Fall Semester Review - Physics [Regular]



Name:________________________ Period:___ Fall Semester Review – Regular Physics 2012USE YOUR FORMULA CHART!!!!!!!!!!!!Multiple ChoiceIdentify the letter of the choice that best completes the statement or answers the question._D___1.If some measurements agree closely with each other but differ widely from the actual value, these measurements area.neither precise nor accurate.b.accurate but not precise.c.acceptable as a new standard of accuracy.d.precise but not accurate.__C__2.These values were obtained as the mass of a bar of metal: 8.83 g; 8.84 g; 8.82 g. The known mass is 10.68 g. The values area.accurate.b.both accurate and precise.c.precise.d.neither accurate nor precise.__A__3.An object is described to have 0.0 m/s at rest. Which of these terms does that describe? a. zero speedb. free fall accelerationc. zero accelerationd. zero net force____7.__C___ 4. What does the graph above illustrate about velocity and acceleration?a.The velocity is positive: the acceleration is positive.b.The velocity is constant: the acceleration is constant.c.The velocity is positive: the acceleration is zero.d.There is not enough information to answer.___A__ 5. Which of the following statements best describes the motion of this object. a. The object is moving at a constant velocity. b. The object is changing direction. c. The object is speeding up. d. The object is slowing down. __C__ 6. The time rate changing velocity is known as what? a. velocity b. force c. acceleration d. displacement__A__ 7.The slope of a position versus time graph represents ? a.Velocityb.forcec.Accelerationd.Displacement1386840116205000__D__8.Find the time it takes for a car with initial velocity of 10 m/s, to acceleration to 5 m/s2 and reach a speed of 25 m/s? a.0 sb.2 sc.1 sd.3 sDiagram: (Given + Unknowns)Horizontala = 5 m/s2vf = 25 m/s vi = 10 m/s t = ???? s Δd = ??? m Equation: a = (vf – vi) / t Substitute: t = (vf – vi) t = (25 – 10) = 15 = 3 s / a 5 5 Solve: t = 3 s __D__9.What is the displacement of a car that moves from x = +5.0 m to x = -10.0 m in 3 seconds? (a = 0)a.+10 metersb.-5 meters c.-5 meters d.-15 meters ____10.__A____ 10. What does the graph above illustrate about acceleration?a.The acceleration varies.b.The acceleration is zero.c.The acceleration is constant.d.The acceleration increases then becomes constant.__C__11.A curious kitten pushes a ball of yarn with its nose. It displaces the ball of yarn, from rest, a displacement of 17.5 cm in 2.00 s. What is the acceleration of the ball of yarn? a.0.11 m/s2b.0.144 m/s2c.0.0875 m/s2d.0.0438 m/s2Diagram: (Given + Unknowns)Horizontala = ????? m/s2vf = ????? m/s vi = 0 m/s t = 2 s Δd = 17.5 cm = 0.175 m Equation: Δd = (vi)t + (1/2) a (t)2Substitute:0.175 = (0 m/s)(2s) + (1/2)(a) (2 s)2 0.175 = 0 + 2a 0.175 = 2a0.0875 m/s2 = aSolve: 0.0875 m/s2 = a__D__12.A race car accelerates from 5.0 m/s to 35.0 m/s with an acceleration 3 m/s2. What is the vehicle's displacement?a.10 mb.150 m c. 87.5 md. 200 mDiagram: (Given + Unknowns)Horizontala = 3 m/s2vi = 5 m/s vf = 35 m/s t = ??? s Δd = ???? m Equation: a = (vf2 – vi2) / 2ΔdSubstitute:a = (vf2 – vi2) Δd = (vf2 – vi2) Δd = (352 – 52) 2Δd 2a 2(3)Δd = 1200 = 200 m 6 Solve: Δd = 200 m __C__13.A car traveling at 3 m/s accelerates uniformly for 6.0 s in order to pass a snow plow. The car travels 108 m during the 6.0 s interval that it was accelerating. What is the car’s speed at the end of the acceleration period. a. 27 m/sb. 30 m/sc. 33 m/sd. 42 m/sDiagram: (Given + Unknowns)Horizontala = ????? m/s2vf = ????? m/s vi = 3 m/s t = 6.0 s Δd = 108 m Equation: Δd = (vi)t + (1/2) a (t)2 a = (vf – vi) / tSubstitute:108 = (3 m/s)(6s) + (1/2)(a) (6 s)2 108 = 18 + 18a 90 = 18a5 m/s2 = aa = (vf – vi) 5 = (vf – 3) t 630 = (vf – 3)33 m/s = vf Solve: vf = 33 m/s__D__14.The value 9.8 m/s2 would represent which of the followinga.free fall velocityb.acceleration due to gravity c.free fall acceleration d.Both B and C_B___15.A 0.180 kg arrow is shot vertically straight up with an initial velocity of 12 m/s. What is the maximum height it will go?a. 5.98 mb. 7.34 mc. 11.96 md. 21.94 mDiagram: (Given + Unknowns)Verticala = -9.8 m/s2vf = 0 m/s vi = 12 m/s t = ???? s Δd = ???? m Equation: a = (vf2 – vi2) / 2Δd Substitute: a = (vf2 – vi2) Δd = (vf2 – vi2) Δd = (02 – 122) 2Δd 2a 2(-9.8)? dy = -144/-19.6 ? dy = 7.34 mSolve: ? dy = 7.34 m__B__16.A 0.180 kg arrow is shot vertically straight up with an initial velocity of 12 m/s. How long does it take to reach it maximum height?a. 0.81 sb. 1.22 sc. 2.16 sd. 3.22 sDiagram: (Given + Unknowns)Verticala = -9.8 m/s2vf = 0 m/s vi = 12 m/s t = ???? sΔd = ???? m Equation: a = (vf – vi) / t Substitute: t = (vf – vi) t = (0 – 12) = -12 = 1.22 s / a -9.8 -9.8 t = 1.22 s Solve: t = 1.22 s _C___17.After winning the Super Bowl, Aaron Rodgers throws his helmet vertically upward and catches it in the same spot as returns to his hand. At the top of its path, the helmet is experiencing a.nonzero velocity and nonzero acceleration b.nonzero velocity and zero acceleration c.zero velocity and nonzero acceleration d.zero velocity and zero acceleration__A__18.When there is no air resistance, objects of different masses fall _____________.a.with equal accelerations b.with different accelerations c.according to massd.sideways__B___ 19.A boy on a wagon travels 6.0 m west and 8.0 m south. The magnitude of the resultant (vector sum ) of the displacement would be?a. 0 mb. 10 m c. 12 md. 14 mDiagram: (Given + Unknowns)Horizontal Vertical Δdx = 6.0 m Δdy = 8.0 mCan’t directly add horizontal with vertical. Equation: a2 + b2 = c2Substitute: (6 m)2 + (8 m)2 = c2 36 + 64 = c2 100 = c2 10 m = cSolve: Δdnet = 10 m__C__20.Which of the following is a physical quantity that has a magnitude but no direction?a.Vectorb.resultantc.Scalard.frame of reference_A___21.Which of the following is a physical quantity that has both magnitude and direction?a.Vectorb.resultantc.Scalard.frame of reference__A__22.A lightning bug flies at a velocity of 0.25 m/s due east toward another lightning bug seen off in the distance. A light easterly breeze blows on the bug at a velocity of 0.25 m/s. What is the resultant velocity of the lightning bug?a.0.50 m/s eastb.0.75 m/s eastc.0.00 m/s eastd.0.25 m/s east_D___23.A quarterback takes the ball from the line of scrimmage and runs backward for 10 m then sideways parallel to the line of scrimmage for 15 m. The ball is thrown forward 50 m perpendicular to the line of scrimmage. The receiver is tackled immediately. What is the total distance that ball traveled?a.0 mb.50 mc. 25 md.75 m_B___24.What is the trajectory of a projectile?a.a wavy lineb.a parabolac.a hyperbolad.projectiles do not follow a predictable path.__D_____ 25.A lead cylinder is dropped from a cliff. How far did it fall in 3.0 s? 9.8 mb. 29.4 mc. 14.7 m d. 44.1 mDiagram: (Given + Unknowns) Vertical a = - 9.8 m/s2vf = ???? m/s vi = 0 m/s t = 3.0 s Δdy = ????? m Equation: Vertical ? dy = vi t + (1/2)at2 Substitute ? dy = vi t + (1/2)at2 ?dy = (0 m/s)(3.0s) + (1/2)(-9.8)(3.0)2 ?dy = 0 + - 44.1 ?dy = - 44.1 mSolve: ?dy = - 44.1 m (answer choice should be negative)_C___26.A 70.0 kg lead projectile is shot horizontally at 40 m/sec from the top of a 50 m high tower. How far from the base of the tower does it hit the ground?a. 3.19 mb. 86.4 mc. 127.6 md. 200.0 mDiagram: (Given + Unknowns)Horizontal Verticala = 0 m/s2 a = - 9.8 m/s2vf = 40 m/s vf = ??? m/s vi = 40 m/s vi = 0 m/st = ??? s t = ???? sΔdx = ????? m Δdy = 50 m Mass is not needed for this problem.Ultimately, looking for Δdx = ???? m Equation: Vertical: ? dy = vi t +(1/2)at2 Horizontal ? dx =vi t+(1/2)at2 209584697559Substitute: (Vertical) (Horizontal)? dy = vi t + (1/2)at2 ? dx =vi t+(1/2)at2 50 m = (0 )(t) + (1/2)(-9.8)(t)2 ? dx = (40)(3.19)+(1/2)(0)(3.19)2 50 m = 0 + -4.9 t2 ? dx = 127.6 + 0 50 m = -4.9t2 ? dx = 127.6 m-10.2 = t23.19 s = tSolve: ? dx = 127.6 m_B___27.A stone is thrown horizontally from the top edge of a cliff with an initial speed of 12 m/s. A stopwatch measures the stone’s trajectory time from the top of the cliff to the bottom at 5.6 s. What is the height of the cliff? (Disregard air resistance. )a.58 mc.120 mb.154 md.180 mDiagram: (Given + Unknowns)Horizontal Verticala = 0 m/s2 a = - 9.8 m/s2vf = 12 m/s vf = ??? m/s vi = 12 m/s vi = 0 m/st = 5.6 s t = 5.6 sΔdx = ????? m Δdy = ???? m Looking for Δdy = ???? m Equation: Vertical ? dy = vi t + (1/2)at2 Substitute: ? dy = vi t + (1/2)at2 ?dy = (0 m/s)(5.6s) + (1/2)(-9.8)(5.6)2 ?dy = 0 + -154 ?dy = -154 mSolve: ?dy = 154 m (height is scalar, must be positive)__B__28.Which of the following is the cause of an acceleration?a.Speedb.Forcec.Inertiad.Velocity__A__29.What is the correct name for speeding up, slowing down, or changing direction?a.Accelerationb.Inertiac.Velocityd.Force____31.___D__ 30. In the free-body diagram shown, which of the following is the gravitational force acting on the car? a. 775 N b. 5800 N c. 13,690 N d. 14,700 N _____ 31. ___B__ 31. In the free-body diagram shown, what is the net force in the x –direction? a. 1,010 N b. 5,025 N (left) c. 6,575 N d. 20,390 N__A___ 32. In the free body diagram shown, what is the net for in the y-direction? a. 1,010 N (down) b. 5,025 N c. 6,575 N d. 20,390 N__A__33.A go-cart with a weight of 500 N is accelerated across a flat surface at 0.7 m/s2. How much net force acts on the go-cart?a. 36 Nb. 73 N c. 350 Nd. 714 NDiagram: (Given + Unknowns)(Horizontal) (Vertical) Fa = ???? N Fg = 500 N m = ??? kg m = ???? kg a = 0.7 m/s2 a = 9.8 m/s2Equation: (Vertical) Fg= ma (Horizontal) Fa = maSubstitute:(Vertical)Fg= ma 500 N = m (9.8 m/s2) m = (500 N)/ (9.8m/s2) m = 51.0 kg (Horizontal) -------------------------------------------------------------Fa = ma F = (51.0 kg)(0.7m/s2) Fa= 35.7 NSolve: Fa= 35.7 N (closest to 36 N)__C__34.Which of the following is the tendency of an object to maintain its state of motion?a.Accelerationb.Forcec.Inertiad.Velocity__B__35.If a nonzero net force is acting on an object, then the object is definitely _________.a.at restb.Acceleratingc.moving with a constant velocityd.losing mass__C__36.A wagon with a weight of 300.0 N is accelerated across a level surface at 0.5 m/s2. What net force acts on the wagon? a.9.0 Nb.150 Nc.15 Nd.610 NDiagram: (Given + Unknowns)(Horizontal) (Vertical) Fa = ???? N Fg = 300 N m = ??? kg m = ???? kg a = 0.5 m/s2 a = 9.8 m/s2Equation: (vertical) Fg= ma (horizontal) Fa = maSubstitute: (Vertical)Fg= ma 300 N = m (9.8m/s2) m = (300 N)/ (9.8m/s2) m = 30.58 kg (Horizontal) -----------------------------------------------------------------Fa = ma F = (30.58 kg)(0.5m/s2) Fa= 15.29 NSolve: Fa= 15.29 N (closest to 15 N)_B___37.According to Newton’s second law, F=ma, when the same force is applied to two objects of different masses,a.the object with greater mass will experience a great acceleration and the object with less mass will experience an even greater acceleration.b.the object with greater mass will experience a smaller acceleration and the object with less mass will experience a greater acceleration.c.the object with greater mass will experience a greater acceleration and the object with less mass will experience a smaller acceleration.d.the object with greater mass will experience a small acceleration and the object with less mass will experience an even smaller acceleration._D___38.According to Newton’s 1st Law of motion, a moving object acted on by a net force of zero, __________.Remains at restb. increases its inertiac. Decelerates at a constant rated. move at a constant rate__A__39.A hockey stick hits a puck on the ice. Identify an action-reaction pair.a.The stick exerts a force on the puck; the puck exerts a force on the stick.b.The stick exerts a force on the puck; the puck exerts a force on the ice.c.The puck exerts a force on the stick; the stick exerts a force on the ice.d.The stick exerts a force on the ice; the ice exerts a force on the puck._B___40.The statement by Newton that for every action there is an equal but opposite reaction is which of his laws of motion?a.firstb.thirdc.secondd.fourth_B___41. What is the net force on the object pictured to the right ? a. 5 N to the left b. 10 N the right c. 35 N to the right d. 60 N to the right__D__42. What is the net acceleration on this object pictured to the right ? a. 0 m/s2 b. 5 m/s2 the right c. 10 m/s2 to the right d. 20 m/s2 to the right79438515494000817880683260001899494410106191974140966415 N0015 N1896110294005008318528702000291979963825 N0025 N18617962603520 N0020 N10311122326110.5 kg000.5 kg__C__43.The magnitude of the force of gravity acting on an object is called?a. frictional forceb. inertia c. weightd. mass__A__44.A sled weighing 1.0 x 102 N sits on level ground. What is the normal force of the slope acting on the sled? (Draw a FBD!!!!!)a.100 Nb.0.098 Nc.10 Nd.9.8 N__B__45.There are six books in a stack, and each book weighs 5 N. A constant force of 2.0 N causes the stack to move with constant velocity. What is the frictional force acting on the books?a.30 Nb.2 Nc.5 Nd.32 N__C__46.The moon revolves around the earth once every year. Which object experiences a great gravitational force?a. the moonb. the earthc. they are the samed. not enough information to tell __C__47.A set of physics textbooks weighs 294 N on earth, what would they weigh on a planet that has ? the mass and ? the radius of earth?a. 147 Nb. 294 Nc. 588 Nd. 1176 N_D___48.After a cannon ball is fired into frictionless space, the amount of force needed to keep it moving equals ____.a.twice the force with which is was fireb.the same amount of force with which it was firedc.one half the force with which it was firedd.zero, since no force is necessary to keep it moving_E___49.Equilibrium occurs when ____.a.all the forces acting on an object are balancedb.the sum of the force acting right equal the sum of the forces acting leftc.the sum of the force acting up equal the sum of the forces acting downd.the net force is zeroe.all of the above__B__50.A force does work on an object if the force _________.a.is perpendicular to the displacement of the objectb.is parallel to the displacement of the objectc.perpendicular to the displacement of the object moves the object along a path that returns the object to its starting position__D__51.Which of the following energy forms is the sum of kinetic energy and all forms of potential energy?a. total energyb. summative energyc. non mechanical energy d. mechanical energy ___B_52.The more powerful the motor is, the __________.a. longer the time interval for doing the work isb. shorter the time interval for doing the work isc. greater the ability to do the work isd. shorter the workload is_B___53.A horizontal force of 200 N is applied to move a 55 kg television set across a 10 m level surface. What is the work done by the 200 N force on the television set?a.4000 Jb.2000 Jc.5000 Jd.6000 JDiagram: (Given + Unknowns) F = 200 N m = 55 kg d = 10 mW = ???? JMass is not needed for this problem.Equation: W = f(d)Substitute: W = 200 N (10 m) W = 2000 JSolve: W = 2000 J_C___54.Which of the following energy forms is involved as a pencil falls from a desk?a.kinetic energy onlyb.non-mechanical energyc.gravitational potential energy and kinetic energyd.elastic potential energy and kinetic energy695502687200at 4 m69518826999239 kg00239 kgPEi = 9,369 J PE =mghKEi = 0 J377198109124-7409410750702202881333504.0 m004.0 m 1 m at 1m PE = 2,342 J KE = 7,026 J842734443762 at 0 m PEf = 0 J KEf = 9,369 J KE = 1/2mv2Use the following scenario for the following questions. The picture to the right shows a 239 kg barrel of toxic waste sitting on the ledge of 4.0 m high shelf.__A__55. What is the kinetic energy of the barrel as it sits on the top shelf? a. 0 J b. 2,342 J c. 7,026 J d. 9,369 J __D__56. What is the potential energy of the barrel as it sits on the top shelf? a. 0 J b. 2,342 J c. 7,026 J d. 9,369 J __D__57. What is the mechanical energy of the barrel as it sits on the top shelf? a. 0 J b. 2,342 J c. 7,026 J d. 9,369 J __D__ 58. When the barrel falls, what is the kinetic energy right before it hits the ground? a. 0 J b. 2,342 J c. 7,026 J d. 9,369 J __2342 J__ 59. What is the potential energy of the falling barrel if the height is 1.0 m above the ground? a. 0 J b. 2,342 J c. 7,026 J d. 9,369 J _7026 J___ 60. What is the kinetic energy of the falling barrel if the height is 1.0 m above the ground? a. 0 J b. 2,342 J c. 7,026 J d. 9,369 J _7.7 m/s___61. What is the velocity of the falling barrel if the height is 1.0 m above the ground? a. 0 m/s b. 4.0 m/s c. 7.7 m/s d. 8.9 m/s __D__62.A 3.00 kg toy falls from a height of 10.0 m. Just before it hits the ground, what is its kinetic energy? a.98.0 Jb.29.4 Jc.0.98 Jd.294 JDiagram: (Given + Unknowns)m = 3.00 kgg =9.8 m/s2 h = 10 mtop PE = mgh KE = 0 Jbottom PE = 0 J KE = ? mv2Equation: KEi + PEi = KEf + PEfSubstitute: KEi + PEi = KEf + PEf O J + mgh = KEf + O J mgh = KEf (3.00kg )(9.8 m/s2)(10m) = KEf 294 J = KEf Solve: KE = 294 J__B__63.What is the kinetic energy of a 0.135 kg baseball thrown at 40.0 m/s?a.54.0 Jb.108 Jc.87.0 Jd.216 JDiagram: (Given + Unknowns)m = .135 kg v = 40 m/sKE = ???? Jgy Equation: KE = (1/2)mv2Substitute: KE = (1/2)(.135kg)(40m/s)2Solve: KE = 108 J_D___64.Old Faithful geyser in Yellowstone National Park shoots water every hour to a height of 40.0 m. With what velocity does the water leave the ground? (Disregard air resistance.) a.7.00 m/sb.19.8 m/sc.14.0 m/sd.28.0 m/sDiagram: (Given + Unknowns)g =9.8 m/s2 h = 40 mKEi=?KEf = 0top PE = mgh KE = 0 Jbottom PE = 0 J KE = ?mv2Equation: KEi + PEi = KEf + PEfSubstitute: KEi + PEi = KEf + PEf ? mv2 + 0 = O J + mgh ? mv2 = mgh m cancels ? v2 = gh ? v2 = (9.8 m/s2)(40m) v2 = 784 v = 28 m/s Solve: 28 m/s = v__D__65.An ice skater (mass of 75 kg), initially moving at 10.0 m/s, coasts to a halt in 1.0 x?102 m on a smooth ice surface. Find the force of friction acting on the skater using the work-kinetic energy theorem.a.0.50 Nb.37.5 Nc.-0.50 Nd.-37.5 NDiagram: (Given + Unknowns)m = 75 kg v =10 m/s d = 100 mforce F = ??? N use the work-kinetic energy theoremEquation: ?KE = work (1/2)mv2 = fdSubstitute: -(1/2)(75kg)(10 m/s)2 = F (100m) -3750 = F (100 m)-37.5 N = FSolve: -37.5 N = Fsince the object is slowing down._C___66.What is the average power supplied by a 60.0 kg student running up a flight of stairs rising vertically 4.0 m in 4.2 s?a.380 Wb.610 Wc.560 Wd.670 WDiagram: (Given + Unknowns)m = 60.0 kg F = ??? Na =9.8 m/s2 d = 4 m W = ??? Jt=4.2 s P = ???? WattEquation: F = ma W = fd P = W/tSubstitute: F = ma F = 60.0 kg (9.8m/s2) = 588 N W = f(d) W= 588 N (4 m) = 2352 J P = W /t P= 2352 J / 4.2 s = 560 Watt Solve: P = 560 Watt__C__67.What is the average power output of a weight lifter who can lift 250 kg in 2.0 m in 2.0 s?a.5000 Wb.4900 Wc.2500 Wd.9800 WDiagram: (Given + Unknowns)m = 250 kg F = ??? Na =9.8m/s2 d = 2 m W = ??? Jt=2.0 s P = ???? WattEquation: F = ma W = fd P = W/tSubstitute: F = ma F = 250 kg (9.8m/s2) = 2450 N W = f(d) W= 2450N (2 m) = 4900 J P = W /t P= 4900 J / 2 s = 2450 Watt Solve: P = 2450 Watts closest to 2500 Watts __D__68.A roller coaster loaded with passengers has a mass of 2.0 x 103 kg; the radius of curvature of the track at the lowest point of the track is 24 m. If the vehicle has a tangential speed of 18 m/s at this point, what force is exerted on the vehicle by the track?a.2.3 x 104 Nb.3.0 x 104 Nc.4.7 x 104 Nd.2.7 x 104 NDiagram: (Given + Unknowns)CircleFc = ???? Nm = 2000 kg ac = ???? m/s2vt =18 m/s r = 24 m Equation: ac = (Vt)2 / r F= macSubstitute: ac = (18m/s)2 / 24 m = 13.5 m/s2 Fc = ma = 2000 kg(13.5 m/s2) Solve: Fc = 27000 N or 2.7 x 104N __B__69.What is the gravitational force between two trucks, each with a mass of 2.0 x 104 kg, that are 2.0 m apart? a.5.7 x 10–2 Nb.6.7 x 10–3 Nc.1.3 x 10–2 Nd.1.2 x 10–7 NDiagram: (Given + Unknowns)Gravitational ForceFg = ???? N G = 6.67 x 10-11 N m2 / kg2m1 = 20000 kg m2 = 20000 d = 2 m Equation: Fg = G m1m2 / d2 Substitute: Fg = (6.67 x 10-11)(20000)(20000)/22SOLVE: Fg = 6.7 x 10-3 NSolve: Fg = 6.7 x 10-3 N or 0.0067 N_D___70.A 80.0 kg passenger is seated 12.0 m from the center of the loop of a roller coaster. What centripetal force does the passenger experience when the roller coaster reaches an tangential speed of 37.7 m/s?a.1.7 x 103 Nb.7.2 x 103 Nc.6.9 x 103 Nd.9.5 x 103 NDiagram: (Given + Unknowns)CircleFc = ???? Nm = 80 kg ac = ???? m/s2vt =37.7 m/s r = 12 m Equation: ac = (Vt)2 / r F= macSubstitute: ac = (37.7m/s)2 / 12 m = 118.4 m/s2 Fc = ma = 80 kg(118.4 m/s2) Solve: Fc = 9500 N or 9.5 x 103N _B___71.If the distance from the center of a 10,000 kg merry-go-round to the edge is 1.2 m, what centripetal acceleration does a passenger experience when the merry-go-round rotates at a speed of 1.5 m/s? (DRAW A PICTURE AND USE GUESS METHOD TO SOLVE)a.1.7 m/s2b.1.9 m/s2c.0.9 m/s2d.0.6 m/s2Diagram: (Given + Unknowns)CircleFc = ???? N m = 10,000 kg ac = ????? m/s2vt = 1.5 m/s r = 1.2 m Do not need to find centripetal forceEquation: ac = (vt)2 / rSubstitute: ac = (1.5m/s)2 / 1.2 m = 1.9 m/s2 Solve: ac = 1.9 m/s2 _A___72.When an object is traveled in a circular motion, the force is directed ___________________, the acceleration is directed ________________, and the velocity is directed ___________________________.a.towards the center : towards the center: along a tangent of the circle b.away from the center : away from the center: towards the centerc.along a tangent line : away from the center: towards the center d. towards the center: along a tangent of the circle: away from the center 73. On the following diagram, label the centripetal acceleration, centripetal force, and the tangential velocity for the following whirling object. 3625851574800074. Fill in the following table. Quantity (Physics Term) Symbol as used in formulasSI UnitQuantity (Physics Term)Symbol as used in formulasSI UnitDistancedmCentripetal Accelerationacm/s2DisplacementΔdmCentripetal ForceFcNTimetsTangential Velocityvtm/sVelocity vm/sRadiusRmSpeedvm/sTorqueτJInitial Velocityvim/sGravitational Potential EnergyPEgJFinal Velocityvfm/sElastic Potential EnergyPEelasticJAccelerationam/s2Kinetic EnergyKEJForceFN Mechanical EnergyMEJWeightFgNWorkWJMassm/ssticd its bone v 334.4 Jsss produced.mKgEnergyEJ75. What is the weight of a 200 N puggle? ________200 N_____________________76. What is the force of gravity on a 200 N puggle? _______200 N__________________77. What is the mass of 200 N puggle? ___________20.4 kg___________ ................
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