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### c13s1

Course: MATH Cal I- Cal, Spring 2008
School: TN Tech
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Calculus Stewart ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves 1. The component functions t , t 1 , and 5 t are all defined when t 1 0 t 5 , so the domain of r(t) is 1,5 . 2. The component functions 2 t 1 and 5 t 0 t 2 2 2 , sin t , and ln (9 t ) are all defined when t 2 and 9 t &gt;0 t+2 3&lt;t&lt;3 , so the domain of r(t) is ( 3, 2 ) ( 2,3) . ln t 1/t 3. lim cos...

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Calculus Stewart ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves 1. The component functions t , t 1 , and 5 t are all defined when t 1 0 t 5 , so the domain of r(t) is 1,5 . 2. The component functions 2 t 1 and 5 t 0 t 2 2 2 , sin t , and ln (9 t ) are all defined when t 2 and 9 t >0 t+2 3<t<3 , so the domain of r(t) is ( 3, 2 ) ( 2,3) . ln t 1/t 3. lim cos t=cos 0=1 , lim sin t=sin 0=0 , lim tln t=lim =lim =lim t=0 . Thus 2 + + + + 1/t + + t 0 t 0 t 0 t 0 t 0 1/t t 0 lim cos t,sin t,tln t = t 0 + lim cos t,lim sin t,lim tln t t 0 + = 1,0,0 . t 0 + t 0 + e 1 e 4. lim =lim =1 [ using l'Hospital's Rule], t 0 t t 0 1 lim t 0 t t 1+t 1 =lim t t 0 1+t 1 t 1+t +1 1+t +1 1, =lim t 0 1 1 3 = , lim =3 . 1+t +1 2 t 0 1+t 1 ,3 . 2 t 1 1 1 5. lim t+3 =2 , lim 2 =lim = , lim t 1 t 1 t 1 t 1 t+1 2 t 1 1 Thus the given limit equals 2, ,tan 1 . 2 Thus the given limit equals 6. lim arctant= t tan t t =tan 1 . 2 , lim e =0 , lim t 2t t 2t ln t 1/t =lim =0 [ by l'Hospital's Rule]. t t 1 ,0,0 . 4 Thus lim t arctant,e , ln t t = 2 7. The corresponding parametric equations for this curve are x=t +1 , y=t . We can make a table of values, or we can eliminate the parameter: t=y x=y +1 , with y R . By comparing different values of t , we find the direction in which t increases as indicated in the graph. 4 8. The corresponding parametric equations for this curve are x=t ,y=t . We can make a table of values, or we can eliminate the parameter: x=t 3 3 2 t= x 3 y=t = 2 ( 3 x ) 2=x2/3 , with t R x R . By 1 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves comparing different values of t , we find the direction in which t increases as indicated in the graph. 9. The corresponding parametric equations are x=t , y=cos 2t , z=sin 2t . Note that y +z =cos 2t+sin 2t=1 , so the curve lies on the circular cylinder y +z =1 . Since x=t , the curve is a helix. 2 2 2 2 2 2 10. The corresponding parametric equations are x=1+t , y=3t , z= t , which are parametric equations of a line through the point ( 1,0,0 ) and with direction vector 1,3, 1 . 11. The parametric equations give x +z =sin t+cos t=1 , y=3 , which is a circle of radius 1 , center ( 0,3,0 ) in the plane y=3 . 2 2 2 2 12. The parametric equations are x=t , y=t , z=cos t . Thus x=y , so the curve must lie in the plane x=y . Combine this with z=cos t to determine that the curve traces out the cosine curve in the vertical plane x=y . 13. The parametric equations are x=t , y=t , z=t . These are positive for t 0 and 0 when t=0 . So the curve lies entirely in the first quadrant. The projection of the graph onto the xy plane is y=x , y>0 , a half parabola. On the xz plane z=x , z>0 , a half cubic, and the yz plane, y =z . 3 3 2 2 2 4 6 14. The parametric equations give x +y +z =2sin t+2cos t=2 , so the curve lies on the sphere with radius 2 and center ( 0,0,0 ) . Furthermore x=y=sin t , so the curve is the intersection of this sphere with the plane x=y , that is, the curve is the circle of radius 2 , center ( 0,0,0 ) in the plane x=y . 2 2 2 2 2 15. Taking r = 0,0,0 and r = 1,2,3 , we have from Equation 13.5.4 0 1 r(t)=(1 t) r +t r =(1 t) 0,0,0 +t 1,2,3 , 0 t 0 1 1 or r(t)= t,2t,3t , 0 t 1. 1. 2 Parametric equations are x=t , y=2t , z=3t , 0 t Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves 16. Taking r = 1,0,1 and r = 2,3,1 , we have from Equation 13.5.4 0 1 r(t)=(1 t) r +t r =(1 t) 1,0,1 +t 2,3,1 , 0 t 0 1 1 or r(t)= 1+t,3t,1 , 0 t 1. 0 1 1. Parametric equations are x=1+t , y=3t , z=1 , 0 t 0 1 17. Taking r = 1, 1,2 and r = 4,1,7 , we have r(t)=(1 t) r +t r =(1 t) 1, 1,2 +t 4,1,7 , 0 t or r(t)= 1+3t, 1+2t,2+5t , 0 t 0 1 1 1. 1 . Parametric equations are x=1+3t , y= 1+2t , z=2+5t , 0 t 0 1 18. Taking r = 2,4,0 and r = 6, 1,2 , we have r(t)=(1 t) r +t r = ( 1 t ) 0 t 0 t 1 or r(t)= 2+8t,4 5t,2t , 0 t 1. 2,4,0 +t 6, 1,2 , 1 . Parametric equations are x= 2+8t , y=4 5t , z=2t , 19. x=cos 4t , y=t , z=sin 4t . At any point ( x,y,z ) on the curve, x +z =cos 4t+sin 4t=1 . So the curve lies on a circular cylinder with axis the y axis. Since y=t , this is a helix. So the graph is VI. 20. x=t , y=t , z=e . At any point on the curve, y=x . So the curve lies on the parabolic cylinder y=x . Note that y and z are positive for all t , and the point ( 0,0,1 ) is on the curve (when t=0 ). As t , ( x,y,z ) ( , ,0 ) , while as t , ( x,y,z ) ( , , ) , so the graph must be II. 21. x=t , y=1/(1+t ) , z=t . Note that y and z are positive for all t . The curve passes through ( 0,1,0 ) when t=0 . As t , ( x,y,z ) ( ,0, ) , and as t , ( x,y,z ) ( ,0, ) . So the graph is IV. 22. x=e cos 10t , y=e sin 10t , z=e . x +y =e cos 10t+e sin 10t=e (cos 10t+sin 10t)=e =z , so the curve lies on the cone x +y =z . Also, z is always positive; the graph must be I. 23. x=cos t , y=sin t , z=sin 5t . x +y =cos t+sin t=1 , so the curve lies on a circular cylinder with axis the z axis. Each of x , y and z is periodic, and at t=0 and t=2 the curve passes through the same point, so the curve repeats itself and the graph is V. 24. x=cos t , y=sin t , z=ln t. x +y =cos t+sin t=1 , so the curve lies on a circular cylinder with axis , so the graph is III. the z axis. As t 0 , z 25. If x=tcos t , y=tsin t , and z=t , then x +y =t cos t+t sin t=t =z , so the curve lies on the cone z =x +y . Since z=t , the curve is a spiral on this cone. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2t 2 2t 2 2t 2 2 2t 2 2 2 2 t t t 2 2 2 2 t 2 2 2 2 2 3 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves 26. Here x =sin t=z and x +y =sin t+cos t=1 , so the curve is the intersection of the parabolic cylinder z=x with the circular cylinder x +y =1 . 2 2 2 2 2 2 2 2 2 27. r(t)= sin t,cos t,t 2 28. r(t)= t t +1,t,t 4 2 2 29. r(t)= t , t 1 , 5 t 2 30. We have the computer plot the parametric equations x=sin t , y=sin 2t , z=sin 3t , 0 t 2 . The 4 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves shape of the curve is not clear from just one viewpoint, so we include a second plot drawn from a different angle. 31. x=(1+cos 16t)cos t , y=(1+cos 16t)sin t , z=1+cos 16t . At any point on the graph, x +y =(1+cos 16t) cos t+(1+cos 16t) sin t =(1+cos 16t) =z , so the graph lies on the cone x +y =z . From the graph at left, we see that this curve looks like the projection of a leaved two dimensional curve onto a cone. 32. 2 2 2 2 2 2 2 2 2 2 2 x= 1 0.25cos 10t cos t , y= 1 0.25cos 10t sin t , z=0.5cos 10t . At any point on the graph, 5 2 2 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves x +y +z =(1 0.25cos 10t)cos t +(1 0.25cos 10t)sin t+0.25cos t =1 0.25cos 10t+0.25cos 10t=1 , so the graph lies on the sphere x +y +z =1 , and since z=0.5cos 10t the graph resembles a trigonometric curve with ten peaks projected onto the sphere. The graph is generated by t 2 2 2 2 2 2 2 2 2 2 2 2 2 0,2 . 33. If t= 1 , then x=1,y=4,z=0 , so the curve passes through the point ( 1,4,0 ) . If t=3 , then x=9,y= 8,z=28 , so the curve passes through the point ( 9, 8,28 ) . For the point ( 4,7, 6 ) to be on the curve, we require y=1 3t=7 t= 2. But then z=1+( 2) = 7 3 6 , so ( 4,7, 6 ) is not on the curve. 2 2 34. The projection of the curve C of intersection onto the xy plane is the circle x +y =4,z=0 . Then we can write x=2cos t,y=2sin t,0 t 2 . Since C also lies on the surface z=xy , we have z=xy=(2cos t)(2sin t)=4cos tsin t , or 2sin (2t) . Then parametric equations for C are x=2cos t,y=2sin t,z=2sin (2t),0 t 2 , and the corresponding vector function is r(t)=2cos t i+2sin t j+2sin (2t) k , 0 t 2 . 35. Both equations are solved for z , so we can substitute to eliminate z : x +y =1+y 1 2 2 2 2 2 x +y =1+2y+y x =1+2y y= (x 1) . We can form parametric equations for the curve C of 2 1 2 1 2 1 2 intersection by choosing a parameter x=t , then y= (t 1) and z=1+y=1+ (t 1)= (t +1) . Thus a 2 2 2 1 2 1 2 vector function representing C is r(t)=t i+ (t 1) j+ (t +1) k . 2 2 36. The projection of the curve C of intersection onto the xy plane is the parabola y=x ,z=0 . Then we can choose the parameter x=t 2 2 2 22 2 2 2 2 2 y=t . Since C also lies on the surface +y z=4x , we have 2 2 4 4 2 2 2 z=4x +y =4t +(t ) . Then parametric equations for C are x=t , y=t , z=4t +t , and the corresponding vector function is r(t)=t i+t j+(4t +t ) k . 37. 6 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves The projection of the curve C of intersection onto the xy plane is the circle x +y =4,z=0 . Then we can write x=2cos t , y=2sin t , 0 t 2 2 2 2 2 2 . Since C also lies on the surface z=x , we have 2 2 z=x =(2cos t) =4cos t . Then parametric equations for C are x=2cos t , y=2sin t , z=4cos t,0 t . 38. 2 x=t y=t 2 4z =16 x 4y =16 t 4t 2 2 2 2 4 z= 4 1 t 2 2 4 t . Note that z is positive because the 2 intersection is with the top half of the ellipsoid. Hence the curve is given by x=t , y=t , 1 2 4 z= 4 t t . 4 39. For the particles to collide, we require r (t)=r (t) 2 2 1 2 2 t ,7t 12,t 2 2 = 4t 3,t ,5t 6 . Equating 2 2 components gives t =4t 3 , 7t 12=t , and t =5t 6 . From the first equation, t 4t+3=0 (t 3)(t 1)=0 so t=1 or t=3 . t=1 does not satisfy the other two equations, but t=3 does. The particles collide when t=3 , at the point ( 9,9,9 ) . 40. The particles collide provided r (t)=r (t) 1 2 t,t ,t 2 3 = 1+2t,1+6t,1+14t . Equating components 7 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves gives t=1+2t , t =1+6t , and t =1+14t . The first equation gives t= 1 , but this does not satisfy the other equations, so the particles do not collide. For the paths to intersect, we need to find a value for t and a value for s where r (t)=r (s) 1 2 2 3 2 3 t,t ,t 2 3 = 1+2s,1+6s,1+14s . Equating components, t=1+2s , 2 t =1+6s , and t =1+14s . Substituting the first equation into the second gives ( 1+2s ) =1+6s 1 1 2 4s 2s=0 2s(2s 1)=0 s=0 or s= . From the first equation, s=0 t=1 and s= t=2 . 2 2 Checking, we see that both pairs of values satisfy the third equation. Thus the paths intersect twice, at 1 the point ( 1,1,1 ) when s=0 and t=1 , and at ( 2,4,8 ) when s= and t=2 . 2 41. Let u(t)= u (t),u (t),u (t) and v(t)= v (t),v (t),v (t) . In each part of this problem the basic 1 2 3 1 2 3 procedure is to use Equation 1 and then analyze the individual component functions using the limit properties we have already developed for real valued functions. (a) lim u(t)+lim v(t)= lim u (t),lim u (t),lim u (t) + lim v (t),lim v (t),lim v (t) t a t a t a 1 t a 2 t a 3 t a 1 t a 2 t a 3 and the limits of these component functions must each exist since the vector functions both possess limits as t a . Then adding the two vectors and using the addition property of limits for real valued functions, we have that lim u(t)+lim v(t) = t a t a = lim u (t)+lim v (t),lim u (t)+lim v (t),lim u (t)+lim v (t) t a 1 t a 1 t a 2 t a 2 t a 3 t a 3 lim t a 1 u (t)+v (t) ,lim 1 1 t 2 a 2 u (t)+v (t) ,lim 2 2 t 3 a u (t)+v (t) 3 3 = lim t u (t)+v (t),u (t)+v (t),u (t)+v (t) [using(1)backward 1 3 a = lima u(t)+v(t) t (b) lim cu(t) =lim t a t a cu (t),cu (t),cu (t) = lim cu (t),lim cu (t),lim cu (t) 1 2 3 t a 1 t a 2 t a 3 = clim u (t),clim u (t),clim u (t) =c t a 1 t 2 a 3 2 t a 3 lim u (t),lim u (t),lim u (t) t a 1 t a 2 t a 3 =clim t a u (t),u (t),u (t) =clim u(t) 1 t a 8 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves (c) lim u(t) lim v(t) = lim u (t),lim u (t),lim u (t) t a t a t a 1 t a 2 t 1 a 3 lim v (t),lim v (t),lim v (t) t 2 a 1 t 2 a 2 t a 3 3 = lim u (t) t t t a a a 1 1 1 lim v (t) + lim u (t) t t 1 a a 2 t 2 t 3 a 2 3 lim v (t) + lim u (t) t 3 a t a lim v (t) t a 3 =lim u (t)v (t)+lim u (t)v (t)+lim u (t)v (t) a 3 =lim (d) u (t)v (t)+u (t)v (t)+u (t)v (t) =lim u(t) v(t) 1 2 t a lim u(t) lim v(t) = lim u (t),lim u (t),lim u (t) t a t a t a 1 t a 2 t a 3 lim v (t),lim v (t),lim v (t) t a 1 t a 2 t a 3 = lim u (t) t 3 a 2 lim v (t) t 1 a 3 lim u (t) t 1 a 3 lim v (t) , t 3 a 2 lim u (t) t a lim v (t) t a lim u (t) t a lim v (t) , t a lim u (t) t a 1 lim v (t) t a 2 lim u (t) t a 2 lim v (t) t a 1 = lim t a u (t)v (t) u (t)v (t) ,lim 2 3 3 2 t a u (t)v (t) u (t)v (t) , 3 1 1 3 lim t a u (t)v (t) u (t)v (t) 1 2 2 1 =lim t 1 a u (t)v (t) u (t)v (t),u ( t ) v (t) u (t)v (t), 2 3 3 2 3 1 1 3 2 2 1 u (t)v (t) u (t)v (t) =lim u(t) v(t) t a 42. The projection of the curve onto the xy plane is given by the parametric equations x=(2+cos 1.5t)cos t , y=(2+cos 1.5t)sin t . If we convert to polar coordinates, we have 2 2 2 2 r = x +y = (2+cos 1.5t)cos t + (2+cos 1.5t)sin t 2 = (2+cos 1.5t)2(cos 2t+sin 2t) = (2+cos 1.5t)2 9 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves y (2+cos 1.5t)sin t = =tan t =t . x (2+cos 1.5t)cos t Thus the polar equation of the curve is r=2+cos 1.5 . At =0 , we have r=3 , and r decreases to 1 as 2 2 4 increases to . For , r increases to 3 ; r decreases to 1 again at =2 , increases 3 3 3 8 10 to 3 at = , decreases to 1 at = , and completes the closed curve by increasing to 3 at 3 3 =4 . We sketch an approximate graph as shown in the figure. r=2+cos 1.5t . Also, tan = We can determine how the curve passes over itself by investigating the maximum and minimum 5 values of z for t= 0,4 . Since z=sin 1.5t , z is maximized where sin 1.5t=1 1.5t= , , or 2 2 9 2 5 3 7 11 7 t= , , or 3 . z is minimized where sin 1.5t= 1 1.5t= , , or t= , , or 3 3 2 2 2 3 11 . Note that these are precisely the values for which cos 1.5t=0 r=2 , and on the graph of the 3 projection, these six points appear to be at the three self intersections we see. Comparing the maximum and minimum values of z at these intersections, we can determine where the curve passes over itself, as indicated in the figure. We show a computer drawn graph of the curve from above, as well as views from the front and from the right side. 10 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves Top view Front view Side view The top view graph shows a more accurate representation of the projection of the trefoil knot on the xy plane (the axes are rotated 90 ). Notice the indentations the graph exhibits at the points corresponding to r=1 . Finally, we graph several additional viewpoints of the trefoil knot, along with two plots showing a tube of radius 0.2 around the curve. 11 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves 43. Let r(t)= f ( t ) ,g ( t ) ,h ( t ) and b= b ,b ,b b=lim r(t)= lim f (t),lim g(t),lim h(t) t t a a 2 t a t a t a 3 t a 1 2 3 . If lim r(t)=b , then lim r(t) exists, so by (1), t a t a t a 1 . By the definition of equal vectors we have lim f (t)=b , lim g(t)=b and lim h(t)=b . But these are limits of real valued functions, so by the definition of limits, for every >0 there exists whenever 0< t a < 2 1 >0 , 2 >0 , 3 3 >0 so f (t) b < /3 whenever 0< t a < 1 3 1 , g(t) b < /3 2 , and h(t) b < /3 whenever 0< t a < 1 2 3 . Letting = minimum of { = 1 2 3 , , } 1 , we have f (t) b + g(t) b + h(t) b < /3+ /3+ /3= whenever 0< t a < . But f (t) b ,g(t) b ,h(t) b 1 2 2 2 2 3 3 2 r(t) b = ( f (t) b ) +(g(t) b ) +(h(t) b ) 1 2 3 [ f (t) b ] + 1 2 [g(t) b ] + 2 2 [h(t) b ] 3 2 = f (t) b + g(t) b + h(t) b r(t) b 1 2 . Thus for every >0 there exists >0 such that 3 f (t) b + g(t) b + h(t) b < whenever 0< t a < . Conversely, if for every >0 , 2 2 2 there exists >0 such that r(t) b < whenever 0< t a < , then f (t) b ,g(t) b ,h(t) b 1 2 2 2 1 2 3 < 2 2 2 3 [ f (t) b ] +[g(t) b ] +[h(t) b ] < 1 2 3 [ f (t) b ] +[g(t) b ] +[h(t) b ] < 1 whenever 0< t a < . But each term on the left side of this 2 inequality is positive so [ f (t) b ] < , [g(t) b ] < 2 2 2 and [h(t) b ] < 3 2 2 whenever 0< t a < , or 1 2 taking the square root of both sides in each of the above we have f (t) b < , g(t) b < and h(t) b < whenever 0< t a < . And by definition of limits of real valued functions we have 3 12 Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves lim f (t)=b , lim g(t)=b and lim h(t)=b . But by (1), lim r(t)= lim f (t),lim g(t),lim h(t) t a 1 t a 2 t a 3 t a t a t a t a , so lim r(t)= b ,b ,b t a 1 2 3 =b . 13
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Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books IX-XIISpring 2008Book IX. War finally breaks out, the full-scale battles spoken of in Book VII. The book divides into three sections: (1) Turnus and the Rutulians attack the Trojan ship
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Classics 2123: The Roman Way Lecture 5. Roman Imperialism and Expansion Readings: BHR 44-67; Shelton 291-293 I. Roman Imperialism - definition of `imperialism' - older notion of `defensive imperialism': how realistic? - was Rome more warlike/aggressi
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Classics 2123: The Roman Way Roman Names A Roman had three names: praenomen (first name) nomen (name of the gens or clan) cognomen (family branch) Thus for: Publius praenomen Cornelius nomen Scipio cognomenSpring 2008his given name was &quot;Publius,&quot;
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AC CIRCUITS35.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency w. Solve: (a) Refemng to the phasor in Figure Ex35.1, the phase angle isU? = 180'n rad - 30&quot; = 150 x -= 2.618 rad180&quot;w=2*618ra
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15.1. Solve: The density of the liquid is=m 0.120 kg 0.120 kg = = = 1200 kg m 3 V 100 mL 100 10 -3 10 -3 m 3Assess: The liquid's density is more than that of water (1000 kg/m3) and is a reasonable number.15.2. Solve: The volume of the helium
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16.1. Solve: The mass of lead mPb = Pb VPb = (11,300 kg m 3 )(2.0 m 3 ) = 22,600 kg . For water to have thesame mass its volume must beVwater =mwater 22,600 kg = = 22.6 m 3 water 1000 kg m 316.2. Solve: The volume of the uranium nucleus isV
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17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth =Kmicro. Solve: The number of atoms isN=M 0.0020 kg = = 3.01 10 23 m 6.64 10 -27 kgBecause helium atoms have an atomic mass number A
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18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):1.013 10 5 Pa N p = 2.69 10 25 m -3 = = V kB T (1.38 10 -23 J K )(273 K )()18.2. Solve: Nitrogen is a diatomic molecule, so r 1.0 10-1
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19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycleconsists of three individual processes. Visualize: Please refer to Figure Ex19.1. Solve: (a) The work done by the heat engine per cycle is the a
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20.1. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second.Visualize: Please refer to Figure Ex20.1. The snapshot graph shows the wave at all points on the x-axis at t = 0 s. You can see that nothing is h
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21.1. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:The graph at t = 1 s differs from the graph at t = 0 s in that the left wave has moved to the right by 1 m and the right wave has moved to the left by
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22.1. Visualize: Please refer to Figure Ex22.1.Solve: (a)(b) The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates
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23.1. Model: Light rays travel in straight lines.Solve: (a) The time ist=x 1.0 m = = 3.33 10 -9 s = 3.33 ns c 3 10 8 m / s(b) The refractive indices for water, glass, and zircon are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, l
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24.1. Model: Balmer's formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Substituting into the formula for the Balmer series,=91.18 nm 91.18 nm = = 410.3 nm 1 1 1 1 - 2 - 2 2 22 n 2 6where n = 3, 4, 5, 6, . and wher
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ELECTROMAGNETIC AND WAVES FIELDSw.1. Model: The net magnetic flux over a closed surface is zero. Visualize: Please refer to Ex34.1. Solve: Because we can't enclose a &quot;net pole&quot; within a surface, Q, = f B . d i = 0 . Since the magnetic field isunif
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CLA 2123: The Roman Way First Exam. February 8, 2007Name _Please read all directions carefully; no credit will be given for doing more than is required in each section. Part I. Identifications (35 points). Choose FIVE of the following and identif
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14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, henceT= 1 1 = = 2.27 10 -3 s = 2.27 ms f 440 Hz14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your hear
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1.1.Solve:1.2.Solve:Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim
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2.1.Solve:Model: The car is represented by the particle model as a dot. (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0(b)2.2. Solve:Diagram (a) (b) (c)Position Negative Negative PositiveVelocity
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3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the
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4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a
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5.1.Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:Solve:Written in component form, Newton's first law is( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 NT1 x = - T1T1y = 0 N Using Newton's first l
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6.1. Model: We will assume motion under constant-acceleration kinematics in a plane.Visualize:Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector f
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7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4 rad to -2 rad. Therefore, = -2 - ( +4 ) = -6 rad in one sec, or = -6 rad s . From t = 1 s to t = 2 s, = 0 rad/s. From t = 2 s to t = 4 s the partic
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8.1. Visualize:Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact forc
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Solve: (a) The momentum p = mv = (1500 kg)(10 m /s) = 1.5 10 4 kg m /s . (b) The momentum p = mv = (0.2 kg)( 40 m /s) = 8.0 kg m /s .9.1. Model: Model the car and the baseball as particles.9.2. Model: Model the bicycle and its rider as a particl
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10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).Visualize:Solve:For the bullet,KB =For the bowling ball,1 1 2 mB vB = (0.01 kg)(500 m /s) 2 = 1250 J 2 2 1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2
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11.1. Visualize:r Please refer to Figure Ex11.1. rSolve: (b) (c)(a) A B = AB cos = ( 4)(5)cos 40 = 15.3. r r C D = CD cos = (2)( 4)cos120 = -4.0. r r E F = EF cos = (3)( 4)cos 90 = 0.11.2. Visualize:r Please refer to Figure Ex11.2. rSolve
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12.1.Solve: (b)Model: Model the sun (s), the earth (e), and the moon (m) as spherical. (a)Fs on e =Gms me (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg)(5.98 10 24 kg) = 3.53 10 22 N = (1.50 1011 m ) 2 rs2 e -Fm on e =GMm Me (6.67 10 -1
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13.1. Model: The crankshaft is a rotating rigid body.Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up
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Classics 2123: The Roman Way Lecture 3. The Early History of Rome and Roman Republican Government Readings: BHR 15-41; Shelton 2-4, 7-8, 251-3, 255-259, 262, 264-5Spring 2008Early History of Rome Regal period: dimly known, mostly through legends;
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Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 6. Internal Disorders; Roman Slavery Readings: BHR 82-92; Shelton 207-209, 219, 227-229, 317-318Spring 2008I. Establishment of Roman provincial government Provincia: sphere of action of a magistrate with imper
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Chapter 10Standard CostingAccounting 21210 - 1Learning Objective 1 Describe standard costing and indicate why standard costing is important.Accounting 212 10 - 2Why is Standard Costing Used?A standard is a preestablished benchmark for des
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Classics 2123: The Roman Way Outline for Lecture 7. Roman Religion Readings: BHR 41-44; Shelton nos. 402-419, 423-428Spring 2008Problems Studying Ancient Religious Systems - time and culture (modern politics separated from religion) - vocabulary
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Classics 2123: The Roman Way Lecture 8. The Strains of Empire (I) Readings: BHR 72-77, 92-110; Shelton 187-189, 266, 317-318Spring 2008I. Continuation of Roman Imperialism Macedonian Wars - Rome fights four wars in Greece, first against the Maced
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Classics 2123: The Roman Way Lecture 9. The Strains of Empire (II) Readings: BHR 111-128, 132-140Spring 2008I. New Developments in the Roman State in the Late Republic Breakdown of concordia in the late Republic the result of many things, includi
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Chapter 9The Operating Budget2004 Prentice Hall Business Publishing Introduction to Management Accounting , 2/e Werner/Jones9-1Learning Objective 1Describe some of the benefits of the operating budget.2004 Prentice Hall Business Publishing
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Classics 2123: The Roman Way Lecture 10. The Fall of the Roman Republic Readings: BHR 124-179.Fall 2008I. Career of Pompey (Gn. Pompeius Magnus) (106-48 BCE) - begins as supporter of Sulla: raises legions from his father's troops (client army); a
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Chapter 11Evaluating PerformanceAccounting 21211 - 1Learning Objective 1Describe centralized and decentralized management styles.Accounting 212 11 - 2Centralized ManagementTop management makes most of the decisions.The most experience
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Classics 2123: The Roman Way Lecture 11. Augustus' `Restored' Republic Readings: BHR 167-199; Shelton 38-40, 77-78, 267, 271, 274-276, 294-305.Spring 2008I. From Octavian to `Augustus' - 31 BCE victory over Antony and Cleopatra at Actium; after h
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Earth 1. When was Big bang and when did our Earth form? 13.5 billion years ago big bang. 4.53 billion years. 2. Know earth's radius/diameter, thicknesses of crusts and lithosphere 6,370km radius, crust is 40 km, lithosphere is 100 km 3. Know the phys
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HAPTER 1MANAGEMENT ACCOUNTING: ITS ENVIRONMENT AND FUTURESOLUTIONS TO CHAPTER 1 QUICK QUIZ 1. 2. 3. 4. 5. C B D C D 6. 7. 8. 9. 10. C B D B D 2004 Prentice Hall, Inc.M1 - 2Chapter 1 Management Accounting: Its Environment and FutureQUICK