Answers to test yourself questions

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1 nswers to test yourself questions Topic Measurement in physics 5 Taking the diameter of a proton to be order m 5 we find = 3 = 3 s 7 The mass of the Earth is about kg and the mass of a hydrogen atom about kg = 3 so we need = heartbeat lasts or s so = 5 7 There are 3 g of water in the glass and hence 3 3 = 5 moles of water Hence the number of molecules is 5 = 9 8 There are g of water in the body and hence number of molecules is 3 = The mass is about 7 kg and the radius about 7 7 π ( ) = 5 = 5 kg m = 3 moles of water Hence the m so the density is = 3 s yr 9 9 a E = 5 = J 8 b E = 8 9 = 5 ev V = (8 ) = m /3 3 3 a = (588 ) = 838 m physics for the IB Diploma Cambridge University Press 5 NSWERS TO TEST YOURSELF QUESTIONS

2 a g b kg c g 3 5 The mass is about kg 3 and the radius is 9 kg m π 3 ( ) m so the density is of about 3 In SI units the acceleration is 3 = = 5 m s 7g 7 ssuming a mass of 7 kg made out of water we have 7 g of water in the body and hence = 5 moles of water Hence the number of molecules is = 3 3 Each molecule contains electrons from hydrogen and 8 from oxygen 7 8 for a total of 3 electrons 8 The ratio is F F e g 9 9 ke 9 ( ) = = Gm 7 (9 ) f = cm x k y The units of m is kg ie M and those of k are N kg m s = = kg s = M T Hence m m x y x+ y y T = M (M T ) = M T From this we deduce that x + y = y = y = x = k Thus, f = c m P = = 7 W The answer must be given to sf and so P = = 7 W 5 E K = 5 5 = 393 J The answer must be given to 3 sf and so EK = 39 J a 3 5 = b = c = d (8 ) e 9 ( ) NSWERS TO TEST YOURSELF QUESTIONS physics for the IB Diploma Cambridge University Press 5

3 Uncertainties and errors 3 sum = (8 ± 8) N = (8 ± 8) N dif = ( ± 8) N = ( ± 8) N a a Q = b = = ; Q a b = + = + = 5 Q = 5 = 3 Hence Q = ± 3 Q a b b Q = = 85 ; Q = = 3 Hence Q = 85 ± 3 (85 ± ) c Q = 5 = ; Q = + = 3 Hence Q = ± 3 d Q = ; Q a 3 = = = Q = = Q a Hence Q = ± = ( ± ) e Q Q a b 5 = = 5; = + = + = 3 Q = 5 3 = 75 8 Q a b Hence Q = 5 ± F = = 8 N 8 F m v r = + + = + + = 3 F = 8 3 = 37 N F m v r 8 8 Hence F = (8 ± ) N (7 ± ) N a = πr = 89 cm R = = = 833 = = 5 cm R Hence = (89 ± ) cm (8 ± ) cm S R b S = π R = 58 cm = = = 7 S = 58 7 = 8 cm S R Hence S = (58 ± 8) cm (5 ± ) cm 7 = ab = 37 cm 8 9 a b 3 = + = + = 879 = = 3 3 cm a b 85 Hence = (37 ± 3) cm (37 ± 3) cm P = ( a + b) = 58 cm P = a + b = + 3 = cm Hence P = (58 ± ) cm ( ± ) cm T L T = (assuming g is accurately known) Hence = % = % T L T V R h = + = % + % = % V R h physics for the IB Diploma Cambridge University Press 5 NSWERS TO TEST YOURSELF QUESTIONS 3

4 3 The line of best-fit does not go through the origin There is a vertical intercept of about m Lines of maximum and minimum slope give intercepts of about and 9 m implying an error in the intercept of about m The intercept is thus ( ± ) m This just barely includes the origin so the conclusion has to be that they can be proportional Current / m 8 linear fit for: data set: current / m y = mx + b m(slope): 9 b(y-intercept): 5 correlation: x 3 The vertical intercept is about m No straight line can be made to pass through the origin and the error bars unless a systematic error of about m in the current is invoked Current / m 8 linear fit for: data set: current / m y = mv + b m(slope): 9 b(y-intercept): correlation: Voltage / mv However, a line of best fit that is a curve can also be fitted through the data and that does go through the origin (However, it may be objected that this particular functional form is chosen at low voltages we might expect a straight line (Ohm s law) So a different functional form may have to be tried) auto fit for: data set: current / m y = V^B : 58 + / 759 B: / 378 RMSE: 53 Current / m Voltage / mv NSWERS TO TEST YOURSELF QUESTIONS physics for the IB Diploma Cambridge University Press 5

5 P 3 Let P the common perimeter Then the radius of the circle satisfies πr = P R = and the side of the π P P P square a = P a = The circle area is then c = π = π π The square area is P P s = = and is smaller 33 a The initial voltage V is such that lnv = V = e = 55 V b When V V = 7 V, lnv = ln 7 39 From the graph when lnv 39 we find t 7 s c Since V Ve t / = RC t, taking logs, lnv = lnv so a graph of ln V versus time gives a straight RC line with slope equal to The slope of the given graph is approximately = Hence RC = R = = = Ω RC C 5 3 We expect L = km lnl α and so ln L = ln k + α ln M graph of ln L versus ln M is shown below The slope is a 5 lnm Drawing a best-fit line gives: lnl lnm Measuring the slope gives α = 3 physics for the IB Diploma Cambridge University Press 5 NSWERS TO TEST YOURSELF QUESTIONS 5

6 3 Vectors and scalars 35 3 a + B : length 9 cm F 8 N q 9 + B B B b B : length 5 cm F 9 N Θ below horizontal c B : length cm F N Θ 5 below horizontal scale: cm N B B NSWERS TO TEST YOURSELF QUESTIONS physics for the IB Diploma Cambridge University Press 5

7 37 The components are: x = cos 3 = 39 = sin 3 = y B Hence a ( + B) x = = 799 ( + B) = = 3878 y x y = 8 cos 8 = 389 = 8 sin 8 = The vector + B has magnitude = 8 and is directed at an angle θ = arctan = 9 to the horizontal 799 b ( B) x = = 9 ( B) y = 7878 = 878 The vector B has magnitude = 99 and is directed at an angle 878 θ = arctan = 8 (below) the horizontal 9 c ( B) x = = 7 ( B) y = 7878 = 975 The vector B has magnitude = and is directed at an angle 9 75 θ = arctan = 5 (below) the horizontal 7 38 a + = 5 cm in a direction θ = 8 + arctan = 5 b + 58 = km in a direction θ = arctan 58 = 5 c + 5 = 5 m at θ = 7 or θ = 9 d 8 + = 8 N at θ = 39 a = 3 at θ = arctan = 5 3 b = 539 at θ = 8 arctan = c + 8 = 8 at θ = 9 d + = 7 at θ = arctan = e + = 8 at θ = arctan = 9 The displacement has components rx = = and ry = 8 = diagram is: change initial final physics for the IB Diploma Cambridge University Press 5 NSWERS TO TEST YOURSELF QUESTIONS 7

8 The magnitude of the change n the velocity vector is + = m s The vector makes an angle of 5 with the horizontal as shown in the diagram diagram is: change initial 3 final The other two angles of the triangle are each ( ) = Using the sine rule we find p p sin 3 = p = p = 58 p 5 p sin 3 sin 75 sin 75 3 The components of the velocity vector at the various points are: : vx = m s and vy = B: vbx = + m s and vby = C: vcx = and vcy = m s Hence a From to B the change in the velocity vector has components vbx vx = + ( ) = 8 m s and vby vy = = b From B to C the change in the velocity vector has components vcx vbx = = m s and vcy vby = = m s c From to C the change in the velocity vector has components vcx vx = ( ) = + m s and vcy vy = = m s The change in the vector from to C is the sum of the change from to B plus the change from B to C x = cos = 7 and y = sin = + 3 B x = cos 35 = 8 9 and y = sin 35 = 5 7 C x = + cos 8 = and y = sin 8 = 9 7 D x = + cos( 9 8 ) = and y = sin( 9 8 ) = 9 E x = cos( 9 3 ) = 5 and y = sin( 9 3 ) = 8 5 The vector we want is C = ( + B ) The components of and B are: x = cos = + 3 and y = sin = + 5 ; B x = cos = 3 and y = sin = + 5 Hence C x = ( + 3 3) = and C y = ( ) = The magnitude of the vector C therefore is units and is directed along the negative y axis B B C C 8 NSWERS TO TEST YOURSELF QUESTIONS physics for the IB Diploma Cambridge University Press 5

9 a x = cos = + 8 and y = sin = + ; B x = cos 5 = + 9 and y = sin 5 = + 7 Hence the sum has components: Sx = = 8 and Sy = = 8 The magnitude of the sum is thus = 5 Its direction is θ = arctan = 3 8 b x = 5 cos 5 = + 9 and y = 5 sin 5 = ; B x = 8 cos 5 = and B y = 8 sin 5 = Hence the sum has components: Sx = 9 = 983 and Sy = = 7 The magnitude of the sum is thus = 3 Its direction is θ = arctan = c x = cos = and y = sin = + 8; B x = 5 cos 3 = + 9 and B y = 5 sin 3 = 9 Hence the sum has components: Sx = = + 9 and Sy = = + 37 The magnitude of the sum is thus = 5 Its direction is θ = arctan 37 = 3 9 physics for the IB Diploma Cambridge University Press 5 NSWERS TO TEST YOURSELF QUESTIONS 9