Course Hero has millions of student submitted documents similar to the one

below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

4 CHAPTER PROBLEM SOLUTIONS
Problem
4.
j An airliner is flying at a velocity of 260 m/s , when a wind gust gives it an acceleration of 0.38 + 0.72^ m/s for a period of 24 s. (a) What is its velocity at the end of that time? (b) By what angle has it been deflected from its original course?
2
Solution
2 j j (a) Equation 4-3 gives v = 260 m/s + (0.38 + 0.72^ )(m/s )(24 s) = (269 + 17.3^) m/s. (b) Since v0 is along the x-axis, the
angular deflection is just tan (v y= x ) = tan (17.3 269 ) = 3.67 . v =
!1
!1
Problem
6. An airplane heads northeastward down a runway, accelerating from rest at the rate of 2.1 m/s . Express the plane's velocity and position at t = 30 s in unit vector notation, using a coordinate system with x-axis eastward and y-axis northward, and with origin at the start of the plane's takeoff roll.
2
Solution
Since the acceleration is constant and the airplane starts from rest (v0 = 0) at the origin (r0 = 0), Equations 4-3 and 4-4 2 2 ^ give v(t) = at and r(t) = 1 at . Both vectors are in the NE direction, parallel to a = (2.1 m/s )( cos 45 + j sin 45) =
2
(1.48 m 2)( + ^ ). Thus, v(30 s) = (1.48 m/s2 )(30 s)( + ^ ) = (44.5 m/s)( + ^), and r(30 s) = /s j j j ^ (668 m)( + ^). (Note: ( + j)= 2 is a unit vector in the NE direction.) j
1 2
(1.48 m/s)(30 s) ( + ^ ) = j
2
Problem
13. Figure 4-27 shows a cathode-ray tube, used to display electrical signals in oscilloscopes and other scientific instruments. 9 Electrons are accelerated by the electron gun, then move down the center of the tube at 2.0 ! 10 cm/s . In the 4.2-cmlong deflecting region they undergo an acceleration directed perpendicular to the long axis of the tube. The acceleration "steers" them to a particular spot on the screen, where they produce a visible glow. (a) What acceleration is needed to deflect the electrons through 15, as shown in the figure? (b) What is the shape of an electron's path in the deflecting region?
Solution
(a) With x-y axes as drawn on Fig. 4-27, the electrons emerge from the deflecting region with velocity v = v 0 + at ^ , j
9 after a time t = x=0 , where x = 4.2 cm and v = 2 ! 10 cm/s. The angle of deflection (direction of v) is 15, so v 0 2 tan 15 = v y= x = at=0 = ax=0 . Thus, a = v0 tan15= = 2.55 ! 101 7 cm/s2 (when values are substituted). (b) Since the v v v2 x acceleration is assumed constant, the electron trajectory is parabolic in the deflecting region.
CHAPTER 4
71
Problem
14. You toss an apple horizontally at 8.7 m/s from a height of 2.6 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground?
Solution
The time of flight for either projectile can be determined from the vertical component of the motion, which is the same for both, since v y = 0. Thus, Equation 4-8 gives t = 0
2( y ! y0 )= = g
2( 2.6 m)= (9.8 m/s2 ) = 0.728 s.
Problem
18. In a chase scene, a movie stuntman is supposed to run right off the flat roof of one city building and land on another roof 1.9 m lower. If the gap between the buildings is 4.5 m wide, how fast must he run?
Solution
The horizontal and vertical distances covered by the stuntman are x ! x0 = v0 t and y0 ! y =
2
1 2
gt 2 (since v = v0 , 0x
and v y = 0). Eliminating t, one finds v 0 = (x ! x 0 ) g= y 0 ! y) = (4.5 m) (9.8 m/s )= 2( 2(1.9 m) = 7.23 m/s. (Note that 0 Equation 4-9 with !0 = 0 and y0 = 0 provides an equivalent solution.)
72 CHAPTER 4
Problem
22. Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed v from a 0 height h.
Solution
This problem is solved by reasoning identical to that in problem 15, where it was shown that x = v0 2 h g. =
Problem
15. A carpenter tosses a shingle off a 8.8-m-high roof, giving it an initially horizontal velocity of 11 m/s. (a) How long does it take to reach the ground? (b) How far does it move horizontally in this time?
Solution
(a) The shingle reaches the ground when y(t) = 0 = y0 !
1 2
gt2 , or
2(8.8 m) (9.8 m/s2 ) = 1.34 s.
t =
2 y0 = g
(b) The horizontal displacement is x = v0 t = (11 m /s)(1.34 s) = 14.7 m.
CHAPTER 4
73
Problem
28. A rescue airplane is flying horizontally at speed v0 at an altitude h above the ocean, attempting to drop a package of medical supplies to a shipwreck victim in a lifeboat. At what line-of-sight angle ! (Fig. 4-29) should the pilot release the package?
Solution
The package, when dropped, is a horizontally launched projectile, like that in Problem 15 (the expressions for t and x are the "1 g same). The line-of-sight angle is ! = tan (h x), where x = v 0 2 h= is the horizontal distance at drop time. Thus, =
! = tan "1
gh= 2 . 2v 0
FIGURE 4-29 Problem 28 Solution.
Problem
29. At a circus, a human cannonball is shot from a cannon at 35 km/h at an angle of 40. If he leaves the cannon 1.0 m off the ground, and lands in a net 2.0 m off the ground, how long is he in the air?
Solution
The time of flight can be calculated from Equation 4-8, where the trajectory begins at y0 = 1.0 m and t = 0, ends at
y(t) = 2.0 m, and v y = v0 sin !0 = (35 m/3.6 s) sin 40 = 6.25 m/s . The equation is a quadratic, 0
( y ! y0 ) = 0, with solutions t = [v 0 y
1 2
gt 2 ! v0y t +
v 2 ! 2 g( y ! y 0 )]=g = [6.25 m/s 0y
(6.25 m/s)2 ! 2(9.8 m/s2 )( 2 m ! 1 m)]
(9.8 m/s2 ) = 0.188 s or 1.09 s. The trajectory crosses the height 2.0 m twice, once going up, at the smaller time of flight, and once going down into the net, at the larger time of flight. The latter is the answer to the question asked here.
74 CHAPTER 4
Problem
33. A projectile launched at an angle !0 to the horizontal reaches a maximum height h. Show that its horizontal range is 4h tan !0 . =
Solution
The intermediate expression for the horizontal range (when the initial and final heights are equal) is x = 2v0 sin! 0 cos ! 0= = 2v0xv0y= (see the equation before Equation 4-10). The components of the initial velocity are related by g g
2
v y=0x = tan !0 . The maximum height, h = y max ! y0 , can be found from Equation 2-11 (when vy = 0) or v2 y = 2 gh. 0 v 0 Then x = 2v0 yv0x= = 2v0 y (v0 y= !0 )= = 2(2gh)= tan !0 = 4h= !0 . (This result reflects a classical geometrical property g tan g g tan of the parabola, namely, that the latus rectum is four times the distance from vertex to focus.)
Problem
34. You're 5.0 m from the left-hand wall of the house shown in Fig. 4-30, and you want to throw a ball to a friend 5.0 m from the right-hand wall. (a) What is the minimum speed that will allow the ball to clear the roof? (b) At what angle should it be thrown? Assume the throw and catch both occur 1.0 m above the ground.
FIGURE 4-30 Problem 34.
Solution
Since the trajectory is symmetrical (begins and ends at the same height), one can use the result of the previous problem with "1 x) h = y max ! y0 = 6 m ! 1 m = 5 m and horizontal range x = 5 m+ 6 m + 5 m = 16 m. Then (b) !0 = tan (4h= = 51.3 , and (a) v 0 = v 0y = ! 0 = sin
2 gh= ! 0 = 12.7 m/s. sin
Problem
42. How fast would a car have to round a turn 75 m in radius in order for its acceleration be to numerically equal to that of gravity?
Solution
For circular motion with constant speed, ar = v = or r
2
v =
ar r =
(9.8 m/s 2 )( 75 m) = 27.1 m/s = 97.6 km/h = 60.7 mi h. /
CHAPTER 4
75
Problem
46. A 12-in-diameter circular saw blade rotates at 3500 revolutions per minute. What is the acceleration of one of the saw teeth? Compare with the acceleration of gravity.
Solution
If the saw blade rotates at 3500 rpm, a point on its circumference has a linear speed of v = (3500)! (12 in)= s = 183 ft/s. 60 The (radial) acceleration is ar = v = = (183 ft/s) = ft= = 6.72 ! 10 ft/s ' 2 ! 10 g, where g = 32.2 ft/s . r (6 12) Alternatively, one could use Equation 4-12 with period equal to (3500)
!1 2 2 4 2 3 2
min .
Problem
54. A car moving at 65 km/h enters a curve that describes a quarter turn of radius 120 m. The driver gently applies the 2 brakes, giving a constant tangential deceleration of magnitude 0.65 m/s . Just before emerging from the turn, what are (a) the magnitude of the car's acceleration and (b) the angle between the acceleration vector and the direction of motion?
Solution
The car's tangential acceleration is constant, a = !0.65 m/ s along the direction of motion, so Equation 2-11 can be used to determine the speed at the end of the turn, v = v0 + 2at s. Here, s = 2! r= is the linear distance around the quarter turn. 4 Then the centripetal acceleration is ac = v = = (v 0 + ! rat )= = (65 m= s) = r r 3.6 120 m + ! ("0.65 m/s ) = 0.675 m/ s . (a) The magnitude of the total acceleration is
"1 2 2 2 2 2
2 2
2
a2 + at2 = 0.937 m/ s2. (b) The angle of the total acceleration with respect to c
the tangent to the curve is ! = tan (ac= t ) = 134. (This angle is in the second quadrant because the tangential acceleration a is opposite to the direction of motion.)
Problem
57. If you can throw a stone straight up to a height of 16 m, how far could you throw it horizontally over level ground? Assume the same throwing speed and optimum launch angle.
Solution
To throw an object vertically to a maximum height of h = 16 m = y max ! y0 requires an initial speed of v = 0 2 g( ymax ! y0 ) = 2 gh . With this value of v and the optimum launch angle !0 = 45, Equation 4-10 gives a maximum 0 horizontal range on level ground of x = v0 = = 2h = 32 m. (The maximum horizontal range on level ground is twice the g maximum height for vertical motion with the same initial speed. This result holds in the approximation of constant g and no air resistance.)
2
76 CHAPTER 4
Problem
63. You toss a chocolate bar to your hiking companion located 8.6 m up a 39 slope, as shown in Fig. 4-36. Determine the initial velocity vector so that the chocolate bar will reach your friend moving horizontally.
FIGURE 4-36 Problem 63.
Solution
The candy bar moves horizontally only at the apex of its trajectory (where vx = v0x and vy = 0). Thus, ymax ! y0 =
2 (8.6 m) sin 39 = 5.41 m, and v y = 2 g( ymax ! y0 ) = 2(9.8 m/s )( 5.41 m) = 10.3 m/s (see Equation 2-11). The time to 0 t reach the apex is t = v0y= so v 0x = (x ! x0 )= = (x ! x0)g=v 0y (see Equations 4-6 and 4-7). The horizontal distance from g,
apex to origin is x ! x0 = (8.6 m) cos 39 = 6.68 m, so v 0x = (6.68 m)(9.8 m/s )= (10.3 m/s) = 6.36 m/s. v0 can be 2 ^ ) m/s, or by its magnitude v 2 + v 0 y = 12.1 m/s and direction ! = expressed in unit vector notation as (6.36 + 10.3j 0x
2
tan!1 (v y=0x ) = 58.3 (CCW from the x-axis). 0 v
CHAPTER 4
77
Problem
69. A monkey is hanging from a branch a height h above the ground. A naturalist stands a horizontal distance d from a point directly below the monkey. The naturalist aims a tranquilizer dart directly at the monkey, but just as she fires the monkey 2 2 2h lets go. Show that the dart will nevertheless hit the monkey, provided its initial speed exceeds (d + h )g= .
Solution
Gravity accelerates the dart and the monkey equally, so both fall the same vertical distance from the point of aim (the monkey's original position) resulting in a hit, provided the initial speed of the dart is sufficient to reach the monkey before the monkey reaches the ground. To prove this assertion, let the dart be fired from ground level ( y = 0) with speed v0 and
"1 direction !0 = tan (h d) (line of sight from naturalist N to monkey M), while the monkey drops from height h at t = 0. The =
vertical height of each is ymonkey = h !
1 2
gt2 and ydart = v 0 yt !
1 2
2 2 gt 2 , where v 0 y = v 0 sin ! 0 = v 0 h= d + h . (The term
!
1 2
gt2 represents the effect of gravity, which appears the same way in both y-coordinate equations.) The dart strikes the
monkey when ymonkey = ydart , which implies h = v0 yt , or t = required for the monkey to fall to the ground, which is
d 2 + h2 =v 0 . This must be less than the time
d 2 + h2 = 0 < v 2h=g or
2h= (from ymonkey = 0). Thus g
g(d + h )= (This condition can also be understood from the horizontal range formula, Equation 4-10. The 2h. 2 2 2 range of the dart has to be greater than the horizontal distance to the monkey, d < v 0 sin 2! 0=g = v 0 2hd= + h ) g. (d2 ) v0 >
2
2
Problem 69 Solution.
78 CHAPTER 4
Problem
70. A child tosses a ball over a flat-roofed house 3.2 m high and 7.4 m wide, so it just clears the corners on both sides, as shown in Fig. 4-37. If the child stands 2.1 m from the wall, what are the ball's initial speed and launch angle? Assume the ball is launched essentially from ground level.
Solution
Choose the coordinate system shown in Fig. 4-37, such that the four given points on the trajectory are the origin (x0 , y0 ) = (0, 0), the first corner (x1, y1 ) = (2.1 m, 3.2 m), the second corner (x2 , y2 ) = (9.5 m, 3.2 m), and ground level at the right (x3 , y3 ) = (11.6 m, 0). Equation 4-9 for the trajectory, evaluated at any two points other than the origin, provides two equations which can be solved for v and !0 . Before substituting values, we may divide Equation 4-9 by x, and let 0
2v2 2 a = tan ! 0 and b = g= 0 cos !0 . For example, selecting the two corner points, we obtain y1=x1 = a ! x1 b, and y2= 2 = a ! x2 b, with solutions (x2 ! x1)a = ( x2 y1 =x1) ! (x1 y2 =x2 ), and (x2 ! x1)b = ( y1 =x1) ! (y2 =x2). In terms of a x
and b, !0 = tan
"1
a , and v 0 =
g(1 + a2 )= b, so with the given numerical values, one finds 2
1 % " 9.5 2.1% " 1 7.4a = $ ! ' 3.2, and 7.4b = $ ! ' 3.2, # 2.1 9.5& # 2.1 m 9.5 m&
or !0 = tan
"1
1.86 = 61.7, and v 0 =
(9.8 m/s2)( 1 + 1.862 )= 2(0.160 m!1) = 11.7 m/s.
FIGURE 4-37 Problem 70 Solution.
Problem
72. In your calculus class, you may have learned that you can find the maximum or minimum of a function by differentiating and setting the result to zero. Do this for Equation 4-10, differentiating with respect to !0 , and thus verify that the maximum range occurs for !0 = 45.
Solution
The derivative of Equation 4-10 with respect to !0 is dx=d!0 = 2(v 0 =g) cos 2! 0 . This is zero when 2! 0 = 90, or !0 = 45 as stated. (This is the only maximum, since launch angles are restricted to the range 0 < !0 < 90, and d x=d!0 < 0.)
2 2 2

**Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.**

Below is a small sample set of documents:

UCSD - PHYS - 2A

CHAPTER 5 PROBLEM SOLUTIONSProblem4. A car leaves the road traveling at 110 km/h and hits a tree, coming to a complete stop in 0.14 s. What average force does a seatbelt exert on a 60-kg passenger during this collision?SolutionAssume that the se

UCSD - PHYS - 2A

CHAPTER 6Problem3. A 3700-kg barge is being pulled along a canal by two mules, as shown in Fig. 6-59. The tension in each tow rope is 1100 N, and the ropes make 25 angles with the forward direction. What force does the water exert on the barge (a)

UCSD - PHYS - 2A

Chapter 7Problem3.Problem SolutionsA crane lifts a 650-kg beam vertically upward 23 m, then swings it eastward 18 m. How much work does the crane do? Neglect friction, and assume the beam moves with constant speed.SolutionLifting the beam at

UCSD - PHYS - 2A

Chapter 8Problem2.Problem SolutionsNow take Fig. 8-26 to lie in a vertical plane, and find the work done by the gravitational force as an object moves from point 1 to point 2 over each of the paths shown.SolutionTake the origin at point 1 in

UCSD - PHYS - 2A

Physics 2A Olga Dudko UCSD PhysicsLecture 5Today: Motion in more than one dimension. Projectile motion.The Monkey and the Hunter.Kinematics Equations The kinematics equations that we learned for one dimensional motion (when a = const) still h

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 6Today: Projectile trajectories. Circular motion.Projectile trajectories Very often we are interested in mathematically describing the trajectory of a projectile. If it indeed follows a parabolic path,

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 9Today: Newton's Third Law.Donkey-and-cart dilemmaApplying Newton's Laws.Newton's Third LawNewton's Third Law: If an object A exerts a force on object B, then object B exerts an oppositely directed

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 11Today: Pulleys.Atwood's MachineCircular motion. Centripetal force.ExamplePulley (Atwood's Machine) Two objects with masses 2.00kg (m1) and 6.00kg (m2) are connected by a light string that passe

UCSD - PHYS - 2A

Physics 2A Lecture 13Olga Dudko UCSD PhysicsToday: Work, Energy, PowerVariable Work Until now, we have dealt with work equations where the forces were constant. When the force varied with position, you have to integrate the force over the dist

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 14Today: Conservation of Energy Conservative and nonconservative forcesPotential Energy Kinetic energy quantifies motion. Another type of energy is potential energy. Potential energy (U) is the amount

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 15Today: Conservation of mechanical energySwinging pendulum - should I duck as it swings back?Potential energy curvesMechanical Energy Example A 10.0kg block is released from point A in the figure bel

UCSD - PHYS - 2A

Physics 2ALecture 17"If you can find a path with no obstacles, it probably doesn't lead anywhere." -Frank A. ClarkNewton's Laws COMSince the center of mass is now calculable, we can use this to describe the motion of complicated objects. We wil

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 18Today: Kinetic energy in many-particle systems Physics of collisionsKinetic energy in many-particle systems We've seen that the momentum of a many-particle system is determined entirely by the motion

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 19Today: Physics of collisions and explosions Rotational motionTotally inelastic collision: p conserved two colliding objects stick together; conservation of momentum: m1v1 + m2v2 = (m1+m2)vf r r r m1v

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 20Today: Relating Rotational and Linear motion Rotational vectors and Right Hand RuleEquations for constant angular accelerationconstant-acceleration equations apply with the proper substitutions:Equat

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 22Today: Rotational kinetic energyGreat race of shapes!Conservation of angular momentumArt of figure skatingWork in rotational motion Just like a force can perform work over a distance, torque can p

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 23Today: Static Equilibrium Center of GravityTips to avoid tipping overWhat keeps the rock and the bridge from falling?Equilibrium A body is in equilibrium when the net external force and torque on t

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 24Today: Static equilibriumThe tragedy of Romeo and JulietStability of EquilibriaStatic EquilibriumRomeo is trying to reach Juliet by climbing an 8.00m, 200N uniform ladder which rests against a smoo

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 27Today: Gravitation. Gravitational energy.Gravitational Force Fg = mg only works near the surface of the Earth! Newton: in general, the magnitude of the gravitational force is mEarth Fg = mgEarthm1

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 28Today: Escape speed. Kepler's laws of planetary motion.Escape speeddown. Is this always true? Throw a ball straight up, and eventually it comes Not if the ball has enough energy!Etot = E3 can esc

UCSD - PHYS - 2A

Physics 2AOlga Dudko UCSD PhysicsLecture 29Today: Class review. Final exam information.Things We Have Learned1. Math Techniques and Problem Solving: Know how to pick an appropriate coordinate system. Know how to use dimensional analysis and

UCSD - PHYS - 2A

Spring 2008DEPARTMENT OF PHYSICS Physics 2A Physics MechanicsMarch 27, 2008Web page: INSTRUCTOR:http:/physics.ucsd.edu/students/courses/spring2008/managed/physics2a/ 2A Prof. Olga Dudko dudko@physics.ucsd.edu Office: 7234 Urey Hall Office Ho

UCSD - PHYS - 2A

Chapter 10 Solution Problem1. A 28-kg child sits at one end of a 3.5-m-long seesaw. Where should her 65-kg father sit so the center of mass will be at the center of the seesaw?SolutionTake the x-axis along the seesaw in the direction of the fathe

UCSD - PHYS - 2A

Chapter 11 SolutionsProblem1. What is the impulse associated with a 650-N force acting for 80 ms?SolutionThe impulse for a constant force (from Equation 11-1) is I = ! F dt = (650 N)(0.08 s) = 52 N " s , in the direction of the force.Problem6

UCSD - PHYS - 2A

CHAPTER 12 SOLUTIONSProblem5. A wheel turns through 2.0 revolutions while being accelerated from rest at 18 rpm/s. (a) What is the final angular speed? (b) How long does it take to turn the 2.0 revolutions?SolutionFor constant angular accelerati

UCSD - PHYS - 2A

CHAPTER 13 SolutionsProblem3. A wheel is spinning at 45 rpm with its spin axis vertical. After 15 s, it's spinning at 60 rpm with its axis horizontal. Find (a) the magnitude of its average angular acceleration and (b) the angle the average angular

UCSD - PHYS - 2A

CHAPTER 14 SolutionsProblem2. A body is subject to three forces: F = 2 + 2^ N, applied at the point x = 2 m, y = 0 m; F = !2 ! 3^ N, applied j j 1 2 ^ at x = !1 m, y = 0; and F = 1j N, applied at x = !7 m, y = 1 m. (a) Show explicitly that the ne

UCSD - PHYS - 2A

CHAPTER 15 SOLUTIONSProblem3. The vibration frequency of a hydrogen chloride molecule is 8.66 ! 10 complete one oscillation?13Hz. How long does it take the molecule toSolutionT = 1=f = 1= (8.66 ! 101 3 Hz) = 1.15 ! 10"1 4 s = 11.5 fs (Equatio

UCSD - CHEM - 140B

Additional NMR ProblemsCHEM 140B Burkart Winter, 20041.C4H8O3H3H2H2.C3H8O6H1H1H3.C5H10O33H 2H4.C8H14O33H 2H2H5.C3H6ClBr1H 1H 1HAdditional NMR Problem SolutionsCHEM 140B Burkart Winter, 20041. IR: 1705

A.T. Still University - ENG - sem212

CHAPTER 5DIFFUSIONPROBLEM SOLUTIONSIntroduction5.1 Self-diffusion is atomic migration in pure metals-i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal.Diffusion Mechan

A.T. Still University - ENG - See103

SEE103 Electronics in 2006 Who ? When & Where ? What ? Why ?1- Dr Hieu Trinh Room: ka5.305 Phone: (03) 5227 2030 Email: hmt@deakin.edu.au Fax: (03) 5227 2167Unit ChairWho ?Lecturers- Dr Alan Wong (first 4 weeks, module 1) - Dr H

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

ASSIGNMENT 5 (10 marks) Complete the following questions in the text book: 8.3, 8.4 (the first part only, ie. don't have to do the comparison), and 8.9.

A.T. Still University - ENG - sem314

Topic 1Introduction and basic definitionsThroughout this topic you will need to refer to chapter 1 in your textbook.IntroductionThermodynamics is the science of energy in its different forms, and methods of converting one form of energy into a

A.T. Still University - ENG - sem314

Topic 2Energy conservation: The First Law of ThermodynamicsThroughout this topic you will need to refer to sections 1.61.8 in your textbook.Conservation of energyThe First Law of Thermodynamics simply introduces the concept of `conservation of

A.T. Still University - ENG - sem314

Topic 3Working fluidsThroughout this topic you will need to refer to chapter 2 in your textbook. Topic 1 covered definitions/concepts such as `system', `working fluid', and `system boundary'. Topic 2 introduced the First Law of Thermodynamics, th

A.T. Still University - ENG - sem314

Topic 4Typical thermodynamic processesThroughout this topic you will need to refer to chapter 3 in your textbook. Working fluid (in a system) can undergo two types of thermodynamic processes which can be broadly classified as reversible and non-r

A.T. Still University - ENG - sem314

Topic 5The Second Law of ThermodynamicsThroughout this topic you will need to refer to chapter 4 in your textbook.IntroductionThe following section will help you to deal with the Second Law of Thermodynamics, entropy, and irreversibility. Befo

A.T. Still University - ENG - sem314

Topic 6The typical heat engine cyclesThroughout this topic you will need to refer to chapter 5 in your textbook. We are going to discuss some typical heat engine cycles for energy conversion (heat to work). We shall only be concerned with convers

A.T. Still University - ENG - sem314

STUDY TOPIC: STEAM PLANTReference: Chapter 8 in the unit Text Book. Previously, you learned the basics of thermodynamics including basic definition, first and second laws of thermodynamics heat engine cycles. From these study, we know that the most

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem314

A.T. Still University - ENG - sem212

Gaussian Error Function z 0.00 0.03 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 erf(z) 0 0.03 0.06 0.11 0.17 0.22 0.28 0.33

A.T. Still University - ENG - sem212

6.35 This problem asks us to demonstrate that true strain may also be represented byA T = ln 0 A iRearrangement of Equation 6.17 leads toli l0 A0 Ai=Thus, Equation 6.16 takes the forml A T = ln i = ln 0 l A 0 i A The express

A.T. Still University - ENG - sem212

6.38 We are asked to compute how much elongation a metal specimen will experience when a true stress of 415 MPa is applied, given the value of n and that a given true stress produces a specific true strain. Solution of this problem requires that we u

A.T. Still University - ENG - sem212

6.40 For this problem we first need to convert engineering stresses and strains to true stresses and strains so that the constants K and n in Equation 6.19 may be determined. Since T = (1 + ) then,T = (315 MPa)(1 + 0.105) = 348 MPa1T2= (340

A.T. Still University - ENG - sem212

7.7 Below is shown the atomic packing for a BCC {110}-type plane. The arrows indicate two different <111> type directions.7.21 Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in

A.T. Still University - ENG - sem212

7-67.6 (a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.11 asPD110 (FCC) = 1 4 R2 2 = 0.177 R2Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.53

A.T. Still University - ENG - sem212

8-48.4 The maximum allowable surface crack length for MgO may be determined using Equation 8.3; taking 225 GPa as the modulus of elasticity (Table 12.5), and solving for a, leads to2 E s2 ca==(2) (225 x 10 9 N / m2 ) (1.0 N / m) () (13.5 x

A.T. Still University - ENG - sem212

9.9 It is possible to have a Cu-Ag alloy, which at equilibrium consists of a phase of composition 92 wt% Ag-8 wt% Cu and a liquid phase of composition 77 wt% Ag-23 wt% Cu. From Figure 9.7 a horizontal tie line can be constructed across the + L phas