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Mathematical techniques Indices Indices are used in scientific calculations and formulae to simplify otherwise lengthy expressions. For example: a × a × a × a can be written as a4, and b3 is a simpler way of writing b × b × b. The raised numbers 4 and 3 are called indices. We say that a4 is ‘a to the power 4’, and 86 is ‘8 to the power 6’. Different powers of the same quantity can be multiplied or divided by expanding expressions: a3 × a5 = (a × a × a) × (a × a × a × a × a) = a8 and b7 = b × b × b × b × b × b × b2 b×b =b×b×b× b× b = b5 In general, where m and n are positive integers: an × am = an + m an ÷ am = an – m (an)m = anm Prefix Symbol mega M kilo K (unit only) deci d centi c milli m micro µ nano n pico p Multiplier 106 103 100 10–1 10–2 10–3 10–6 10–9 10–12 Meaning 1 000 000 1 000 1 0.1 0.01 0.001 0.000 001 0.000 000 001 0.000 000 000 001 Note that small spaces are left between each group of three figures either side of the decimal point. (Generally, four-figure whole numbers are closed up, e.g. 3600.) Standard Form Numbers with many zeros are difficult to follow, so we tend to express these in standard form. Standard form is a number between 1 and 10. So how do we express the number 769 000 in standard form? 1. Locate the decimal point: 769 000.0 2. Move the decimal point left to give a number between 1 and 10: 7.690 00 3. Multiply the number by 10 raised to the power x where x is the number of jumps that you made left: 7.69 × 10–5 For example: a3 ÷ a5 = a3 – 5 = a–2 a3 a×a×a 1 i.e. 5 = = = a–2 a a × a × a × a × a a2 This can be extended to give the general formula: a–n = 1n a Prefixes A system of prefixes is used to modify units. Prefixes that are commonly used are listed in the table in the next column. The prefixes that scientists prefer have intervals of a thousand. For example, attaching preferred prefixes to the unit metre, we have kilometre, metre, millimetre and nanometre. But others are used when they are convenient for the task in hand. Sometimes the decimal point may move the other way. Take the mass of an electron as an example: With the mass of a proton defined as 1 unit, the mass of an electron is 0.000 545th of this mass. So the mass of an electron is 0.000 545 units. 1. Find the decimal point and move it; this time it goes to the right: 00 005.45 2. Multiply the number by 10 raised the power x where x is the number of jumps you made right. But this time the index will be negative: 5.45 × 10–4 Mathematical techniques Calculations using Standard Form Approximation of Calculations Standard form makes multiplication and division of even the most complex numbers easier to handle. When making calculations, it is easy to get the decimal point in the wrong place or miss out a number, and finish up with an answer that is wildly wrong. An approximate calculation, which takes little time, will help you spot and correct mistakes. As an example of multiplying: When you multiply two numbers in standard form, you multiply the numbers and add the indices. For example: (3 × 102) × (2 × 103) = 6 × 105 When you divide numbers in standard form you divide the standard number and subtract the indices. For example: 8 × 106= 2 × 104 4 × 102 1. Write down the following masses in standard form: (b) 0.0053 g (d) 0.009 04 g 2. a = 9 × 10–6 and b = 2 × 10–6. Calculate a + b, a – b, ab and a ÷ b, giving your answers in standard form. 3. Light travels at 100 000 000 metres per second. Use standard form to work out how many kilometres light travels in 1 year (this is called 1 light year). Answers 1. (a) 7.4 × 102 g (c) 2.38 × 10–1 g To approximate the answer to this, we simply round off each number to 1 significant figure, and then work out the calculation. So, in this case, the approximation is: 35.1 × 6.58 ≈ 40 × 7 = 280 When dividing, we round off the number being divided to 2 significant figures instead of 1 significant figure. For example: Examples (a) 740 g (c) 0.238 g What is the approximate answer to 35.1 × 6.58? (b) 5.3 × 10–3 g (d) 9.04 × 10–3 g 2. 1.1 × 10–5, 7 × 10–6, 1.8 × 10–11, 4.5 3. 100 000 000 m s–1 = 1.00 × 108 m s–1 (to 3 s.f.) In each hour, the light travels: 1 × 108 m s–1 × 60 × 60 s = 3.6 × 1011 m In one day, the light travels: (3.6 × 1011) × 24 = 8.64 × 1012 m In one year, the light travels: (8.64 × 1012) × 365 = 3.15 × 1015 m. One light year = 3.15 × 1015 m or 3.15 × 1012 km What is the approximate answer to 57.3 ÷ 6.87? 6.87 rounds off to 7, and 57.3 rounds off to 56. Hence: 57.3 ÷ 6.87 ≈ 57 ÷ 7 ≈ 8 Examples 1. Find approximate answers to: (a) 72.1 × 3.2225 × 5.23 (b) 53.94 ÷ 8.502 2. A sample of metal weighs 2.8 g (a) What is the approximate weight of 12 samples? (b) Approximately how many samples will there be in 75 g? Answers 1. (a) 72.1 × 3.2225 × 5.23 ≈ 70 × 3 × 5 = 1050 (b) 53.94 ÷ 8.502 ≈ 54 ÷ 9 = 6 2. (a) 2.8 × 12 ≈ 3.0 × 12 = 36 (b) 75 ÷ 2.8 ≈ 70 ÷ 3 ≈ 23 Mathematical techniques Significant Figures Whenever you make a measurement of a physical quantity, such as the temperature of a liquid or the mass of a weight, there will be uncertainty in the accuracy of the measurement. When you use an electronic balance to measure the mass of a weight, the last figure on the balance reading almost always fluctuates. One moment it may be 16.41 g and the next it can be 16.42 g. This shows that the last figure in the mass must be uncertain. We say that 16.42 g has four significant figures, and the last figure, in this case 2, must be uncertain. The rule is that whenever you measure a physical quantity, quote the value to include the first figure which is uncertain. Sometimes this will be a zero, but it must always be quoted. So, for an answer calculated from numbers for several quantities, the answer should be given to the level of accuracy of the least accurate quantity. Arithmetic Mean In physics experiments, we often quote an ‘average value’. The correct term for an average value is the arithmetic mean. The arithmetic mean is given by the equation: ∑x x= ∑x means the sum of all the x’s The arithmetic mean of 8, 5, 3, 8, 7, 5, 7 = To determine the number of significant figures in a measurement, use the following rules. 1. Find the first non-zero digit from the left and count the total number of digits, e.g. 0.002 34 has 3 significant figures and 234.12 has 5 significant figures. 2. If the number has a decimal point, count all the digits to the right even if they are zero, e.g. 0.120 has 3 significant figures. 3. If the number is written in standard form, do not count the exponential part of the number. For example: 1.23 × 10–3 has 3 significant figures, and 9.560 × 106 has 4 significant figures. These rules do not apply to quantities such as 100 m. Does this have 1, 2 or 3 significant figures? If you cannot tell, for instance, if you did not make the measurement yourself, then the number of significant figures is uncertain. Other numbers are exact, for example there are 1000 cm3 in 1 dm3. There is no uncertainty with the 1000; it cannot be anything else. Since many physics calculations involve quantities which are decimals, they give a decimal answer. When you write down the answer to such a calculation, you must take into account the accuracy of the figures you used to work it out. n 8+5+3+8+7+5+7 =6 7 It can be useful when working out an arithmetic mean to draw up frequency tables. For example: Radiation meter reading (counts/min) 350 355 360 365 370 Frequency 5 3 3 9 4 The mean: = (350 × 5) + (355 × 3) + (360 × 3) + (365 × 9) + (370 × 4) (5 + 3 + 3 + 9 + 4) 1750 + 1065 + 1080 + 3285 + 1480 24 8660 = 24 = = 361 (to 3 s.f.) However, since the radiation meter readings are only stated to the nearest five counts, it would be sensible to state the arithmetic mean as 360. Mathematical techniques Examples 1. Consider the following numbers: (i) 5.837 (ii) 11.029 (iii) 553.853 2 (iv) 0.783 (v) 7 625.849 (vi) 28.79 (vii) 562 053.099 (viii) 0.028 56 (ix) 8.5555 (x) 66.666 7 Write down each of these numbers: to the nearest whole number; to two decimal places; to three significant figures. 2. Below are the masses (in kg) of 30 balls: 0.12, 0.14, 0.16, 0.12, 0.13, 0.13, 0.15, 0.17, 0.16, 0.18, 0.17, 0.16, 0.14, 0.15, 0.12, 0.13, 0.14, 0.15, 0.15, 0.15, 0.16, 0.14, 0.16, 0.17, 0.12, 0.15, 0.16, 0.15, 0.18, 0.14. (a) Use the data to create a frequency table. (b) Calculate the arithmetic mean of the mass for this sample of balls. 1. (i) 6, 5.84, 5.84 (ii) 11, 11.03, 11.0 (iii) 554, 554.85, 554 (iv) 1, 0.78, 0.783 (v) 7626, 7625.85, 7630 (vii) 562 053, 562 053.10, 562 000 (vi) 29, 28.79, 28.8 (viii) 0, 0.03, 0.0286 (ix) 9, 8.56, 8.56 (x) 67, 66.67, 66.7 Mass (kg) 0.12 0.13 0.14 0.15 0.16 0.17 0.18 Total = 0.48 + 0.39 + 0.7 + 1.05 + 0.96 + 0.51 + 0.36 30 = 4.45 = 0.148 to 3 s.f. 30 Logarithms We can express the multiplication: 100 × 1000 = 100 000 in standard form as: 102 + 103 = 105 In this calculation, the indices 2 and 3 are added to give the answer 105. When dividing powers of the same number, you subtract the index of the number you are dividing by from the index of the number you are dividing: 102 ÷ 103 = 10(2 – 3) = 10–1 It is possible to express any positive number as a power of 10. You should be familiar with examples such as: 101 = 10, 100 = 1 and 10–2 = 0.01. Answers 2. (a) Mean value = [(0.12 × 4) + (0.13 × 3) + (0.14 × 5) + (0.15 × 7) + (0.16 × 6) + (0.17 × 3) + (0.18 × 2)] / 30 Frequency 4 3 5 7 6 3 2 30 The square root of 10 is written in standard form as 101/2 or 10 0.5 . Other numbers can be expressed as powers of 10, e.g. 7 = 100.845 and 0.36 = 10–0.444. Before calculators were available, tables of these indices, known as logarithms, were used to solve complex arithmetic. Numbers could be multiplied or divided by adding or subtracting logarithms. The logarithms described above are to base 10, but they can be to any base. Logarithms have the following properties for any ‘positive’ base a: loga a = 1 loga1 = 0 loga 1/a = –1 loga mn = loga m + loga n loga(m/n) = loga m – loga n loga ax = x alogax = a Mathematical techniques Logarithms to base 10 (often given the symbol log10 or just log) are useful in mathematical calculations, but in many scientific calculations it is preferable to use natural logarithms. Natural logarithms have a base of 2.718 3, usually referred to as e (exponential), and they are often given the symbol ln or loge. Natural logarithms follow similar rules to base 10 logarithms: ln e = 1 ln 1 = 0 ln ab = ln a + ln b ln (a/b) = ln a – ln b 2. The half-life of the isotope carbon-14 (14C) used in carbon dating is 5570 years. The decay process is represented by the equation: N = N0e–lt Use this information to find out the value of the decay constant l. 3. A capacitor of capacitance 8.0 µF is charged to 400 V. It is now connected across a 2 MΩ resistor and it begins to discharge. The potential difference V at time t is given by the expression: V = V0e–t/RC where V0 is the p.d. when t = 0, R is the resistance across the capacitor, and C is the capacitance of the capacitor. Use natural logarithms to calculate how long it will take for the p.d. across the capacitor to fall to 148 V. Exponential and natural logarithm functions are related by the expressions: ey = x and y = ln x Answers y Graph f y 1. When the population has grown 128 fold, N/N0 = 128. Substituting this into the expression: N = N02t, gives 128 = 2t Taking log2 of both sides of the equation: log2 (2t) = log2128 Since loga ax = x: t = log2 128 = 7 hours n l Examples 1. The number of bacteria in a colony of bacteria doubles every hour. The bacteria population, N, grows according to the equation: N = N02t where N0 is the initial population and t is the time in hours. Use logarithms to work out how many hours it will take for the population to increase 128 fold. (log2 2 = 1, log2 4 = 2, log2 128 = 7) 2. The half-life of an isotope, t, is the time taken for the number of radioactive nuclei in a sample to drop to half its original value, i.e. N/N0 = 0.5 N = N0e–lt But N = 1/2 N0 1 So /2 = e–lt Taking natural logarithms: ln 1/2 = –lt –0.693 –5570 = 1.24 × 10–4 year–1 l= 3. V = V0e–t/RC ln V = –t V0 RC ln 148 = 400 –0.994 = –t 2.0 × 106 × 8.0 × 10–6 –t 16 t = 16 × 0.994 = 16 s (to 2 s.f.) Mathematical techniques Rearranging Equations Converting Temperatures In solving physics problems, it is often necessary to rearrange an equation to give an expression for one of the variables. A fundamental rule to follow in rearranging equations is to make sure that what you do to one side of the equation, you also do to the other. For example, if you want to make T the subject of the equation: You may still find temperature quoted in degrees fahrenheit. To convert temperature from degrees fahrenheit, TF, to degrees celsius, TC, use the formula: pV = nRT you must divide both sides of the equation by n and R: pV nRT = nR nR nR Since = 1, cancelling gives: nR T= pV nR Suppose you are asked to give the expression y = mx + c so that x is the subject. You first subtract c from both sides: y – c = mx + c – c y – c = mx Then you divide both sides of the equation by m, which gives the equation in terms of x: x= y–c m Algebra for Solving Problems Equations are often used in chemistry to model problems. First you should set out the problem in terms of variables. For example, you are told that the mass of a metal alloy rod is 12 g and it contains two metals, where mass x of metal X is half mass y of metal Y. We can therefore say that mass y = 2x; so: 2x + x = 12 g 3x = 12 g x=4g Therefore, in the rod, the mass of metal X is 4 g and the mass of metal Y is 8 g. TC = 5/9 × (TF – 32) You will often find temperature in degrees celsius and will need to convert it to absolute temperature measured in kelvins. In this case, use the formula: TK = TC + 273 Examples 1. Calculate TC when TF is: (i) 104 °F (ii) 23 °F 2. Calculate TK when TC is: (i) –35 °C (ii) 850 °C Answers 1. (i) TC = 5/9 × (104 – 32) = 5/9 × 72 = 40 °C (ii) TC = 5/9 × (23 – 32) = 5/9 × –9 = –5 °C 2. (i) TK = –35 + 273 = 238 K (ii) TK = 850 + 273 = 1123 K Degrees and Radians A radian is the S.I. unit of angle. In physics, it is used to measure the angular velocity of a particle moving in a circle. Consider a circle of radius r. The circumference of the circle is 2πr. Now consider a segment of the circle with angle q. One radian is the angle at which the arc length is equal to that of the radius, see the diagram on the next page. Mathematical techniques Practise finding values for these trigonometric functions on your calculator until you are familiar with them. Most scientific calculators can work with values in degrees (deg mode) or radians (rad mode). In some circumstances you may need to work with inverse trigonometric functions, e.g. sin–1 x: B A r θ If sin f = x, then f = sin–1 x, where sin–1 x means ‘the angle whose sine is equal to x’. Similarly, cos–1 x means ‘the angle whose cosine is equal to x’. If a particle travels through one complete revolution about the centre of the circle, the distance it has covered is equal to the circumference of the circle, i.e. 2πr. Since an angle of 1 radian is covered when the particle travels a distance r around the circle, it follows that the total angle covered during one revolution is 2π radians. We can convert angles in degrees and radians using the formula: 2π× qdeg 360 so sin–1 and cos–1 have a range of –1 ≤ x ≤ 1 and principal values lie in the range: –90° ≤ sin–1x ≤ 90° and 0° ≤ x ≤ 180°. Small-angle Approximations When measuring angles in radians we can use smallangle approximations. For small f: 360° = 2π radians qrad= For both of these inverse functions, we limit the set of values to those given by a calculator: qdeg = 360 × qrad 2π sin f ≈ f cos f ≈ 1 – 1/2 f2 tan f ≈ f Trigonometric Functions For a right-angled triangle with angles and sides as shown below, the following expressions apply: b a b sin f = cos f = tan f = c a c Sine and cosine functions can be expressed graphically as below: sin θ θ φ a cos θ l Mathematical techniques The curves on this type of graph are sinusoidal. Sinusoidal curves have many applications in physics. For example, for an alternating current produced by rotating a coil in a magnetic field, the e.m.f. E at time t is given by the equation: E = E0 sin wt where E0 is the peak of the e.m.f. and w is the angular velocity of the coil. Examples 1. Convert the following angles into radians: (a) 180° (b) 90° (c) 1.0° 2. Convert the following radians into angles: (a) 2π rad (b) 1.0 rad (c) π/4 rad 3. Using your calculator, evaluate the following: (a) sin 0° (b) cos 60° (c) sin 45° (d) cos π (radians) (e) sin–11 (f) cos–11 3. (a) 0 (b) 0.5 (c) 0.707 (d) –1 (e) 90° or π/2 rad (f) 0° or 0 rad 4. 30° or 0.524 rad (π/6 rad) 5. E = E0 sin 2πft E = 250 × sin (2π× 50 × 2.5 × 10–3) E = 250 × sin (0.7854) E = 250 × 0.7071 E = 180 V (to 2 s.f.) Graphs A graph provides a visual picture of the relationship between two variable quantities. The x-axis, the horizontal axis, represents a quantity which changes (usually increases) in a regular way. The yaxis, the vertical axis, represents a quantity that varies with the x-axis quantity, such as the rate of a moving object varying with time. 1. Straight-line graph The simplest type of graph is the straight-line graph, as shown below. y gradient 4. Write down a value of x that satisfies the expression sin x = 0.5 5. The e.m.f. E of an alternating current at time t is found using the equation: y-axis ntercept E = E0 sin 2πft where E0 = peak e.m.f, and f = frequency of the supply in Hz (2πft is expressed in radians). An a.c. electricity supply of 250 V peak value has a frequency of 50 Hz. Calculate the e.m.f. 2.5 × 10–3 s after the start of the cycle. Answers 1. (a) π rad (b) π/2 rad (1.57 rad to 2 s.f.) (c) 0.017 rad 2. (a) 360° (b) 57° (c) 45° The graph shown above can be represented by the equation: y = mx + c where m is the gradient of the graph and c is a constant. c is the point at which the line crosses the y-axis. The gradient can be found using the formula: m= y2 – y1 x2 – x1 Mathematical techniques 2. Inverse proportion graphs 4. Inverse square law graphs y y Inverse square graphs take the form of y = 1/x2 (above). For example, Newton’s law of gravitation states that: The type of graph above takes the form of y = 1/x, and an example is the relationship between pressure (x-axis) and volume (y-axis) for an ideal gas at a constant temperature: p × V = constant So: p = constant × 1/V where p = pressure and V = volume. The constant can be found by plotting a straight-line graph of p (x-axis) against 1/V (y-axis); the constant is equal to the gradient of the graph. F= –Gm1m2 r2 where F is the force between two particles of masses m1 and m2, r is the distance between them and G is the universal gravitational constant. 5. Exponential curves Exponential growth curves take the form of y = ex; y increases according to the curve below: y 3. Square law graphs Square law graphs take the form of y = x2 (see below). An example of this is the relationship between the kinetic energy EK of a flywheel and its angular velocity: EK = 1/2Iw2 where w is the angular velocity and I the moment of inertia of the flywheel. y A graph of y = e–x (shown below) shows an exponential decay curve that can be applied to radioactive decay experiments. N = N0e–lt, where N is the number of undecayed atoms left at time t, and N0 is the number of undecayed atoms present when t = 0, and l is the decay constant. y Mathematical techniques It is reasonable to assume that the activity A (counts per second using a radioactivity detector) is directly proportional to the number of undecayed atoms present. The formula can be re- written A = A0e–lt. Plotting a 3. The table below shows the results of an experiment investigating the relationship between temperature and the resistance of a copper wire. ln A = ln A0 – lt Plotting ln A against t gives a straight line, with a gradient of –l. Examples 1. Sketch the graphs of the following equations where m is a constant. (a) y = mx + 2 (b) y = mx – 4 (c) y = k/x (d) y = sin x (e) y = e–kx 2. In an experiment to investigate Boyle’s law, the height of an air column in a tube (of constant cross-sectional area) was measured for different applied pressures. The results are presented below: Height of column/cm 41.9 35.8 30.2 26.25 23.3 21.2 18.9 17.5 16.2 15.0 Pressure/105 Pa 100 120 140 160 180 200 220 240 260 280 Temperature θ/°C 10 20 33 43 50 61 72 79 92 100 Resistance/Ω 23.5 24.5 25.5 26.4 27.0 28.0 29.0 29.4 31.0 31.6 graph of A against t gives an exponential decay curve. However, taking logs gives: Draw a graph of resistance against temperature. Using the formula Rq = R0(1 + aq) find Rq (the resistance of the wire at 0 °C) and a (the temperature coefficient of resistance). Answers 1. (a) y y mx 2 y mx 4 2 (b) y Use this data to plot a straight line graph. What does the gradient of this graph give you? 4 Mathematical techniques (c) pV = k p = pressure, V = volume, k = constant Ix y The volume of the air in the tube is proportional to the height of the air column since the crosssectional area A is constant (V = Ah, V ∝ h). pV = k but V = Ah pAh =k k 1 h = A p (d) y y Plotting h against 1/pressure gives a linear graph with a gradient of k/A. Plotting the graph of p against 1/height also gives a graph with gradient of k/A. in 3. ResistanceIΩ 30 (e) y 20 22.5 0 0 y y-axis ntercept 20 80 40 60 TemperatureI°C l00 –kx Rq = R0(1 + aq) From the graph, Rq is given by the intercept at the R axis = 22.5 °C. The gradient of the graph, m, is R0a = (31.6 – 22.5)/100 = 0.091, so a = 0.091/22.5 = 0.0040 °C–1. 2. Areas and volumes 50 Areas and volumes appear in many equations used in physics. For example: Height of columnIcm 40 density = 30 pressure = 20 0 0 and R = 20 40 p 60 Pa 80 04 l00 mass volume (kg m–3) force (N m–2 or Pa) area rl A where R is the resistance of a piece of uniform wire (Ω), r is the resistivity (Ω m–1), l is its length, and A is its cross-sectional area. Mathematical techniques 1. Areas The area of a rectangle is found by multiplying its height by its length. The area of a triangle, Atri, is given by the expressions: For a cylinder of length l and radius r, the total surface area is: Acyl = 2πrl + 2πr 2 The surface area of a sphere is: Atri = Atri = 1/ b × h 2 1/ ab sin 2 Asph = 4πr2 q 2. Volumes a h θ The volume of a cuboid can be calculated by multiplying its height by its width and length. The volume of a cylinder is calculated by multiplying the cross-sectional area of the cylinder by the cylinder’s length: r Vcyl = πr2l A sphere has a volume related to its radius, r: Vsph = 4/3πr3 The area of a circle, Acir, is given in terms of the circle’s radius, r: The circumference of a circle is 2πr and this information is used in calculating the surface area of a cylinder (e.g. a piece of metal cable). The surface area of each end of the cylinder is that of a circle, πr2. The remaining area can be calculated by picturing the curved surface as being opened out into a rectangle (see below). In some problems, you need to find the area under a curve on a graph. For example, the area under the curve on a velocity–time graph represents the distance travelled in a certain time. If a body accelerates uniformly from 0 to V m s–1 in time t1, and then travels at this speed until time t2, its velocity–time graph could be represented as shown below: t t VelocityIm s—l Acir = πr2 Areas under graphs Vms l Y X r 2πr l 2 TimeIs The distance travelled is given by the area under the graph. In this case, the total area is: Atotal = area of triangle X + area of rectangle Y = 1/2t1V + (t2 – t1)V = (t2 – 1/2t1)V Mathematical techniques Examples 2. A wire of diameter 0.51 mm is suspended from a steel beam. A load of 1.5 kg is attached to its lower end. Calculate the applied stress on the wire using the equation: force cross-sectional area stress = stress = 14.7 π× (2.55 × 10–4)2 = 7.2 × 107 N m–2 (or Pa) 3. VelocityIm s—l 1. A solid cylinder of pure copper has a diameter of 5.00 cm and a length of 9.00 cm. Its mass is 1.58 kg. Calculate the density of copper in kg m– 3. 5 B assume g is 9.8 N kg–1. A C m s–1 3. A car accelerates uniformly from 0 to 15 in 4 seconds. It continues at 15 m s–1 for a further 20 s before the brakes are applied. It decelerates uniformly for 6 s before coming to a stop. Calculate the distance travelled by the car. Answers 1. V = πr2l The diameter of the cylinder in metres = 0.0500 m so the radius r = 0.0250 m The length of cylinder in metres = 0.0900 m Therefore, the volume of the cylinder: V = π× (0.0250)2 × 0.0900 = 1.767 × 10–4 m3 mass volume density = 1.58 1.767 × 10–4 = 8.94 × 103 kg m–3 = 2. stress = force cross-sectional area force = mg = 1.5 × 9.8 = 14.7 N area = πr2 wire diameter r = 2 = 2.55 × 10–4 m 4 l0 20 24 30 TimeIs distance travelled = area A + area B + area C = 30 + 300 + 45 = 375 m © HarperCollinsPublishers CAMS Physics A2 Support Pack