Solutions to the Exercises of Chapter 9

Transcription

1 9A. Vectors an Forces Solutions to the Exercises of Chapter 9. F = 5 sin 5.9 an F = 5 cos a. By the Pythagorean theorem, the length of the vector from to (, ) is + = 5. So the magnitue of the force that it represents is 5. In the same way, the magnitue of the force represente by the vector from to (, ) is + =. The horizontal components of these two vectors are, respectively, the vectors from to on the x-axis an from to on the x-axis. So the resultant of the two horizontal components is the vector from to on the x-axis. The vertical components of the two vectors are the vectors from to on the y-axis an from to on the y-axis. So the resultant of the two vertical components is the vector from to on the y-axis. The resultant is obtaine by combining the horizontal an vertical resultants. Doing so yiels the vector from to (, ). b. The magnitue of the vector from to (, ) remains to be etermine. This is + = by the Pythagorean theorem. The horizontal component is the vector from to on the x-axis an the vertical component is the vector from to on the y-axis. It follows from the iscussion in (a) that the resultant of the three horizontal components is the vector from to on the x-axis, an that the resultant of the three vertical components is the vector, from to on the y-axis. It follows that the resultant of the three vectors is the vector from to (, ).. Because the system is in equilibrium, F cos θ = F cos θ. So cos = F cos 6. Therefore, cos F = = cos 6 = 7. pouns. The sum of the magnitues of the vertical components of F an F must be equal to F. So F = F sin θ +F sin θ = sin + sin 6 = + = 5+5 = pouns. 4. By Archimees s Law of Hyrostatics, the maximal buoyant force is the weight of the water that the entire hull isplaces. This is ()(6.5) =,5 pouns. So this is the largest weight that the boat plus its cargo can have. The maximum weight for the cargo is,5 8, = 4,5 lbs If the block weighs 5 pouns, it isplaces =.8 cubic feet of water. So if h is the amount 6.5 in feet of the block that is submerge, then (.6)(.6)h =.8. So h. feet. If the block weighs 5 pouns, then in the same way, (.6)(.6)h = 5. So h.67 feet. Because this 6.5 is greater than the height of the block, it means that the block is completely submerge. It follows that the weight of the block is greater than the weight of the water that it isplaces. So the block will sink.

2 Correction: In the introuction to Exercises 6 an 7 replace contain by ask. In aition, line of Exercise 7 shoul rea Exercises 8K not Exercises 8G. 6. As was the case with the floating block of woo of Section 9.B, the weight of the basketball is equal to the weight of the water that it isplaces. To compute the weight of the water that r h is isplace, we nee to compute the volume of the water that is isplace. Notice that this volume is obtaine by rotating one revolution about the x axis the part of the circle of raius r centere at the origin that lies over the interval [ r, r + h] or, equivalently, over the interval y y -r -r+h x r h r x [r h, r]. See the two figures above. By Chapter 5.6B this is equal to r+h r πf(x) x an r h r πf(x) x respectively, with f(x) = r x, the function whose graph is the top half of the circle. We will evaluate the first of these integrals an invite the stuent to work with the secon. Observe that r+h r π(r x )x = r+h r (πr πx )x = πr x πx r+h r

3 = πr ( r + h) π ( r + h) (πr ( r) π ( r) ) = πr + πr h π ( r +r h rh + h )+πr π r = πr h πr h + πrh π h = πrh π h. It follows that the weight of the volume of water that the ball isplaces is (πrh π 6.5 pouns h ) foot = 6.5(πrh π h ) pouns. Since this is the weight of the basketball, we get that 6.5(πrh π h )=. or, equivalently, that πrh π h =.8. Inserting r =.9 an rouning off to three ecimal accuracy gives us.47h.5h +. =. 7. To apply Newton s Metho, we nee the erivative f (x) =.4x.45x. of f(x). With c =.9, we get Next, c =.9 f(.9) f (.9) c =.7956 f(.7956) f (.7956) c 4 =.47 f(.47) f (.47) = = = =.47, an = =.95. Plugging x =.95 into f(x) gives us f(.95) =.5. Rouning off to the two significant figures that parallels the ata gives h = With CB = c =, an CF = a =8, we get EC = x = a 4c (a + a +8c )= 8 4 ( ) Notice that cos θ By the inverse cosine button on your calculator (the theory of inverse functions will be taken up in Chapter.5), we get θ =.8. Observe next that EF + EC = CF. So F Because tan θ θ Finally,.7, we get by the inverse tan button of a calculator, that BF = BE + EF ( x) + (.84) (.5) + (.84) 4.4. So BF.798. The ata supplie suggests that all these answers shoul be roune off to two significant figures.

4 9. From a review of Section 9.B, T = W = pouns, an T cos θ = T cos θ. After inserting the results of Exercise 8, we get T 7.9 pouns. cos 48.4 cos.8. By the symmetry of the situation, the tensions in the wires AC an BC are equal. If T is this tension, then its vertical component is T sin α. Hence T sin α =. So T = = 47 sin 5 pouns.. Let T AC an T BC be the tensions in the wires AC an BC respectively. From the figure, A α β C B cos α (sin α)(t AC ) + (sin β)(t BC ) = an T AC cos α = T BC cos β. So T BC = T AC. A look cos β at the graph of the cosine over [, π ] (refer to Figure 4.5) tells us that if α > β, then cos α cos α < cos β, so that T AC > T BC. Because T AC sin α + T AC sin β =, we get, cos β T AC (sin α + cos α tan β) =, an hence T AC =. In the same way, T sin α + cos α tan β BC =. sin β + cos β tanα. Let T be the tension in the string. Because the angle that the string makes with the horizontal is π α, the horizontal an vertical components of T are T cos ( π α) an T sin ( π α) respectively. i. The vertical component T sin ( π α) has to counterbalance the weight of the sphere. So T sin ( π α) = 5 an hence T = 5 sin ( 5 = π α) cos α. ii. This force F is the horizontal component T cos ( π α) of the tension T. It is equal to F = 5 cos ( π α) sin ( π α) = 5 tan ( = 5 tan α. π α) iii. T = 5 cos 5 5 pouns, an F = 5 tan 5 pouns. The fact that the iameter of the sphere is 8 inches oes not play a role in this problem. 9B. Archimees an the Crown. Let s first solve the problem by assuming that w = w. If w pouns of gol loses f pouns, then w pouns of gol will lose f. Because w = w, we see similarly that the w = w pouns of silver will lose f pouns. It follows that f = f + f. We can now check whether the formula is correct. On the one han, w w f f = f f f f f f + f = f 4 = w = =, an on the other, w (f f ) (f f ) =.

5 To solve the problem in general, start with w = cw. Because w = w + w = cw + w, we get that w = ( c)w. Now simply repeat the argument above with c in place of an c in place of. 4. By Archimees s law of hyrostatics f = 6.5v, f = 6.5v, an f = 6.5v. Plug these values into the formula of Exercise an cancel C. Suspension Briges 5. Notice that w = 9,+,5 =,7 = 675, = = , an s = 8. So T 4 4 = w s pouns, an T = w + ( s) pouns. The angle α satisfies tan α = s =.. Using the inverse tan button of a calculator, α 7.8. The horizontal component of the tension must be T cos α pouns. 6. Notice that w = 9,+7,6 = 6,7 = 659.5, = 6 = 8, an s = 77. So T 4 4 = w + ( ) s. 7 pouns. The angle α satisfies tan α = s.44. Using the inverse tan button of a calculator, we get α.9. Let T S be the tension in a cable over the sie span at the tower. Since T cos.9 = T S cos.7, we fin that T S.9 7 pouns. It follows that the total compression on one tower is approximately 4(. 7 ) sin.9 + 4(.9 7 ) sin pouns. 7. Observe that w = 4,+, = 875, = 75, an s = 6. So T 4 = w + ( s).6 7 pouns. Since each wire has cross-sectional area π(.95 ).99 square inches, each wire has an ultimate strength of, pouns pouns. So the cable inch has an ultimate strength of (7)(56)(657) 6. 7 pouns. Therefore the safety factor is Notice that w =,+4, = 5, =,65, =,, an s = 47. So T = w s pouns, an T = w + ( s) pouns. As in Exercise 7, the ultimate strength of each wire is (,)(π(.96 ) ) pouns an that of the cable is (6)(45)((,)π(.98).8 8 pouns. The safety factor is Notice that w = 8,68+4, = 64, = 4, an s = 8. So T = w + ( s).9 7 pouns. The angle α satisfies tan α = s =.4. Using the inverse tan button of a calculator, we get α.8. Each cable over the center span generates a compression of T sin α pouns.. Notice that w = 7,+4,8 =,45, =, an s = 85. Thus, T 4 = w + ( s) pouns. The angle α satisfies tan α = s =.6, so that α The compression prouce by all four cables is 4 T sin α pouns.. Begin with a review of the units involve by consulting Exercises 7E. In this problem, = 995 5

6 meters an s = meters. As in the previous exercises, the important formula is ( T = w s ) +. The compression that one cable prouces in the tower is T sin α, so it is T sin α for both. As an estimate, take the compression generate by the two cables over the sie span to be T sin α as well. Therefore, 4T sin α 98 6 newtons. Now recall that tan α = s. So tan α = 4 =.44. Using the inverse tan button, we get α It follows that T newtons. Because w = (ea + live loa), it remains to estimate w. From w ( s) , we get 656w So w newtons/meter. Make use of Exercise 7E of Chapter 7 an check that this correspons to approximately,7 pouns per foot. 9D. About the Cables Correction: Delete The strans are arrange in one of... have iscusse at the en of the statement of Exercise. This point was mae earlier in the Exercise.. Following the pattern inicate in Figure 9.5, a to the horizontal iagonal of 9 circles: two rows of 8 circles each, one above an one below the iagonal; two rows of 7 circles each above an below those; two rows of 6 circles each above an below; an, finally, two rows of 5 each above an below those. The resulting array of circles is a hexagonal array with a total of = 6 circles.. Consier the ifferences between consecutive hexagonal numbers. Notice that 7 = 6; 9 7 = ; 7 9 = 8; 6 7 = 4; an that the numbers 6,, 8, an 4 are consecutive multiples of 6. The next multiple of 6 is, so 6 + = 9 is the next hexagonal number. The hexagonal number = 7 is next, followe by = 69. Since each circle in the hexagonal configuration of Figure 9.5 represents a stran, observe that there are gaps between the strans in a cable constructe as escribe in Exercise. The avantage is that the hexagonal strans of the cable of the Akashi Straits Brige can be (an 6

7 are) packe without gaps as shown. Incientally, the pattern that provie the 5 th hexagonal number 6 (with iagonal 9) suggests a way to evelop a formula for the n-th hexagonal number. Give it a try! Start with any o number n an use Gauss s aition process of Note 4 of Chapter 4 to get that the n-th hexagonal number is equal to n n +. Correction: A, where c is a constant, after +cx in the statement of Exercise The comparison as well as the approximation of +x was alreay iscusse in Chapter 6.. To get the approximation of +cx simply replace x by cx in the approximation for +x. 5. Refer to Figures 9.5 an 9.6 an recall that the curve representing the cable is the graph of f(x) = s x. Note that f() =s. By Chapter 5.6C, the length of the cable from its lowest point at (, ) to the point (, s) where it meets the tower is given by +f (x) x. Because f (x) = s x, this equals ( ) s + x x = + (s) x x. 4 Correction: In the statement of Exercise 6: Insert in Exercises 5- after consiere. 6. For the George Washington Brige, s = 7 feet an = 75 feet. So ( s ) ( x ) ( ) ( ) s 654 < For the Brooklyn Brige (see Exercise 5), s = 8 feet an = 798 feet. So ( s ) ( x ) ( ) s ( ) 56 < Combining Exercise 4 (with c = ( ) s ) an Exercise 5 we get ( s + ) x + ( s ) x 8 Because an antierivative of this polynomial is F (x) =x + 6 ( s ) x 4 ( s ( s ) 4 x ) 4 x 5 + ( s ( s ) 6 x ) 6 x ( ) 8 s x 8. ( ) 8 s x 9, 7

8 we get ( ) s + x x F () F () = F () = + ( ) s ( ) 4 s 5 + ( ) 6 s 7 5 ( ) 8 s = + s s s 6 7 s For the George Washington, = 75 an s = 7. So s 4.7 an 5 s The remaining terms of the approximation are too small to be relevant. So the length of the cable from the lowest point to the tower is approximately = 79 feet. For the Brooklyn Brige, = 798 an s = 8. In this case, ( s + ) x x + s s feet. 9E. Calculus of Rotation 7. When t =, the angle is θ() = = 8 + =. The average angular velocity between t = an t = is θ() θ() = = raians per secon. The angular velocity at any time t is w(t) =θ (t) = 5 t + raians per secon. At the 5 instant t =, the angular velocity is w() = + =.4 +. =.6 raians per 5 5 secon. The average angular acceleration from t = to t = is w() w() =.6. =.4 raians/sec. The angular acceleration at any time t is α(t) =w (t) = 6 t 5 raians/sec. When t =,α = 6 =.48 raians per 5 secon. Correction: In the statement of Exercise 8 replace Pick a point on the outer iameter. by Pick a point on the tire that comes into contact with the roa. 8. Consier an instant at which the point on the surface of the tire touches the roa an conclue that the point moves with a velocity of, meters per hour. Use of the formula v(t) = rw(t) tells us that the angular velocity of the point is w = r v =,.5 7,69 raians per 8

9 hour. Because π raians correspon to revolution, the tire turns at a rate of 7, π revolutions per hour, or approximately, revolutions per secon Let s A an s B be the rotational spees of pulleys A an B in revolutions per minute. Because revolution correspons to π raians, we see that πs A = w A. Hence the angular velocity w A of a point moving on the perimeter of pulley A is w A =πs A =π() = 4π. Similarly, πs B = w B where w B is the rotational spee of pulley B in raians per minute. Consier the point on the fan belt uring its motion aroun pulley A. Note that v = r A w A where r A is the raius of pulley A. Hence the velocity of a point moving on the fan belt is v = r A w A = 4(4π) = 96π inches per minute. In the same way, v = r B w B where r B is the raius of pulley B. So r A w A = r B w B, an hence r A r B = w B w A = πs B πs A = s B s A. Because r A =4,r B = 6, an s A =, we get s B = 4 = 8 rpm. 6. We will use the formula s B sa = r A rb that was establishe in the solution of Exercise 9. Taking = 8 rpm. It follows that = r D rc, we get s D = r C rd s C = s A =,r A = 4, an r B =, we get s B = r A rb s A = 4 the rotational spee s C of pulley C is 8 rpm also. Because s C sd rpm F. The Rotational Effect of Forces Exercise an Exercises 4 to 6 all make use of Leibniz s strategy of summing up many very small quantities, in other wors, of efinite integrals. A review of Sections 5. to 5.5 might therefore be in orer. Note also that the beams uner iscussion in Exercises to are assume to be horizontal (as illustrate). See the Aitional Exercises for Chapter 9 for a what happens when this is not so.. By Leibniz s strategy the sum of all the little torques 6.5x x as x varies from to 6 is given by the efinite integral 6 6.5x x. Because 6.5 x is an antierivative of 6.5x, we get 6 6.5x x =6.5 x 6 = 8 foot-pouns as require. Note: Suppose more generally that a beam of length L weighs p pouns. It then weighs p pouns L per foot. It follows, as in the iscussion above, that the torque it prouces aroun one of its enpoints is L p L x x = p L x L = p L.. By the last assertion of the solution of Exercise, the torque prouce by the right sie of the beam aroun A is 6 = 8 foot-pouns an that prouce by the left sie of the 9

10 beam is 5 8 = foot-pouns. The torque on the left sie prouce by the weight of the person is 8x where x is the person s istance from A. It follows that the system will still be in balance when + 8x = 8, or when x = 6 8 = feet. This is the maximum istance that the person can walk beyon A without tilting the beam. As soon as x> feet the beam will begin to tip.. Let W be the weight that has to be place at A. The torque aroun B prouce by the right sie is maximum when the man stans at the right en of the beam. This is the torque that the left sie must balance. Using the last assertion of the solution of Exercise, we get 6W + 6 = (6)() + 4. Therefore, 6W = = 464 foot-pouns. So W = 9 pouns. Any weight greater than 9 pouns will also o the job. 4. The mass of the typical little piece is m times its length x, so it is m x. Its istance from r r the axis of rotation is x. So it follows from the iscussion of Section 9.B that the moment of inertia of the typical piece is ( m x) x. By aing the moments of inertia of all the little r pieces as x ranges from to r, we get that the total moment of inertia is r m r x x = m r x r = mr. The flaw is simply that the istance of the center of mass of an object from the axis of rotation has no irect bearing on the question of moment of inertia. This point is best illustrate by consiering a rotating bicycle wheel. Its center of mass is on the axis of rotation. So its istance from the axis is zero. However, the wheel oes resist rotation an hence has a non-zero moment of inertia. 5. Because the circular piece has raius x, its length is πx. Because it has thickness x, its area an mass are approximately πx x an m (πx x) as asserte. By the iscussion in πr Section 9.B about the moment of inertia of the circle with mass, we get that the moment of inertia of our typical circular piece is mass (raius) = ( m ) πr πx x x = m r x x. Aing up the moments of inertia of all the circular pieces as x varies from to r, we get r m r x x = m 4r x4 r = m r r4 = mr. 6. Summing up the moments of inertia of all the iscs as their x-coorinates range from r to r, we get that the total moment of inertia (i.e. that of the soli sphere) is r r m 8r (r x ) x.

11 This is the moment of inertia of the soli sphere of raius r an mass m. To evaluate this efinite integral, we nee an anti-erivative of m 8r (r x ) = m 8r (r4 r x + x 4 ). ( ) Since this is F (x) = m r 4 x 8r r x + x5, we get 5 r r m 8r (r x ) x = F (r) F ( r) = m (r 5 ) 8r r5 + r5 m ( r 5 + ) 5 8r r5 r5 5 = m (r 5 4 ) 8r r5 + r5 5 = m ( ) + 6 r 5 = 8r 5 5 mr. 9G. Galileo s Experiment 7. The velocity of the ball at the bottom of the incline plane is v = 7 gh where h is the starting height. Because v = rω where r is the raius an ω is the rotational velocity in raians per secon, we get ω = gh. As π raians correspon to revolution, r 7 the rotational velocity in revolutions per secon is In M.K.S., g =9.8, an r =., so πr 7 gh. πr 7 gh = 4. h h..π Taking h =.8,.564,.75, an.94 meters respectively, we get.64, 44.74, 5.67, an revolutions per secon. 8. We know that with a starting height of h meters, the ball is preicte to lan at a istance of R =.49 h meters from the foot of the table. For h =., we get R = (.49)(.49) =.564 meters. Recall that in actuality the ball lane.8,.45,.4, an.6 meters short of the preicte istance. This is an average of about.5 meters. So the expectation is that the ball woul have lane from about.5 meters to about.55 meters from the foot of the table.

12 Correction: It shoul be Stilman Drake not Stilmann Drake in Exescise In terms of the height y of the table, the ball is preicte to lan at a istance of 7 y h from the foot of the table given that it starte its motion at a height of h above the table top. Taking y =.75 meters an h =.8,.564,.75, an.94 meters, respectively, we get the istances of.777,.99,.69, an.49 meters. 7 y h for the istance has no β in it. Therefore R oes not epen on β. 4. The expression R = Neither oes it epen on the gravitational acceleration g. This is why Galileo s experiment woul have prouce essentially the same istance measurements on the Moon, even though - see Exercises 7F - the gravitational acceleration is much less on the Moon than on Earth. 9H. Basic Optics 4. Refer to Figure 9.. The meium containing A is air an that containing B is crown glass. The respective inices of refraction are n A =.9 an n B =.5. So sin β = n A sin α =.9 n B.5 sin (.658)(.5).9. Using the inverse sine button of your calculator (in egree moe) we get β Let α be the angle of refraction of the incoming ray. So by Snell s law sin α v A = sin α v B, where v A is the spee of light in air an v B its spee in the plate glass. Notice that α is also the angle of reflection at the silver coating. So α is also the angle of incience of the ray as it travels through the plate glass towar the interface between the plate glass an the air. Using Snell s law a secon time, we get that sin α v B = sin β v A. Putting the two equations together, we see that sin β = sin α. Because both α an β are between an π, we get that β = α as require. (A look at the graph of the sine, see Figure 4.4, confirms this conclusion.) Corrections: Exercises 4 an 44 require aitional assumptions. To obtain the conclusions sin α = n sin β an sin α = n k sin β k, it is necessary to assume that n A =, in other wors that the light ray approaches the first slab through a vacuum. For the conclusions α = α an α 4 = α, it is require that the inices of refraction of the region above an below the array of plates are equal.

13 On the other han - as the solutions of the exercises will show - it is not necessary to make any assumptions about the meia between the plates. Finally, in this context, the wor translucent might be better than transparent. 4. Suppose that the inex of refraction of the meium above the two slabs is n A an that of the meium below the slabs is n B. Let n be the inex of refraction of the meium between the two slabs, an let α be the angle that the path of the ray between the slabs makes with a perpenicular to the slabs. A look at Figure 9.6 an the repeate use of Snell s law shows that n A sin α = n sin β = n sin α = n sin β = n B sin α. Assuming that the meium above the plates is a vacuum, so that n A =, we get sin α = n sin β. Assuming that n A = n B, we get sin α = sin α, an hence α = α. 44. Let n A be the inex of refraction of the meium above the slabs, an let n B be that of the meium below them. Let n,n,n, an n 4 be the inices of refraction of the first, secon, thir, an fourth slabs. Let α,α, α, an α 4 be the respective angles of refraction of the ray in the first, secon, thir, an fourth slabs. Let n,n, an n be the respective inices of refraction of the three meia between the slabs, an let α,α, an α, be the respective angles that the path of the ray between the slabs makes with a perpenicular to the slabs. By repeately applying Snells law an a look at Figure 9.6, we get n A sin α = n sin β = n sin α = n sin β = n sin α = n sin β = n sin α = n 4 sin β 4 = n B sin α 4. Assuming that the meium above the slabs is a vacuum, so that n A =, we get sin α = n 4 sin β 4. Assuming that n A = n B, we get sin α 4 = sin α, an hence that α 4 = α. 45. By the Lens Maker s equation, ( = (n ) + ) ( = (.66 ) f R R.55 + ).55 (.66)(.8 +.8).4 iopters. So f.4 meters. 46. By the Lens Maker s equation, ( = (n ) + ) ( = (.5 ) f R R. + ).45 (.5)(. +.) (.5)(5.56).9 iopters. So this is a.9 iopter lens.

14 47. By the Lens Maker s equation,.8 = (n ) (. + ). = (n )(). So n.6, an n The answer to the first question follows from the formula s i = s +. By substituting f = an s f =.55, we get s i = +, so that.55 s i.8 + = 8.8. Hence the image is s i =.55 meters, or 5.5 centimeters, from the lens. The fact that s i turne out to be positive means that the image is real. A look at the way the objective lens in Figure 9.4 functions, tells us that if y is the size of the image, then.6.55 = tan ϕ = y.55. So y =.6, an hence the image is.6 centimeters high. 49. The magnification m is approximately m +. Because = 4, we get m + 4f f 4 4 =. Thus, the image of the ant is 7 = 77 millimeters or 7.7 centimeters long. 5. The magnification is F where F an f are the focal lengths of the objective lens an the f eyepiece respectively. Because F = 4 meters an = iopters, we get a magnification of f The length of the telescope ( ) must excee the focal length F of the objective. See Figure 9.4. Because = F = F = 4F, we get F = 5 meters. So the length of the telescope will f f excee 5 meters. 9I. The Spee of Light Correction: Lines 5, 6. Replace Mirror is free to rotate about an axis perpenicular to its plane. by Mirror is free to rotate about an axis perpenicular to the plane of the page. Line 7 shoul rea the reflection of the focal point, not reflection on the focal point. 9J. The Phenomenon of Refraction in Astronomy Correction: On page 96, left column, line. Insert after sine: (see Exercise C of Chapter ). Same column, line from the bottom: Replace.8 by.8, an in the right column, line, elete tanz app. 5. For z app =,ρ (58.7) tan by the formula, an ρ = from the table. For z app = 6, ρ (58.7) tan 6 by the formula. This again agrees with the value from the table. For z app = 8, ρ (58.7) tan 8. 4

15 This is secons more than the value from the table. Because z true = z app +ρ, we get the true zenith istances of + ( ) 6 =.6, 6 + ( ) 6 = 6., an 8 + ( 6) = 8.9. Observe that the effects of refraction for these apparent zenith istances are small. 5. The apparent zenith position of the top of the Sun s isc is 9 9 or 89. It follows from the table that the true zenith position of the top of the Sun s isc is 89 + ( ) 76 6 = = 9 > 9. The true zenith position of the bottom of the Sun s isc is 9 + ( ) 6 = 9 5. So the true vertical iameter of the Sun is or about 5. Note: It turns out that the perceive angular iameter of the Sun varies. There are tables for its angular size for every ay of the year in the Astronomical Almanac. The apparent semi-iameter varies from 5 45 to6 7. 5