MAE162D - Winter2017
This homework assignment will cover the following secion.
Using arrays, clusters, and local variables
This is an individual assignment.
this is an INDIVIDUAL assignment
please scan handwritten por
CAPACITANCE AND DIELECTRICS
Q Vab SET UP: 1 F = 10 -6 F EXECUTE: Q = CVab = (7.28 10-6 F)(25.0 V) = 1.82 10-4 C = 182 C EVALUATE: One plate has charge +Q and the other has charge -Q . PA Q IDENTIFY and SET UP: C = 0 , C = and V = Ed .
ra = 0.150 m rb = (0.250 m) 2 + (0.250 m) 2 rb = 0.3536 m
IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). SET UP: Let the initial position of q2 b
^ E = E cos dA, where is the angle between the normal to the sheet n and the
(a) IDENTIFY and SET UP:
electric field E . EXECUTE: In this problem E and cos are constant over the surface so (b) EVALUATE:
E = E cos dA = E cos A = (14 N
ELECTRIC CHARGE AND ELECTRIC FIELD
(a) IDENTIFY and SET UP: Use the charge of one electron ( -1.602 10-19 C) to find the number of electrons required to produce the net charge. EXECUTE: The number of excess electrons needed to produce net charge
THE SECOND LAW OF THERMODYNAMICS
IDENTIFY: SET UP:
For a heat engine, W = QH - QC . e =
W . QH > 0, QC < 0. QH
W = 2200 J. QC = 4300 J.
EXECUTE: (a) QH = W + QC = 6500 J.
2200 J = 0.34 = 34%. 6500 J EVALUATE: Since the engine operates on a cycle,
THE FIRST LAW OF THERMODYNAMICS
The pV-diagram is sketched in Figure 19.1
(a) IDENTIFY and SET UP: The pressure is constant and the volume increases.
Figure 19.1 (b) W =
Since p is constant, W = p dV = p(V2 - V1 ) The problem gi
THERMAL PROPERTIES OF MATTER
(a) IDENTIFY: We are asked about a single state of the system. SET UP: Use Eq.(18.2) to calculate the number of moles and then apply the ideal-gas equation. m 0.225 kg EXECUTE: n = tot = = 56.2 mol M 4.00 10-3 kg/mol
SOUND AND HEARING
IDENTIFY and SET UP: Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. EXECUTE: (a) = v / f = (344 m/s)/1000 Hz = 0.344 m (b) pmax = BkA and Bk is constan
IDENTIFY: v = f . T = 1/ f is the time for one complete vibration. SET UP: The frequency of the note one octave higher is 1568 Hz. v 344 m/s 1 EXECUTE: (a) = = = 0.439 m . T = = 1.28 ms . f 784 Hz f
v 344 m/s = = 0.219 m .
IDENTIFY: Use Eq.(14.1) to calculate the mass and then use w = mg to calculate the weight. SET UP: EXECUTE:
= m / V so m = V From Table 14.1, = 7.8 103 kg/m3 .
For a cylinder of length L and radius R, V = ( R 2 ) L = (0.01425 m)
IDENTIFY and SET UP: The target variables are the period T and angular frequency . We are given the frequency f, so we can find these using Eqs.(13.1) and (13.2) EXECUTE: (a) f = 220 Hz
T = 1/f = 1/220 Hz = 4.54 10-3 s = 2 f = 2 (
Use the law of gravitation, Eq.(12.1), to determine Fg .
IDENTIFY and SET UP: EXECUTE:
FS on MG
mm mSmM (S = sun, M = moon); FE on M = G E2 M (E = earth) 2 rSM r EM
2 FS on M mSmM r EM mS rEM = G 2 = FE on M rSM GmE mM mE rSM rEM ,
EQUILIBRIUM AND ELASTICITY
IDENTIFY: Use Eq.(11.3) to calculate xcm . The center of gravity of the bar is at its center and it can be treated as a point mass at that point. SET UP: Use coordinates with the origin at the left end of the bar and th
DYNAMICS OF ROTATIONAL MOTION
EXECUTE: = Fl l = r sin = (4.00 m)sin 90 l = 4.00 m = (10.0 N)(4.00 m) = 40.0 N m
IDENTIFY: Use Eq.(10.2) to calculate the magnitude of the torque and use the right-hand rule illustrated in Fig.(10.4) to calculate th
ROTATION OF RIGID BODIES
IDENTIFY: s = r , with in radians. SET UP: rad = 180 . s 1.50 m = 0.600 rad = 34.4 EXECUTE: (a) = = r 2.50 m s 14.0 cm (b) r = = = 6.27 cm (128)( rad /180) (c) s = r = (1.50 m)(0.700 rad) = 1.05 m EVALUATE: An angle is
CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE
IDENTIFY: I = Q / t . SET UP: 1.0 h = 3600 s EXECUTE: Q = It = (3.6 A)(3.0)(3600 s) = 3.89 104 C. EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amount of char
IDENTIFY: The newly-formed wire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance R/3. The circle is two R/6 resistors in parallel. EXECUTE: The resis
MAGNETIC FIELD AND MAGNETIC FORCES
IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. ^ j EXECUTE: v = ( +4.19 10 4 m/s ) i + ( -3.85 10 4 m/s ) ^
^ (a) B = (1.40 T ) i ^ ^ F = qv B = (
(a) (b) (c)
28 14 85 37
Si has 14 protons and 14 neutrons. Rb has 37 protons and 48 neutrons. Tl has 81 protons and 124 neutrons.
(a) Using R = (1.2 fm)A1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. (b) Usin
L = l (l + 1) . Lz = ml . l = 0, 1, 2,., n - 1. ml = 0, 1, 2,., l . cos = Lz / L .
IDENTIFY and SET UP:
EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2 , Lz = ,0, - . l = 2 : L = 6 , Lz = 2 , ,0, - , -2 . (b) In each case cos
n2h 2 . 8mL2
IDENTIFY and SET UP: The energy levels for a particle in a box are given by En = EXECUTE: (a) The lowest level is for n = 1, and E1 =
(1)(6.626 10-34 J s) 2 = 1.2 10-67 J. 8(0.20 kg)(1.5 m) 2
1 2E 2(1.2 10-67 J) (b)
THE WAVE NATURE OF PARTICLES
IDENTIFY and SET UP: EXECUTE: (a) =
h h = . For an electron, m = 9.11 10 -31 kg . For a proton, m = 1.67 10 -27 kg . p mv
6.63 10-34 J s = 1.55 10-10 m = 0.155 nm (9.11 10-31 kg)(4.70 106 m/s)
m 9.11 10 -31 kg 1
PHOTONS, ELECTRONS, AND ATOMS
h f - . The e e
IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 = slope of V0 versus f is h/e. The value fth of f when V0 = 0 is related to by = hf th .
EXECUTE: (a) From
IDENTIFY and SET UP: Consider the distance A to O and B to O as observed by an observer on the ground (Figure 37.1).
(b) d = vt = (0.900) (3.00 108 m s) (5.05 10-6 s) = 1.36 103 m = 1.36 km. 37.3.
1 IDENTIFY and SET UP: The
IDENTIFY: Use y = x tan to calculate the angular position of the first minimum. The minima are located by m , m = 1, 2,. First minimum means m = 1 and sin 1 = / a and = a sin 1. Use this Eq.(36.2): sin = a equation to calculate . SET
IDENTIFY: Compare the path difference to the wavelength. SET UP: The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. EXECUTE: (a) For constructive interfer
y = 4.85 cm
IDENTIFY and SET UP: Plane mirror: s = - s (Eq.34.1) and m = y / y = - s / s = +1 (Eq.34.2). We are given s and y and are asked to find s and y. EXECUTE: The object and image are shown in Figure 34.1. s = - s = -39.2