHOD 2 Find the components of each vector : Substitute values into formula : * B IAICOs O - IBIsinG * B. = AxBx + AyBy X- COMPONENT (4 00) cos 530. (5 00). sin 400 = (2 41)( 3 21). -. + ~ 2 41. = - 3 21. (3 20)(3 83).. ~ 4 52. 11 sin O IBICOS O Y COMPONENT (4 00)sin530. (5 007 cos400. Both methods give about ~ 3 20. ~ 3 83. the same answer.  - The cross product , on the other hand , yields a vector value , denoted by # XB. (vector product) The cross product describes two results : The vector product is This concept can be visualized by perpendicular to both vectors. the Right Hand Rule. B · 1 * ~ *x 7 *x 1 > 7 > · The magnitude/length of the vector product is also the area of the parallelogram formed by mirroring the vectors. Remember : As = bh 11(1 # /SinG) = IF/IBIsinG ~ * * *x -, 111/IIIIII · t E & is is Since CP gives the area We get h by getting the giving us the formula ! of the parallelogram... y component - of A via SinG... We can determine the direction of this vector via RHR giving us the final , formula 111B/sinOn where n is the unit vector perpendicular to both vectors. , , However there's , easier method of finding the vector product via a matrix an. using the components of the vector. Given vectors # Axi + AyY = + Ack and B = BxY + Byj + Bak , we can create a 3 x 3 matrix. #x 11 = Then we find the determinant of this matrix. I] #x - [ * -[*]5 + [*] Then we find the determinant of each 2x2 matrix : [u- "Cross multiply" + [] (AyB2-AzBy)T - (AxB2-AzBx)y + (AxBy- AyBx)π giving us the formula This derivation is the easiest way to know the formula.. Cunfortunately) I we flip the equation , the resulting vector also flips This is evident in. the Right Hand Rule. B i Thus this process is NOT E · commutative. Instead. (# X B) = - (BX). Example Vector # has amagnitude of 6 units and is in the direction of the Xaxis. Vector B has magnitude ofa units and lies in the Xy-plane , making anangle of 300 with the X-axis. Find the vector product of A and B. METHOD 1 We can use lAllBIsinOn to find the vector : #xi = 1AllBIsinOn = (6)(4) sin30on = 12 units h By RHR , the resultant points to the z-axis. Therefore , #xB = 12 units. METHOD 2 First find the components of each vector ,. - * B M X- COMPONENT 6 units IBICOSO 300 25 I = Y- COMPONENT O units (ulBIsin 300 w Bx = 2 Then substitute into the formula : (AyB2-AzBy)T - (AxB2-AzBx)y + (AxBy- AyBx)π = [(0)(0) - (0)(2)]4 - [(6)(0) - (0)(253)]5 + [(6)(2) (O)(253)] - 12 1 = 04 -05 + = 11 units k  KINEMATICS · Kinematics is a branch of classical mechanics which describes the motion of bodies and systems without consideration of outside forces. · Motion is relative - meaning it depends on the observer or the reference point/frame. - This reference point may be inertial (stationary or moving at constant , velocity) or non-inertial (moving/changing speed). DoE - -S - J- g -. 0 - 0.0 - -S & In the boy's POV the girl is moving.. , I... but in the cat's POV , She's not moving ! · An object in motion involves displacement , velocity , acceleration , and time. Distance(X) is a scalar quantity that describes & the total path covered by an object. VS. Displacement (*) is a vector quantity describing ⑧ the direct length between two points. The average speed (v) is a scalar The average velocity (v) is a vector that describes the change in that describes the change in VS distance over time. displacement over time. finitial value -- = = finala X - Xo V = Fto Instantaneous speed and velocity is the quantity at a specific time. We can get this through calculus. We can take the limit of the velocity formula as the change in time approaches O (At-0) which gives us the derivative of position , with respect to time : -= in Acceleration (a) is defined change in velocity over time. as the = An object is accelerating if it's : --- s speeding up r· -i Y -- changing directions slowing down Having negative acceleration (deceleration) means that the object's speed changes from a higher to lower valu e. Like instantaneous speed/velocity we can take the limit of the acceleration , formula as the change in time approaches O (At-0) which gives us the , derivative of velocity with respect to time : =in > Acceleration < - Y ithe 2nd derivative of Acceleration is the 1st DifferentiateWith Instantaneous displacement derivative of velocity respect to time velocity formula This also gives us the second derivative of displacement. With this two , types of motion can occur : ↑ > - + -- q 123456 123456 Time (min) Time (min) An object in uniform motion has An object in uniform accelerated a constant speed/velocity and motion has constant acceleration thus , zero acceleration. instead. Pos1 - vei1 v(t) = 2 Pos1 X(t) = x2 Velnv(t) = 2x 2 X(t) = 2X 3 < Time < Time 3 < 3 > Time Time Position Time - Velocity Time - Position Time - Velocity-Time Acc 1 Accel 1 The behavior of 9 (t) = 2 a(t) = 0 quantities can better 3 be seen as a graph & Time < <. Time Acceleration Time - Acceleration Time -  Thus , we get this relationship : Differentiate Differentiate X(t) - v(t) - a(t) Integrate Integrate Position Function Velocity Function Acceleration Function Where we can inputt to get the instantaneous quantity at said time. Example A bus is moving along the X-axis so that its position at time t is 5 given by the formula x (t) 3 0t2 + 8 Ot 2 0t = -.... Compute its velocity and acceleration as functions of t Then decide. , which direction (left or right) the bus is moving when t=1 and whether its velocity is increasing or decreasing. J To get the velocity function : POWER RULE 2 5 (x1) = 1x1 - 1 X (t) = 3 0t + 8 0t 2 0+ -... 7- X(t) = v(t) = 3 0(2) +. 2= + 8 0(1) +. - 2. 0(5/2) + 5/2 - 7 = 6 0t. + 8 0. - 5 07 3/2. To get the acceleration function : v'(t) a(t) 6 0(1) + + 7 8 0(0) 5 0(3/2)t 32 1 - - = =. + -.. = 6 0. - 7 57'. Now we can substitute t 1s into both equations : = v (1) = 6 0(1). + 80. - 5. 0(1)3/2 a(1) = 6 0. - 7. 5(1)7/2 = 6 0. 8 0 +. - 5 0. = 60. - 75. = 9 0m/S. = = 1 5 m/s2. Since it's positive the bus , is