sio20
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sio20

Course Number: SIO SIO 20, Winter 2008

College/University: UCSD

Word Count: 1626

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SIO 20 - Midterm Examination 1 v1 Winter 2009 The correct answers are highlighted below. Note that there were four versions of the exam. Each version had the same questions, just in a different order. Circle the letter corresponding to the best answer. 1. Over the last 50 years, the atmospheric concentration of carbon dioxide (CO2) a. decreased b. increased c. remained constant 2. The cooling of the ground to...

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20 SIO - Midterm Examination 1 v1 Winter 2009 The correct answers are highlighted below. Note that there were four versions of the exam. Each version had the same questions, just in a different order. Circle the letter corresponding to the best answer. 1. Over the last 50 years, the atmospheric concentration of carbon dioxide (CO2) a. decreased b. increased c. remained constant 2. The cooling of the ground to produce dew is mainly the result of: a. convection b. radiational cooling c. advection mixing d. 3. Radiation fog forms best: a. during the day when the surface is warming b. during the night when the surface is cooling c. when the air near the surface is initially very dry d. when clouds emit infrared radiation towards the surface 4. When fog burns off it: a. evaporates from the ground up b. dissipates by lifting away from the ground c. settles to the ground in the form of rain or drizzle 5. Low clouds slow surface cooling at night compared to clear skies because: a.

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STATICS AND MECHANICS OF MATERIALS, 2nd EditionRILEY, STURGES AND MORRISChapter 11-1 Calculate the mass m of a body that weighs 600 lb at the surface of the earth. SOLUTIONm1-2W g600 18.63 slug . Ans. 32.2Calculate the weight W of a bod
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ELECTRIC CHARGES AND FORCES25.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other han
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THEELECTRIC FIELD26.1. Model: The electric field is that of the two charges placed on the y-axis.Visualize: Please refer to Figure Ex26.1. We denote the upper charge by q1 and the lower charge by q2. Becauseboth the charges are positive, their el
Stevens - PEP - 112
GAUSS'S LAW27.1. V i a l i e :As discussed in Section 27.1, 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 co
Stevens - PEP - 112
CURRENT AND CONDUCTIVITY0 28.1. Solve: The wire's cross-sectional area is A = mz = ~ ( 1 . x 10" m)' = 3.1415 x lo4 m2, and the electron current through this wire is 2.0 X loi9 s-l . Using Table 28.1 for the electron density of iron and Equation 28
Stevens - PEP - 112
THEELECTRIC POTENTIAL29.1. Model: The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniformelectric field. Visualize:After Before*. .,v=oE`*II01.ox2.0The figure shows the before-and-afte
Stevens - PEP - 112
POTENTIAL AND FIELD30.1. Solve: The potential difference AV between two points in space is9AV = V(xf) - V(x,) = - I E , dxx,where x is the position along a line from point i to point f. When the electric field is uniform,xrAV = - E x j d .
Stevens - PEP - 112
FUNDAMENTALS OF CIRCUITS31.1. Solve: From Table 30.1, the resistivity of carbon is p = 3.5 xof lead from a mechanical pencil isR m. From Equation 31.3, the resistance= 5.5 Rp~ p~ R=-=-= A m '(3.5 x lo-'R m)(0.06 m)n(0.35~10-'m)'31.2.
Stevens - PEP - 112
THE MAGNETIC FIELD32.1. Model: A magnetic field is caused by an electric current.Visualize: Please refer to Figure Ex32.1. Solve: Because the north poles of the magnets point counterclockwise, the magnetic force is counterclockwise. When you point
Stevens - PEP - 112
ELECTROMAGNETIC INDUCTION33.1. Model: Assume the magnetic field is uniform.Visualize: Please refer to Figure Ex33.1. Since a motional emf was developed the field must be perpendicular to V .The positive charges experienced a magnetic force to the
Stevens - PEP - 112
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 "net pole" within a surface, Q, = f B . d i = 0 . Since the magnetic field isunif
Stevens - PEP - 112
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" = 150 x -= 2.618 rad180"w=2*618ra
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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
Stevens - PEP - 112
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