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Textbook: Physics for Scientists & Engineers with Modern Physics Edition: 4rd Author: Douglas Giancoli Publisher: Pearson; ISBN-13: 978-0131495081

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Problem 1 a.) A 1000 kg car climbs a 5.0° slope at a constant velocity of 80.0 km/h. Assuming that air resistance may be neglected, at what rate must the engine deliver energy to the drive wheels of the car? Express your answer in kW. (10 pts) b.) Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0° and the coefficient of friction is 0.40. Find the length of a ramp that will stop a 15,000 kg truck that enters the ramp at 35 m/s (about 75 mph). (10 pts) Problem 2 This problem is related to the Air Track Collisions lab. Suppose two gliders of equal masses, m, move toward each other at equal velocities, v, as shown. One of the gliders has a massless spring attached to it with spring constant k so that the gliders bounce off each other. Assume there is no friction. a.) In terms of m, v and k, compute Δx, the maximum amount of compression that occurs in the spring. (4 pts) b.) In terms of m, v, and k, what is the maximum force experienced by glider 1 (magnitude and direction)? (4 pts) c.) What is the magnitude and direction of the impulse felt by glider 1? (4 pts) d.) Make an approximate plot of the force felt by glider 1 as a function of time. (Don’t try to come up with an equation for F(t) . There’s no need to put units on the time axis.) (4 pts) e.) Estimate the duration, Δt, of the impact. State any assumptions that you make. (4 pts) ____________________ _____________________ ________ 3 A flywheel of mass M = 150 kg and radius Problem R =2.0 m rotates about a vertical axis through its center of mass. The flywheel is in the shape of a hollow disk so its moment of inertia is I = MR2. a) The flywheel is initially rotating with an angular speed of 100 revolutions per minute (rpm). Find the initial kinetic energy of the flywheel. (4 pts.) b.) Find the angular momentum of the flywheel (magnitude and direction). (4 pts) c) The flywheel is stopped by applying a brake pad to the rim with force F. The coefficient of kinetic friction between the brake pad and the flywheel is µk = 0.25 and the wheel stops rotating after 10 s. Find the magnitude of the force F. (4 pts.) ____________________ _____________________ ________ d.) What is the torque (magnitude and direction) on the flywheel? (4 pts) e.) Find the work done by the force on the flywheel (make sure to indicate whether the work is positive or negative) (4 pts.) ____________________ _____________________ ________ Problem 4 Timbie and Chung want to determine whose car is more massive. They are advised by Physics 207 students to find out the answer by colliding the cars together. In experiment 1, Timbie’s car, with mass mT, is at rest when Chung’s car, mC crashes into it with initial velocity vi. After the impact the two cars remain locked together and move with velocity vf. In experiment 2, the cars are pulled apart from each other and collided again, this time with Chung’s car at rest and Timbie’s crashing into it with the same initial velocity, vi, as in experiment 1. The cars remain locked together after the impact and this time move with velocity vf′. a.) In experiment 1, what is the ratio of the final kinetic energy to the initial kinetic energy? Express your answer in terms of mT and mC. (7 pts) b.) Using the results of both experiments, what is the mass of mT in terms of mC, vf, and vf′? (7pts) ____________________ _____________________ ________ Now for a different problem: c.) A 1.6 kg particle moving along the x-axis experiences the net force shown in the figure. The particle's velocity is 4.0 m/s in the positive x direction at x = 0 m. What is its velocity at x = 2 m? (6 pts) ____________________ _____________________ ________ Problem 5 Suppose you want to compute the kinetic energy of a football thrown by Bret Favre. You measure the mass, m, of the football on a two-pan balance in the lab and determine that the mean mass is 0.40 kg with a standard deviation of 0.02 kg. You measure the speed, s, of the football (measured off your TV screen in instant replay mode) multiple times and record the values given in the table below: Measurement # 1 2 3 4 5 6 7 8 9 Speed (m/s) 36 36 35 37 35 37 34 38 36 Assume that each measurement is independent and distributed according to a Gaussian distribution. a.) What is your best estimate of the speed of the football and its standard deviation? (10 pts) b.) Based on these measurements, what is your best estimate of the kinetic energy of the ball and its standard deviation (i.e. the spread in the distribution of this derived quantity)? (10 pts) For full credit, make your work clear to the grader. Show the formulas you use, essential steps, and results with correct units and significant figures. Partial credit is available if your work is clear. Points shown in parenthesis. Use g = 9.80 m/s2 . For TF and MC questions, choose the best answer. 1. (2) T F A collision is considered elastic if it conserves total momentum. 2. (2) T F A collision where two objects stick together is called completely inelastic. 3. (2) T F The force during a collision between two objects will be larger if the time in contact is larger. 4. (2) T F When a 2-kg mass collides with a 5-kg mass, the masses experience equal but opposite impulses. 5. (8) Ralph throws a ball (mass = 145 grams) horizontally, his hand applying a force of 4.0 N horizontally over 0.22 s (ignore gravity here). Find a) (4) The impulse applied to the ball: b) (4) The speed of the ball when it left his hand: 6. (12) A ping-pong ball (mA , very light) is travelling at 24 m/s due east when it collides head-on with a basketball (mB , much much heavier) travelling at 12 m/s due west. Their collision is elastic. a) (4) Before they collide, at what speed are they approaching each other? b) (4) After the collision, at what speed are they separating from each other? c) (4) What is the velocity of the ping-pong ball (magnitude and direction) after the collision? 1 7. (12) A rocket far from earth moving at 92 m/s separates with an explosive charge, into a booster stage (1820 kg) and a capsule (680 kg), leaving the capsule moving at 28 m/s away from the booster. a) (2) After they separate, their center of mass is moving at a. vcm < 92 m/s b. vcm = 92 m/s c. vcm > 92 m/s. b) (2) Which experiences the greater magnitude force due to the explosive charge? a. the booster b. the capsule c. they experience equal magnitude forces. c) (2) Which experiences the greater magnitude impulse due to the explosive charge? a. the booster b. the capsule c. they experience equal magnitude impulses. d) (6) Find the speed of the capsule after they separate, using v20 = v10 + 28 m/s. 8. (16) An engine’s flywheel of radius 28.8 cm accelerates from rest to 2440 rpm in 3.00 s. a) (6) What is the flywheel’s angular acceleration, in rad/s2 ? b) (6) Through how many revolutions did it turn during the 3.0 seconds? c) (4) During the acceleration, what was the tangential acceleration of a point on the edge of the wheel? 2 Name: Rec. Instr.: 2 2 5 mR ), 1 2 2 mR ), Rec. Time: 2 9. (6) A solid sphere (I = a solid cylinder (I = and a hoop (I = mR ) of identical masses and radii are initially all rolling without slipping at 12 m/s on a level surface. Then they come to a gradual incline. a) (2) While on the level surface, which one has the greatest translational kinetic energy? a. sphere b. cylinder c. hoop d. all translational KE’s are equal. b) (2) While on the level surface, which one has the greatest rotational kinetic energy? a. sphere b. cylinder c. hoop d. all rotational KE’s are equal. c) (2) Which one goes the highest up the incline? a. sphere b. cylinder c. hoop d. it’s a 3-way tie. 10. (16) A light (nearly massless) rod has masses m1 = 5.0 kg and m2 = 2.0 kg attached to its ends as shown. Placed on a pivot horizontally, it is initially unbalanced, and starts to rotate. a) (6) How large is the rotational inertia of the rod with masses, about the pivot point as the axis? b) (6) How large is the net torque acting on the rod with masses, and is it clockwise or counterclockwise? c) (4) What is the magnitude of its initial instantaneous angular acceleration? 11. (8) In the diagram (not to scale), an unknown mass m2 is suspended in equilibrium from the end of a uniform 2.00-kg rod (m1 ) of 1.00 m length balanced on a pivot. How large is m2 ? 3 12. (18) The living quarters of a space station is a giant hoop (M = 24000 kg) of outer radius 38 m. To control its rotational speed, three movable masses (m = 18000 kg) are added, whose distance from the center can be changed. Initially, the three masses are set at r = 34 m, and the angular velocity ω is set so people at the outer edge feel 1.00 g of centripetal acceleration. a) (6) What is the initial angular velocity ω, in rpm? b) (8) To perform experiments, the three movable masses are moved to r0 = 4.0 m. What is the new angular velocity ω 0 (in rpm) of the space station? c) (4) After moving the masses to r0 = 4.0 m, how many g’s of centripetal acceleration are the people at the outer edge experiencing? 13. (8) A light fixture that weighs 60.0 N is suspended between ceiling and wall as shown, with θ = 30.0◦ . Find the tensions T1 and T2 in the two supporting cables. 4 Fluids, Waves, Sound Name 1. What force magnitude in newtons does air pressure of 1.00 atm produce on the upper surface of a rectangular 2.00 m × 3.00 m table? 2. One day at a fresh water lake, the pressure in the air is 104 kPa just above the surface of the water. The density of the air above is 1.20 kg/m3 . a) At increasing depth under the water, the pressure a. increases. b. decreases. c. does not change. b) At increasing altitude above the lake, the pressure a. increases. b. decreases. c. does not change. c) Calculate the pressure (in kPa) at a depth of 25.0 m under the water. 3. Iron has a specific gravity of 7.8 . Suppose you have a solid cube-shaped piece of iron and know that its mass is 56 kg. The volume of a cube with edge a is V = a3 . a) How large is the density of the iron, in kg/m3 ? b) What is the volume of the iron cube, in m3 ? 4. The magnitude of the buoyant force equals the weight of the object for a. an object that sinks. c. an object that floats. b. any object submerged partially or completely in a fluid. d. no object submerged to any extent in a fluid. 5. The magnitude of the buoyant force equals the weight of the fluid with the same volume as the object for a. an object completely submerged in a fluid. c. an object that floats. b. any object submerged partially or completely in a fluid. d. no object submerged to any extent in a fluid. 1 6. An object is floating on the surface of a liquid. Some compound is dissolved into the liquid, increasing its density. What happens? a. The object would float submerged less deeply. c. The object might sink to the bottom of the container. b. The object would float submerged more deeply. d. More than one of these outcomes is possible. 7. A 150000-kg undersea research chamber is spherical with an external diameter of 6.8 m. It is anchored to the sea bottom by a cable. The density of the sea water is 1030 kg/m3 . a) How large is the buoyant force FB on the chamber? b) The magnitude of tension in the cable is equal to a. b. c. d. e. the weight mg of the chamber. the buoyant force FB on the chamber. the sum of the weight mg and the buoyant force FB . the difference of weight minus buoyant force, mg − FB . the difference of buoyant force minus weight, FB − mg. 8. Atmospheric pressure of 101.3 kPa can be considered the force per unit area (p = F/A) due to the weight of air above us (per unit area!). What is the total mass in a column of air with a 1.00 cm2 cross-sectional area, extending up through the atmosphere? 9. T F If the length of a pendulum is doubled, its period will also be doubled. 10. T F If the mass of a pendulum is doubled, its period will not change. 11. T F The speed of an oscillating mass on a spring is greatest when passing the equilibrium point. 12. T F The acceleration of an oscillating mass on a spring is greatest when passing the equilibrium point. 13. In some old grandfather clocks, the pendulum has a length of 1.00 m. Suppose the mass is 250 grams. What is the period of the pendulum? 2 14. A spring is attached to the ceiling. It stretches by 3.20 cm when a force of 4.50 N pulls on the free end. a) How large is the spring constant k? b) With what frequency in hertz will a 1.20 kg mass connected to this spring oscillate? c) While oscillating, the mass moves up and down between points at 1.00 m and 1.48 m below the ceiling. How large is the amplitude of the oscillations? d) When the mass is at its lowest point, how large is its acceleration? e) How much total mechanical energy is in the oscillations? 15. The wave speed on a string is 425 m/s. Periodic waves are being generated on that string by an oscillator with a frequency of 20.4 kHz. a) What is the wavelength of these periodic waves on the string? b) The string has a mass per unit length of 8.80 × 10−4 kg/m. What tension is the string under? 3 16. A 78-dB sound wave strikes an eardrum whose area is 4.8 × 10−5 m2 . a) What sound intensity (in W/m2 ) does this sound level correspond to? b) How much sound energy is incident on the eardrum per second? 17. A 0.88 m long string on a musical instrument is vibrating in a two-loop pattern at a frequency of 440 Hz. a) What is the wave speed on this string? b) What is the frequency of the fundamental mode of vibration for this string? c) If the same string were made to vibrate in a five-loop pattern, what would be its frequency? 4 Prefixes a=10−18 , f=10−15 , p=10−12 , n=10−9 , µ = 10−6 , m=10−3 , c=10−2 , k=103 , M=106 , G=109 , T=1012 , P=1015 Physical Constants G = 6.67 × 10−11 N·m2 /kg2 (Gravitational constant) RE = 6380 km (mean radius of Earth) mp = 1.67 × 10−27 kg (proton mass) g = 9.80 m/s2 (gravitational acceleration) ME = 5.98 × 1024 kg (mass of Earth) me = 9.11 × 10−31 kg (electron mass) c = 299792458 m/s (speed of light) Units and Conversions 1 1 1 1 inch = 1 in = 2.54 cm (exactly) mile = 5280 ft m/s = 3.6 km/hour acre = 43560 ft2 = (1 mile)2 /640 1 1 1 1 foot = 1 ft = 12 in = 30.48 cm (exactly) mile = 1609.344 m = 1.609344 km ft/s = 0.6818 mile/hour hectare = 104 m2 Trig summary (opp) sin θ = (hyp) , sin θ = sin(180◦ − θ), (adj) cos θ = (hyp) , cos θ = cos(−θ), tan θ = (opp) (opp)2 + (adj)2 = (hyp)2 . (adj) , tan θ = tan(180◦ + θ), sin2 θ + cos2 θ = 1. Acceleration Equations v̄ = ā = ∆x ∆t , ∆v ∆t , ∆x = x − x0 , slope of x(t) curve = v(t). ∆v = v − v0 , slope of v(t) curve = a(t). For constant acceleration in one-dimension: v̄ = 12 (v0 + v), v = v0 + at, x = x0 + v0 t + 12 at2 , v 2 = v02 + 2a(x − x0 ). Vectors ~ or V, described by magnitude=V , direction=θ or by components (Vx , Vy ). Written V Vx =q V cos θ, Vy = V sin θ, V ~ to x-axis. V = V 2 + V 2 , tan θ = y . θ is the angle from V x y Vx Subtraction: A − B is A + (−B), Addition: A + B, head to tail. Newton’s Second Law: F~net = m~a, means ΣFx = max and ΣFy = may . −B is B reversed. P F~net = F~i , sum over all forces on a mass. Acceleration Equations Centripetal Acceleration: 2 aR = vr , towards the center of the circle. Circular motion: speed v = 2πr T = 2πrf , frequency f = Gravitation: 2 F = G mr1 m ; 2 g= GM r2 , 1 T , where T is the period of one revolution. where G = 6.67 × 10−11 Nm2 /kg2 ; Energy, Force, Power Work & Kinetic & Potential Energies: W = F d cos θ, KE = 12 mv 2 , PEgravity = mgy, Conservation or Transformation of Energy: Work-KE theorem: ∆KE = Wnet = work of all forces. Power: Pave = W t , or use Pave = energy time . PEspring = 12 kx2 . ~ θ = angle btwn F~ and d. General energy-conservation law: ∆KE + ∆PE = WNC = work of non-conservative forces. Linear Momentum Momentum & Impulse: momentum p~ = m~v , impulse ∆~ p = F~ave ∆t. Conservation of Momentum: (2-body collision): mA~vA + mB ~vB = mA v~0 A + mB v~0 B . Center of Mass: 2 x2 +... xcm = m1mx11 +m +m2 +... , vcm = m1 v1 +m2 v2 +... m1 +m2 +... . Rotational Motion Rotational coordinates: 1 rev = 2π radians = 360◦ , ω = 2πf , f= 1 T , Linear coordinates vs. rotation coordinates and radius: l = θr, v = ωr, atan = αr, aR = ω 2 r, Constant angular acceleration: ω = ω0 + αt, θ = θ0 + ω0 t + 12 αt2 , Torque & Dynamics: τ = rF sin θ, I = Σmr2 , Static Equilibrium: ΣFx = ΣFy = ΣFz = 0, ∆θ ∆t , ᾱ = ∆ω ∆t , ∆θ = ω̄∆t. (must use radians in these). ω̄ + 12 (ω0 + ω), τnet = Iα, Στ = 0, ω̄ = L = Iω, ω 2 = ω02 + 2α∆θ. ∆L = τnet ∆t, KErotation = 12 Iω 2 . τ = rF sinθ. Fluids Density: ρ = m/V , SG=ρ/ρH2 O , ρH2 O = 1000 kg/m3 = 1.00 g/cm3 (at 4◦ C). Static Fluids: P = F/A, P2 = P1 + ρgh, ∆P = ρgh, P = Patm. + PG , B = ρgV or FB = ρgV . Pressure Units: 1 Pa = 1 N/m2 , 1 bar = 105 Pa = 100 kPa, 1 mm-Hg = 133.3 Pa. 1.00 atm = 101.3 kPa = 1.013 bar = 760 torr = 760 mm-Hg =14.7 lb/in2 . Moving Fluids: A1 v1 = A2 v2 = a constant, P + 12 ρv 2 + ρgy = a constant. Oscillations and Waves Oscillators, frequency, period, etc.: p F = −kx = ma, f = 1/T , ω = 2πf = 2π/T , ω = k/m, Oscillator energy, speed, etc.: 2 E = 12 mv 2 + 12 kx2 = 12 kA2 = 12 mvmax , vmax = ωA. Waves: q FT λ = vT , v = f λ, v = m/L , I = P/A, I = P/4πr2 . Standing waves: node to node distance = λ/2. ω= p g/L. Equations: Sound Sound: In air, v ≈ (331 + 0.60 T ) m/s, T in ◦ C, v = 343 m/s at 20◦ C, d = vt. Sound Intensity, Level: I = P/A, I = P/4πr2 , β = (10 dB) log II0 , I = I0 10β/(10 dB) , I0 = 10−12 W/m2 .
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1. Pressure= force/area
Force= pressure*area
Pressure=1atm =101325 N/M2
Force=101325*2.00 m × 3.00= 607950 N
2. a) A
b) B
c) Pressure = g h = 1000*9.8* 25= 245000
3. a) specific gravity= density of iron/density of water (1000 kg/m3)
Density of iron= SG * 1000= 7.8* 1000=7800 kg/m3
b) Volume= mass/density = 56/7800 = 0.00718 m3
4. C. an object that floats.
5. A. an object completely submerged in a fluid
6. A. The object would float submerged less deeply
7. a) Fb = ρ g V
V= 4/3πr3, r=6.8/2=3.4
V= 164.64m3
Fb= 1030*9.8*164.64= 1661876.16 N
b) e. the difference of buoyant force minus weight, FB − mg.
8. 101.3 k Pa *1000= 101300 N/m2
F=P*A= 101300 N/m2 *1cm2/10000= 10.13 N
F= mass*g
Mass= F/g =10.13/9....


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