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8 Pages

### c15s6

Course: MATH Cal I- Cal, Spring 2008
School: TN Tech
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Word Count: 1333

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Calculus Stewart ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area 1. Here z= f (x,y)=2+3x+4y and D is the rectangle 0,5 surface is A(S) = D 1,4 , so by Formula 2 the area of the dA D [ f (x,y)] +[ f (x,y)] +1 dA= x y D 2 2 3 +4 +1 dA= 26 2 2 = 26 A(D)= 26 (5)(3)=15 26 2 2 2. z= f (x,y)=10 2x 5y and D is the disk x +y A(S) = D 9 , so by Formula 2 ( 2) +( 5) +1 dA= 30 D 2 2 dA= 30 A(D) =...

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Calculus Stewart ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area 1. Here z= f (x,y)=2+3x+4y and D is the rectangle 0,5 surface is A(S) = D 1,4 , so by Formula 2 the area of the dA D [ f (x,y)] +[ f (x,y)] +1 dA= x y D 2 2 3 +4 +1 dA= 26 2 2 = 26 A(D)= 26 (5)(3)=15 26 2 2 2. z= f (x,y)=10 2x 5y and D is the disk x +y A(S) = D 9 , so by Formula 2 ( 2) +( 5) +1 dA= 30 D 2 2 dA= 30 A(D) = 30 ( 3 )=9 30 2 3. z= f (x,y)=6 3x 2y which intersects the xy plane in the line 3x+2y=6 , so D is the triangular region 3 . Thus given by ( x,y ) 0 x 2,0 y 3 x 2 { D } A(S) = = 14 ( 3) +( 2) +1 dA= 14 D 2 2 dA= 14 A(D) 1 2 3 =3 14 2 2 4. z= f (x,y)=1+3x+2y with 0 x 2y , 0 y 1 . Thus by Formula 2, A(S) = D 1 (3) +(4y) +1 dA= 0 0 2 2 2 1 2y 10+16y dxdy= 0 1 0 2 1 10+16y 2 x x=2y x=0 dy = 2y 0 2 2 10+16y dy= 2 1 2 2 3/2 (10+16y ) 32 3 2 1/2 = 1 3/2 3/2 (26 10 ) 24 5. y +z =9 42 z= 9 y . f =0 , f = y(9 y ) x y 2 A(S) = 0 0 42 0 +[ y(9 y ) 2 2 1/2 2 42 0 0 ] +1 dydx= y 3 y=2 y=0 y 2 2 +1 dydx 2 3 4 9 y dx=3 = 0 0 3 9 y 2 4 dydx=3 0 sin 1 sin 1 x 0 =12sin 1 2 3 1 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area 6. z= f (x,y)=4 x y and D is the projection of the paraboloid z=4 x y onto the xy plane, that is, D= (x,y)| x +y A(S) = D 2 2 2 2 2 { 2 2 4 2 } . So f = 2x , f = 2y x y 2 ( 2x) +( 2y) +1 dA= D 4(x +y )+1 dA= 0 0 2 2 2 2 2 4r +1 r dr d 2 = 0 1 2 3/2 (4r +1) 12 2 2 r=2 r=0 d = 0 2 2 1 ( 17 17 1 ) d = 6 ( 17 17 1 ) 12 7. z= f (x,y)=y x with 1 x +y A(S) = D 2 4 . Then 1+4r r dr d = 2 2 2 1+4x +4y dA= 0 1 2 0 2 2 2 d 0 1 r 1+4r dr 2 = 1 2 3/2 (1+4r ) 12 2 1 = 6 ( 17 17 5 5 ) 8. z= f (x,y)= A(S) = D 1 2 3/2 3/2 1/2 1/2 (x +y ) and D= { (x,y) 0 x 1,0 y 1 } . Then f =x , f =y and x y 3 ( x ) +( 2 y ) +1 dA= 00 y=1 2 11 x+y+1 dydx 1 0 1 0 3/2 3/2 = 0 2 3/2 (x+y+1) 3 2 dx= y=0 3 (x+2) = (x+1) dx 2 2 5/2 2 5/2 (x+2) (x+1) 3 5 5 4 5/2 7/2 = (3 2 +1) 15 = 9. z= f (x,y)=xy with 0 x +y A(S) = D 2 2 2 1 4 5/2 5/2 5/2 (3 2 2 +1) 15 1 , so f =y , f =x x y y +x +1 dA= 0 0 2 2 2 r +1 r dr d = 0 2 2 1 2 3/2 (r +1) 3 r=1 r=0 d = 0 1 2 ( 2 2 1) d = 3 3 (2 2 1) 2 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area 10. Given the sphere x +y +z =4 , when z=1 , we get x +y =3 so D= (x,y)| x +y z= f (x,y)= 4 x y . Thus A(S) = D 2 3 0 2 2 2 2 2 2 2 { 2 2 3 } and [( x)(4 x y ) r 2 2 2 2 1/2 2 ] +[( y)(4 x y ) 2 3 0 2 2 1/2 2 ] +1 dA = 0 2 +1 r dr d = 0 2 2r 4 r 2 dr d 4 r 2(4 r ) = 0 2 2 1/2 r= 3 r=0 d = ( 2+4)d = 2 0 2 0 =4 11. z= a x y , z = x(a x y ) x 2 2 2 2 2 1/2 , z = y(a x y ) y 2 2 2 1/2 , A(S) = D /2 acos x +y 2 2 2 2 2 +1 dA 2 2 a x y = /2 0 /2 acos r 2 +1 r dr d a r ar a r a r 2 2 2 2 = /2 /2 0 dr d r=acos r=0 = /2 a d /2 = /2 a ( a a cos 2 2 2 a d =2a ) 2 /2 0 (1 1 cos 2 )d 3 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area =2a 2 /2 d 0 2a 2 /2 sin 0 2 d =a 2 2a 2 /2 0 sin d =a ( 2 2) 12. To find the region D : z=x +y implies z+z =4z or z 3z=0 . Thus z=0 or z=3 are the planes where the surfaces intersect. But x +y +z =4z implies x +y +(z 2) =4 , so z=3 intersects the upper hemisphere. Thus (z 2) =4 x y or z=2+ 4 x y . Therefore D is the region inside the circle x +y +(3 2) =4 , that is, D= (x,y)| x +y A(S) = D 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 { 2 2 3 } . 2 1/2 2 1+[( x)(4 x y ) r 2 2 2 2 1/2 2 ] +[( y)(4 x y ) 2 3 0 2 ] dA 2 2 1/2 r= 3 r=0 = 0 2 0 1+ r dr d = 0 2 0 2r dr 4 r 2 d = 0 2(4 r ) d 4 r = ( 2+4)d = 2 0 =4 x y 2 2 13. z= f (x,y)=e A(S)= x +y 2 2 x y 2 2 , f = 2xe x x y 2 2 2 , f = 2ye y x y 2 2 2 x y 2 2 . Then 4(x +y )e 2 2 2(x +y ) 2 2 ( 2xe 4 ) +( 2ye ) +1 dA= x +y 2 2 +1 dA . 4 Converting to polar coordinates we have 2 2 A(S) = 0 0 2 4r e 2 2 2r 2 2 2 +1 r dr d = 0 2 d 0 r 4r e 2 2r 2 +1 dr =2 0 r 4r e 2r +1 dr 13.9783 using a calculator. 14. z= f (x,y)=cos (x +y ) , f = 2xsin (x +y ) , f = 2ysin (x +y ) . x y 2 2 2 2 2 2 A(S) = = 4x sin (x +y )+4y sin (x +y )+1 dA 2 2 2 2 2 2 2 2 2 2 x +y 1 4(x +y )sin (x +y )+1 dA 2 2 2 2 2 2 2 x +y 1 4 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area Converting polar to coordinates gives 2 1 A(S) = 0 0 1 4r sin (r )+1 r dr d = 0 2 2 2 2 2 2 2 1 d 0 r 4r sin (r )+1 dr 2 2 2 =2 0 r 4r sin (r )+1 dr 4.1073 using a calculator. 15. (a) The midpoints of the four squares are Here f (x,y)=x +y , so the Midpoint Rule gives A(S) = D 2 2 1 1 , 4 4 2 , 1 3 , 4 4 2 , 3 1 , 4 4 , and 3 3 , 4 4 . [ f (x,y)] +[ f (x,y)] +1 dA= x y D 2 2 2 (2x) +(2y) +1 dA 2 1 4 + = 1 4 2 2 3 +2 2 1 4 3 4 2 + 2 + 2 7 + 2 11 2 1 4 1 4 2 +1 + +1 + 1.8279 2 2 1 4 3 4 2 2 + 2 + 2 3 4 3 4 2 2 +1 +1 (b) A CAS estimates the integral to be A(S)= D 1+(2x) +(2y) dA= 0 0 2 2 1 1 1+4x +4y dydx 1.8616 . 2 2 This agrees with the Midpoint estimate only in the first decimal place. 16. (a) With m=n=2 we have four squares with midpoints 3 3 , 2 2 A(S) = D 1 1 , 2 2 , 1 3 , 2 2 , 3 1 , 2 2 , and . Since z=xy+x +y , the Midpoint Rule gives [ f (x,y)] +[ f (x,y)] +1 dA= x y D 2 2 2 2 (y+2x) +(x+2y) +1 dA 5 2 2 2 2 1 = 3 2 2 + 3 2 2 +1 + + 7 2 2 +1 + 7 2 2 + 5 2 2 +1 + 9 2 2 + 9 2 22 78 78 166 + + + 2 2 2 2 17.619 5 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area (b) Using a CAS, we have A(S)= D (y+2x) +(x+2y) +1 dA= 00 2 2 22 1+(y+2x) +(x+2y) dydx 17.7165 . This is within about 0.1 2 2 of the Midpoint Rule estimate. 17. z=1+2x+3y+4y , so A(S) = D 41 2 1+ z x 2 + 2 z y 2 41 dA= 1 0 1+4+(3+8y) dydx 2 = 1 0 14+48y+64y dydx . Using a CAS, we have 41 1 0 14+48y+64y dydx= or 45 8 14 + 2 45 8 14 + 15 15 ln ( 11 5+3 14 5 ) ln ( 3 5+ 14 5 ) 16 16 11 5+3 70 15 ln . 16 3 5+ 70 2 18. f ( x,y ) =1+x+y+x 1 1 f =1+2x , f =1 . We use a CAS to calculate the integral x y 1 1 2 1 A(S)= 2 1 f + f +1 dydx= x y 1 2 2 (1+2x) +2 dydx=2 2 2 1 4x +4x+3 dx and find that 2 A(S)=3 11 +2sinh 3 2 2 or A ( S ) =3 11 +ln ( 10+3 11 ) . 19. f (x,y)=1+x y 1 1 x 2 2 2 f =2xy , f =2x y . We use a CAS (with precision reduced to five significant x y 1 1 1 x 2 2 2 digits, to speed up the calculation) to estimate the integral A(S)= 1 1 x 2 f + f +1 dydx= x y 1 x 2 2 2 4x y +4x y +1 dydx , and find that A(S) 3.3213 . 2 4 4 2 6 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area 20. Let f (x,y)= 1+x 2 1+y 1 1 |x| 2 . Then f = x 2 x 2 y 2x 1+y 2 , f = 1+x y ( 2 ) 2y ( 1+y ) 2 2 = 2y 1+x ( 2 ) ( 1+y ) 2 2 . We use a CAS to estimate 1 ( 1 |x| ) f + f +1 dydx 2.6959 . In order to graph only the part of the surface above y 1 x as the y range in our plot command. the square, we use ( 1 x ) 21. Here z= f (x,y)=ax+by+c , f (x,y)=a , f (x,y)=b , so x y A(S)= D a +b +1 dA= a +b +1 D 2 2 2 2 dA= a +b +1 A(D) . 2 2 22. Let S be the upper hemisphere. Then z= f (x,y)= a x y , so A(S) = D 2 2 2 [ x(a x y ) x +y D 2 2 2 2 2 2 2 2 1/2 2 ] +[ y(a x y ) 2 t 2 2 2 1/2 2 ] +1 dA = 2 t +1 dA= 0 0 r 2 2 2 +1 r dr d t 0 a x y ar a r 2 a r a =lim t a 0 0 dr d =2 lim 2 2 t a a r 2 2 =2 lim a t a a t 2 2 a =2 ( a)( a)=2 a . Thus the surface area of the entire sphere is 4 a . 23. If we project the surface onto the xz plane, then the surface lies ``above'' the disk x +z the xz plane. We have y= f (x,z)=x +z and, adapting Formula 2, the area of the surface is A(S)= x +z 2 2 2 2 2 25 in 2 2 [ f (x,z)] +[ f (x,z)] +1 dA= x z 25 x +z 2 2 2 2 4x +4z +1 dA 25 7 2 2 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area Converting to polar coordinates x=rcos 2 5 , z=rsin 2 1/2 we have 2 0 A(S) = 0 0 4r +1 r dr d = 0 2 2 5 d 0 r(4r +1) dr= 1 2 3/2 (4r +1) 12 5 0 = 6 ( 101 101 1 ) 24. First we find the area of the face of the surface that intersects the positive y axis. As in Exercise 23 , we can project the face onto the xz plane, so the surface lies ``above'' the disk x +z z= f (x,z)= 1 z and the area is A(S) = [ f (x,z)] +[ f (x,z)] +1 dA= 2 2 2 2 1 . Then 2 2 2 x z 0+ 2 2 z 1 z 2 2 +1 dA x +z 1 2 2 1 1 z 2 x +z 1 = x +z 1 2 2 z 1 1 z 0 2 +1 dA= 1 1 z 2 1 1 z 2 dxdz 1 z 1 1 z =4 0 dxdz [by the symmetry of the surface] 2 This integral is improper (when z=1 ), so t 1 z 0 2 A ( S ) =lim 4 t 1 0 1 2 t dxdz=lim 4 0 1 z 2 2 t dz=lim 4 dz=lim 4t=4 . t 1 t 1 0 t 1 1 z 1 z Since the complete surface consists of four congruent faces, the total surface area is 4(4)=16 . 8
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TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.3 Partial Derivatives1. (a) T / x represents the rate of change of T when we fix y and t and consider T as a function of the single variable x , which describes how quickly the temperatu
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Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates1. The region R is more easily described by polar coordinates: R= { (r, )|0 r2 22,02 } . Thusf (x,y)dA=R 0 0f (rcos ,rsin )r dr d .2. Th
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Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions1.1x21 0 1 0 3(x+2y)dydx =0 0xy+y2 y=x21 0y=0dx=x(x )+(x ) 0 0 dx1 0222= (x +x )dx=41 4 1 5 x+ x 4 5=9 202.22
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.4 Series Solutions1. Let y(x)=n=1n=0 nc x . Then y (x)=nn/n=1nc xnn 1and the given equation, yn=0 n/y=0 , becomesn+1nc xnn 1n=0 n n+1c x
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Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.9 Change of Variables in Multiple Integrals1. x=u+4v , y=3u 2v . ( x,y ) The Jacobian is = ( u,v ) 2. 3.x/ u y/ u x/ v = y/ vx/ v = y/ v 2u 2u1 34 =1( 2) 4(3)= 14 . 2( x,y ) = (
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Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.3 Applications of Second-Order Differential Equations100 is the spring constant and the differential equation is 3 10 10 / / 100 3x + x=0 . The general solution is x(t)=c
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Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.7 Triple Integrals1. xyz dV =0 10 1 2 1 23Bxyz dzdxdy=0 1 x=2 x= 121 2xy1 3 z 3z=3 z=01 2dxdy=1 09xydxdy0 1=09 2 x y 227 27 2 dy= ydy= y 1 4 0 2=27 42
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.6 Directional Derivatives and the Gradient Vector1. First we draw a line passing through Raleigh and the eye of the hurricane. We can approximate the directional derivative at Raleigh in
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.5 The Chain Rule1. z=x y+xy , x=2+t , y=1 t dz z dx z dy 2 = + = 2xy+y dt x dt y dt2243() ( 4t 3) + ( x2+2xy) ( 3t2) =4 ( 2xy+y2) 3 3( x2+2xy) t2 )1/22. z= x +y , x=e , y=
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals1. x=t and y=t , 0 tC22 , so by Formula 32y ds = =20 2t t2dx dt+dy dt2dt= t0 2 02(2t) +(1) dt 1 (17 17 1) 122204t +1 dt=1 2 3/2 (4t +1
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence1. (a) i curl F =2j / y 0F=/ x xyzk / z xy2=( x 0) i ( 2xy xy) j+(0 xz) k2= x i+3xy j xz k (b) div F= 2. (a) curlF = F=2 2F=x(xyz)+y(0)+z(
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Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.6 Parametric Surfaces and Their Areas1. r(u, v)=ucos v i+usin v j+u k , so the corresponding parametric equations for the surface are x=ucos v, y=usin v, z=u . For any point (x, y, z) on the
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TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.8 Stokes' Theorem1. Both H and P are oriented piecewise smooth surfaces that are bounded by the simple, closed, smooth curve x +y =4 , z=0 (which we can take to be oriented positively for bo
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Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem1. The vectors that end near P are longer than the vectors that start near P , so the net flow is inward1 1near P and divF(P ) is negative. The vectors that end ne
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.1 Three-Dimensional Coordinate Systems1. We start at the origin, which has coordinates ( 0,0,0 ) . First we move 4 units along the positive x axis, affecting only the x coo
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.2 Vectors1. (a) The cost of a theater ticket is a scalar, because it has only magnitude. (b) The current in a river is a vector, because it has both magnitude (the speed of
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.3 The Dot Product1. (a) a b is a scalar, and the dot product is defined only for vectors, so ( a b) c has no meaning. (b) ( a b) c is a scalar multiple of a vector, so it d
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.4 The Cross Producti j k 2 0 1 0 1 2 1. a b= 1 2 0 = i j+ k 3 1 0 1 0 3 0 3 1 =(2 0) i (1 0) j+(3 0) k=2 i j+3 k Now (a b) a= 2, 1,3 1,2,0 =2 2+0=0 and (a b) b= 2, 1,3 0,3,
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes1. (a) True; each of the first two lines has a direction vector parallel to the direction vector of the third line, so these vectors are each
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Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves1. 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 functions2t1 and 5 t
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Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books I-IVSpring 2008Virgil's Aeneid, strongly modeled on Homer's Iliad and Odyssey, was considered even in antiquity the great epic of Rome. It tells the story of the Trojan Aeneas who, afte
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books V-VIIISpring 2008Book V provides a transition between the high emotion of Book IV and the sombre majesty of the descent to the underworld in Book VI. Most of the book is taken up with g
FSU - CLA - 2123
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
FSU - CLA - 2123
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;
Stevens - PEP - 111
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
Stevens - PEP - 111
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
Stevens - PEP - 111
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
Stevens - PEP - 111
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
Stevens - PEP - 111
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
Stevens - PEP - 111
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