ENGR 350
Assignment 2
Anne-Marie Eggers
932-229-427
Footprint Comparisons
Traditionally, the more developed the country the larger the footprint, however when
comparing Americas Ecological Footprint to the rest of the developed world it is interesting to

34.1.
Visualize:
To develop a motional emf the magnetic field needs to be perpendicular to both, so lets say its direction is into the page. Solve: This is a straightforward use of Equation 34.3. We have
v=
Assess:
E 1.0 V = = 2.0 104 m/s lB (1.0 m ) ( 5.

33.1. Model: A magnetic field is caused by an electric current.
Visualize: Please refer to Figure EX33.1. Solve: The magnitude of the magnetic field at point 1 is 2.0 mT and its direction can be determined by using the right-hand rule. Grab the current ca

32.1. Solve:
From the circuit in Figure EX32.1, we see that 50 and 100 resistors are connected in series across the battery. Another resistor of 75 is also connected across the battery.
32.2. Solve: In Figure EX32.2, the positive terminal of the battery i

31.1. Solve: The wires cross-sectional area is A = r 2 = (1.0 103 m ) = 3.1415 106 m 2 , and the electron
2
current through this wire is i = 31.3, the drift velocity is
vd =
Ne = 2.0 1019 s 1 . Using Table 31.1 for the electron density of iron and Equatio

30.1. Solve: The potential difference V between two points in space is
V = V ( xf ) V ( xi ) = Ex dx
xi xf
where x is the position along a line from point i to point f. When the electric field is uniform,
V = Ex dx = Ex x = (1000 V/m )( 0.30 m 0.10 m ) =

29.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform
electric field. Visualize:
The figure shows the before-and-after pictorial representation. The proton has an initial speed vi = 0 m/s and a final speed

28.1. Visualize:
As discussed in Section 28.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to th

27.1.
Model: The electric field is that of the two charges placed on the y-axis. Visualize: Please refer to Figure EX27.1. We denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive, their electric fields at P are d

25.1. Model: Balmers 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 11 1 1 2 2 2 22 6 2 n
where n = 3, 4, 5, 6, and where we have used n =

35.1. Model: Apply the Galilean transformation of velocity.
Solve: (a) In the laboratory frame S, the speed of the proton is
v=
(1.4110
6
m/s ) + (1.41 106 m/s ) = 2.0 106 m/s
2 2
The angle the velocity vector makes with the positive y-axis is
= tan 1

36.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency .
Solve: (a) Referring to the phasor in Figure EX36.1, the phase angle is
t = 180 + 30 = 210
(b) The instantaneous value of the emf is
rad
180
= 3.665

ENGR 350
Assignment 3
Eggers, Anne-Marie
932-229-427
Water in the Developing World
The paper, A Water Resources Threshold and its Implications for Food Security
discusses the idea of a threshold, which is 1500 [m3/ (capita year)], (Yang, 2003, p. 3048) an

ENGR 350
Assignment One
Eggers, Anne-Marie
932-339-427
Sustainable Engineering
To even begin to understand what sustainable engineering is it is essential to look at and
understand each word individually. Both sustainability and engineering seem like two

43.1. Model: The nucleus is composed of Z protons and neutrons. Solve: (a) 3H has Z = 1 proton and 3 1 = 2 neutrons. (b) 40Ar has Z = 18 protons and 40 18 = 22 neutrons. (c) 40Ca has Z = 20 protons and 40 20 = 20 neutrons. (d) 239Pu has Z = 94 protons and

42.1.
Solve: (a) A 4p state corresponds to n = 4 and l = 1. From Equation 42.3, the orbital angular
momentum is L = 1(1 + 1) = 2 . (b) In the case of a 5f state, n = 5 and l = 3. So, L = 3 ( 3 + 1) = 12 .
42.2. Solve: (a) Excluding spin, a state is descri

41.1. Model: Model the electron as a particle in a rigid one-dimensional box of length L.
Solve: Absorption occurs from the ground state n = 1. Its reasonable to assume that the transition is from n = 1 to n = 2. The energy levels of an electron in a rigi

40.1. Model: The sum of the probabilities of all possible outcomes must equal 1 (100%).
Solve: The sum of the probabilities is PA + PB + PC + PD = 1. Hence, 0.40 + 0.30 + PC + PD = 1 PC + PD = 0.30 Because PC = 2PD, 2PD + PD = 0.30. This means PD = 0.10 a

39.1. Solve: A steady photoelectric current of 10 A is indicated in the graph. The number of electrons per
second is
10 A = 10
C
s
= 1.0 10 5
C 1 electron = 6.25 1013 electrons/s s 1.6 10 19 C
39.2. Model: Light of frequency f consists of discrete quanta,

38.1. Model: Current is defined as the rate at which charge flows across an area of cross section.
Solve: Since the current is Q / t and Q = N / e , the number of electrons per second is
N 10 nA 1.0 108 C/s = = = 6.25 1010 s 1 6.3 1010 s 1 t e 1.60 1019 C

37.1. Model: S and S are inertial frames that overlap at t = 0. Frame S moves with a speed v = 5.0 m/s
along x-direction relative to frame S. Visualize: the
The figure shows a pictorial representation of the S and S frames at t = 1.0 s and 5.0 s. Solve: F

24.1. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens.
Visualize:
The figure shows the two lenses and a ray-tracing diagram. The ray-tracing shows that the lens combination will produce a real, inverted image

23.1. Model: Light rays travel in straight lines.
Solve: (a) The time is
t= x 1.0 m = = 3.3 109 s = 3.3 ns c 3 108 m/s
(b) The refractive indices for water, glass, and cubic zirconia are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, light will

10.1. Model: We will use the particle model for the bullet (B) and the running student (S).
Visualize:
Solve:
For the bullet,
1 1 2 K B = mBvB = (0.010 kg)(500 m/s) 2 = 1250 J 2 2 For the running student, 1 1 2 KS = mSvS = (75 kg)(5.5 m/s) 2 = 206 J 2 2 T

9.1. Model: Model the car and the baseball as particles.
Solve:
(a) The momentum p = mv = (1500 kg ) (10 m/s ) = 1.5 104 kg m/s.
(b) The momentum p = mv = ( 0.2 kg )( 40 m/s ) = 8.0 kg m/s.
9.2. Model: Model the bicycle and its rider as a particle. Also m

8.1.
Model: The model rocket and the target will be treated as particles. The kinematics equations in two dimensions apply. Visualize:
Solve:
For the rocket, Newtons second law along the y-direction is
( Fnet ) y = FR mg = maR
aR = 1 1 15 N ( 0.8 kg ) (

7.1.
Visualize:
Solve: (a) The weight lifter is holding the barbell in dynamic equilibrium as he stands up, so the net force on the barbell and on the weight lifter must be zero. The barbells have an upward contact force from the weight lifter and the gra

6.1.
Model: Visualize:
We can assume that the ring is a single massless particle in static equilibrium.
Solve:
Written in component form, Newtons first law is
( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 N ( Fnet ) y = Fy = T1 y + T2 y + T3 y = 0 N
Evaluating

5.1. Visualize:
Assess: walls.
Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse
5.2. Visualize:
5.3. Visualize:
5.4. Model: Assume friction is negligible compared to other forces.
Visualize:
5.5. Visu

4.1.
Solve:
(a)
(b) A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the 90 turn the car accelerates back up to 100 mph in the same time it took to slow down.
4.2.
Solve:
(a)
(b) A car drives up