4.10. An object moves from x = -2.1 m, y = 3.7 m to x = 3.3 m, y = -1.1 m in a time of 5.3 s. What is its average velocity? 7.2 m/s, 32° below the positive x-axis 2.5 m/s, 48° below the positive x-axis 1.4 m/s, 42° below the positive x-axis 4.9 m/s, 76° below the positive x-axis 092 m/s 14° below the positive x-axis

4.10. An object moves from x = -2.1 m, y = 3.7 m to x = 3.3 m, y = -1.1 m in a time of 5.3 s. What is its average velocity? 7.2 m/s, 32° below the positive x-axis 2.5 m/s, 48° below the positive x-axis 1.4 m/s, 42° below the positive x-axis 4.9 m/s, 76° below the positive x-axis 092 m/s 14° below the positive x-axis

Name: 4.10. An object moves from x = -2.1 m, y = 3.7 m to x = 3.3 m, y = -1
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: 4.10. An object moves from x = -2.1 m, y = 3.7 m to x = 3.3 m, y = -1.1 m in a time of 5.3 s. What is its averagevelocity?O 7.2 m/s, 32° below the positive x-axis2.5 m/s, 48° below the positive x-axisO 1.4 m/s, 42° below the positive x-axisO 4.9 m/s

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter2: Motion In One Dimension

Section: Chapter Questions

Problem 2.6OQ: An arrow is shot straight up in the air at an initial speed of 15.0 m/s. After how much time is the...

See similar textbooks

Similar questions

A cannonball is fired from the ground with anupward initial velocity of 30 m/s. If the cannonball reaches a height of 5 meters, what is its y-velocity? Use g = 10 m/s^2 a. sqrt of 10 m/s b. 10 m/s c. 12 m/s d. 15 m/s
A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.70 m/s will leave the end of a horizontal channel at the top of a vertical wall h = 2.35 m high, and from there it will fall into a pool (Fig. P3.42). (a) Will the space behind the waterfall be wide enough for a pedestrian walkway? (b) To sell her plan to the city council, the architect wants to build a model to standard scale, which is one-twelfth actual size. How fast should the water flow in the channel in the model? Figure P3.42
A bangus (milkfish) can attain a maximum swimming speed of 6.0 m/s. a). How high can bangus jump vertically out of the water at that speed? a) 1.84 m height above the waterb) 1.84 m height above the groundc) 2.84m the groundd) 2.85 m above the water
A bush baby can leap straight up to a height of 2.1 m off the ground. To reach this height, with what speed would it need to leave the ground, in m/s? Use g = 9.8 m/s2.
The first supersonic flight was performed by Charles Yeager in 1947. During his attempt, he flew from 119 m/s to top speed, accelerating at 7.29 m/s^2 for a distance of 5251 m. What top speed did he achieve?
A ball is thrown with an initial velocity of (30 m/s) at an angle of (60°) above the horizontal, as shown in figure below. Find : (a) The maximum height of the ball? Answer : ( Ymax = 34.5m) (b) The time to rise to the top of the trajectory? Answer:(t,=2.65s) (@) The total time the ball is in the air? Answer: (t:=5.305)
Nate throws a ball straight up to Kayla, who is standing on a balcony 3.8 m above Nate. When she catches it, the ball is still moving upward at a speed of 2.8 m/s. With what initial speed did Nate throw the ball?A. 7.0 m/s B. 12.3 m/sC. 9.1 m/s D. 10.6 m/s
A drowsy cat spots a flowerpot that sails first up and then down past an open window.The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?
A person, standing at the edge of a cliff overlooking the ocean, throws a stone with an initial speed of 20 m/s at an angle of 30o below the horizontal. If the stone is in flight for 2s, the height of the cliff is a.) 19.6 m b.) 14.9 m c.) 4.9 m d.) 5.1 m
Given these two velocities, C = (4î +2ĵ -2k̂) m/s and E= (5î -4ĵ +5k̂), find the magnitude of the velocity, v= 2C –E. Required to answer. Single choice. a. 12.4 m/s b. 17.0 m/s c. 21.0 m/s d. 27.6 m/s
How to solve for the vy(vertical velocity at the maximum height)? Please show solutionsGiven: Vi=15m/s Angle= 30 degrees Range= 9.94m Height= 2.87m Vix= 12.99m/s Viy= 7.5 m/s Total time: 1.53 s T (time to reach the highest point) = 0.77s
if a person can jump a maximum horizontal distance ( by using a 45 degrees projection angle) of 3.66 m on earth, what would be his maximum range on the moon, where the free-fall acceleration is g/6 and g = 9.80 m m/s^2? (a) give the answers in meters.. (b) repeat for mars, where the acceleration due to gravity is 0.38 g.