College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- A teacher drives his car at 3 mi N 40o E, then turns 3.8 mi to the West. What is the magnitude and direction of the teacher’s displacement from the starting position?arrow_forwardMr. Brown takes a walk in the park. He first travels 600. m NW and then turns and walks 800. m @ 20.0° WoS. What is his overall displacement of his walk? Use the triangle method to solve this. Should you use the component method, I will award zero points. I know the answer, I just don’t know how to get it. Answer: 771 m @ 25.1° SoW Thanks!!arrow_forwardA pilot is in his plane with the intention of flying to Nunavut. From the current location, Nunavut is [N 9.0W]. The flight tracker estimates the distance from Toronto to Nunavut is approximate 3.0x103km. The plane hasthe capability to fly at an airspeed of 890 km/hr. The pilot aims the plane at a heading [N], and lands in Nunavutdirectly. If the flight takes 3.33 hrs to complete: a. What is the velocity of the plane relative to the Earth? b. What was the wind velocity present on that day?arrow_forward
- a hockey pucktravels a displacement of 3.4 m/s ( S 30 degrees W). It then struck by a hockey players stick and undergoas a displacement of 4.0 m (E 20 degrees N) a) sketch the two vectors of the hokcey puck motion b) calculate the pucks total displacementarrow_forwardI 2 BE Suppose you walk 13.0 m straight west and then 21.0 m straight north. How far are you from your starting point, 25 m what is angle of the resultant vector, 57.5 O and what is the compass direction of a line connecting your starting point to your final position? 58 X °W of Narrow_forwardA plane flies with the Jet Stream (tailwind) from Chicago to Rome in 9.1 hours, a flight distance of 7740 km. It returns, flying against the Jet Stream (headwind), in 10.8 hours. Calculate the average speed of the plane and the average speed of the wind. Must list assumptions, and draw 1D vector diagram!!! wr Marrow_forward
- A plane initially flies -50 km 30 degrees East of South from start to get to its goal, but due to storm rerouting has ended up 80km at 60 degrees North of East from the base. a) Draw a vector diagram, label the first location rỉ and the second location r as well as the displacement Ar b) What is the magnitude and direction of the planes displacement during this motion? c) If the entire trip took 30 minutes, what is the magnitude and direction of the planes average velocity from initial location to final in km/hr? (Av = ) Atarrow_forwardHelparrow_forwardNeed help figuring out: Using the standard cartesian coordinate system, which of the following vectors have a NEGATIVE X-COMPONENT? Choose all that do.arrow_forward
- Question 20 A helicopter is heading S 43° E (i.e. direction angle of 313°) with an airspeed of 27 mph, and the wind is blowing S 13° E (i.e. direction angle of 283°) at 10 mph. Helicopter's speed Round all numbers in your answers below to 2 places after the decimal point. (a) Find the velocity vector that represents the true heading of the helicopter. Type your answer in component form, (where a and b represent some numbers). Velocity vector of helicopter's true heading: (b) Find the helicopter's speed relative to the ground (in mph). mph (c) Find the helicopter's drift angle, 8. (The drift angle is the number of degrees that the helicopter will end up flying off-course.) 8 = Submit Question degreesarrow_forwardGrade 11 Physics: I haven't yet learned the range formula but don't know how to adjust the angle to hit the target by using the formulas I have used. A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon’s target (which has a radius of 1.0 m) is painted on the asphalt 42.0 m away from the water cannon. Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land and make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. sin??θ=opphyp sin??θ=v⃗yv⃗ v⃗y=v⃗sinθ =(34.0m/s)sin 18° =10.5m/s[up] cosθ=adjhyp cosθ=v⃗xv⃗ v⃗x=v⃗cosθ =(34.0m/s)cos 18° =32.3m/s[forward] Let [up] be positive Δd⃗=0 a⃗=-9.8m/s2[down] v⃗y1=10.5m/s[up] Δt= ? Δd⃗=v⃗y1Δt+12a⃗(Δt)2 0=(10.5m/s)Δt+12(-9.8m/s2)(Δt)2 0=(10.5m/s)Δt-4.9m/s2(Δt)2 0=Δt (10.5m/s-4.9m/s2Δt) Δt=0 or Δt=2.14s v⃗x=32.3m/s [forward] Δt=2.14s Δd⃗x= ?…arrow_forwardGiven the vectors Ā = (-2î + j) m and B = (4î – j) m. The magnitude of the resultant vector R = 2Ã + B is O a. R = 3 m O b.R = 1 m OC.R = V5 m O d.R = V10 marrow_forward
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