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Rules: 1.Everyone must attempt to answer each question. 2.Show your work on a separate sheet of paper. 3.Keep track of your points. 4.Winners receive prize. continue

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MotionProjectilesNewton Forces Vectors 100 200 300 400 500

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Go BACK. You need to click on a number to proceed to the questions. continue Click here

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Really!! Go BACK. You need to click on a number to proceed to the questions. continue Click here

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2)Displacement = 27.4m + 3m -3m Distance = 27.4m + 3m +3m continue

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2) This object in freefall accelerates at 9.8 m/s/s. Every second the velocity changes by 9.8 m/s continue

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A ball rolls down an incline from rest. How far does it travel in 3 seconds if it accelerates at 5 m/s 2 ? Show Work for Credit d = v i t + ½ at 2 d = 0 + ½ (5 m/s 2 ) (3s) 2 d = 22.5 m continue

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4) V avg = (V f + V i )/2 V avg = (10m/s +25m/s)/2 V avg = 17.5 m/s

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1) d = v i t + ½ at 2 -6m = ½ a (2.7s) 2 a = 1.6 m/s/s down continue

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1)The acceleration of a projectile is always 9.81 m/s 2 downward because the only force acting on the object is gravity. continue

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2) d = v i t + ½ at 2 -22m = ½ a (3.0s) 2 a = 4.9 m/s/s down continue

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Show Work for Credit VERTICAL d = v i t + ½ at 2 -45m = 0 + ½ (-9.81m/s 2 ) t 2 t = 3.0 s HORIZONTAL d = v i t + ½ at 2 d = (25 m/s)(3.0 s) + 0 d = 75 m

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2) v vs. t graphs are always linear. Slope = accel. = -9.81 m/s 2 continue

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v f = v i + at 0 = 98 m/s + (-9.81m/s 2 )t t = 10. s continue

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3) g = F g /m

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F net = ma 12 N = (2kg+1kg)a a = 4 m/s 2 continue

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Which graph(s) depict an object that is accelerating? B and C only continue DAILY DOUBLE Select an amount of points you want to wager before going on to the question. You can only wager up to as many points as you currently have A BCD E

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2) An object in equilibrium has an acceleration equal to zero. continue

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Newton’s 3 rd Law: the forces are equal in magnitude, opposite in direction. F nail on hammer = - F hammer on nail continue

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F net = ma 50N – F f = (4 kg)(10 m/s 2 ) F f = 10N continue

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F net = ma F applied – F friction = 0 F applied = F friction F applied = μF N = (0.05)(80kg x 9.81m/s 2 ) F applied = 40N continue

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(C) If the net force is zero, the acceleration must also be zero. The slope of the speed graph = accel. = 0. Constant speed = constant slope of distance graph.

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F net = ma F applied – F friction = 0 F applied = F friction F applied = μF N = (0.67)(60 N) F applied = 40N continue Show Work for Credit

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continue F net = ma = 0 F – F friction – F parallel = 0 F = F friction + F parallel F = F friction + F g sinθ F = 3N + (10N)sin30° F = 3N + 5N F = 8N

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(4) Add vectors head to tail continue B

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List 3 scalar quantities and 3 vector quantities. Include units. Scalar: distance (m), speed (m/s), mass (kg), time (s),temp (°C) Vector: displacement (m), velocity (m/s), acceleration (m/s 2 ), Force (N), impulse (N·s), momentum (kg·m/s) continue DAILY DOUBLE Select an amount of points you want to wager before going on to the question. You can only wager up to as many points as you currently have

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continue (4) F H = Fcosθ F H =(24N)cos35° F H = 20. N F V = Fsinθ F V =(24N)sin35° F V = 14. N

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continue (1) B A R

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continue (4) Arrange 30N vectors to be head-to- tail, resultant starts at the tail of the first and ends at the head of the last. R 30N

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