A particle of mass 0.4 kg moves a long a straight line according to x(t) = 2t + t3, where t is in second and x is in meter. The power (in W) of the resultant force acting on this particle at t = 4.5 s is:

A particle of mass 0.4 kg moves a long a straight line according to x(t) = 2t + t3, where t is in second and x is in meter. The power (in W) of the resultant force acting on this particle at t = 4.5 s is:

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

Chapter39: Relativity

Section: Chapter Questions

Problem 39.1P: The truck in Figure P39.1 is moving at a speed of 10.0 m/s relative to the ground. The person on the...

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Quadratic Equation: x = -6±yb²–4ac 2a AT = i; – ão Vave., drot. At dave.,e Save. || At At Vx = V0x + agt Ax Vont + at? = vox + 2a,Ax Ax (vor + Va)t '0x 1) Wile E. Coyote happens to be chasing after the Roadrunner, who is traveling in the +x-direction with a constant velocity of 8.72 m/s. Wile E. Coyote has a velocity of +3.28 m/s and a constant acceleration of +2.30 m/s? at the moment it is 10.0 m behind the Roadrunner. Take the initial position of the Roadrunner to be the origin (x = 0). a) time. Hint: Your answer should look like xR = fR(t). Express the final position of the Roadrunner (with respect to the origin) as a function of b) time. Hint: Again, your answer should look like xc = Express the final position of Wile E. Coyote (with respect to the origin) as a function of fc(t). c) use the quadratic formula here, you should get two times. Consider which one, if both, of these times makes sense in the context of this problem. How much time elapses before Wile E. Coyote catches the…
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y! Use the graph of t X (56) YouTube Hairdresser Reacts X College Physics by × .A commuter airpl X The eye of a hurric X Homework: PHYS- O File | C:/Users/Faith/Downloads/College%20Physics%20by%20Raymond%20A.%20Serway,%20Chris%20Vuille%20(z-lib.org).pdf Apps College Physics by Raymond A. Serway, Chris Vuille (z-lib.org).pdf 54 / 1050 237% + 11 from the line of scrimmage, runs and then runs sideways parallel to 15.0 yards. At this point, he throws raight downfield, perpendicular to v far is the football from its original 71. A commuter airplane starts from an airport and takes the route shown in Figure Pl1.71. The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it flies for 150. km 20.0° west of north, to city B. Finally, the plane flies 190. km due west, to city C. Find the location of C relative to the location of the starting point. city ent of –25.0 units and a y-compo- у (km) 250 E ne magnitude and direction of the C В W →E ta is 730.…
* 00 Σ %24 %23 O Find a C Classki Cobb C C Clever O Join E Edulast CTLS LE Incite® Battles W A&P Ep + eworksheets.com/workbooks/workbook.asp?t%3D688643&studentworkbookid%38874506×tamp%3D2022-1-27-17-33-27 Part 1: Motion in a straight line. (moving slow, moving faster, and moving very fast) Instructions 1. Move slow: You will walk very slowly through the corridor from start to finish. Record the time (in seconds). 2. Return to start. 3. Move medium: You will walk faster (medium pace) through the corridor from start to finish. Record the time (in seconds). 4. Return to start. 5. Move very fast: You will jog or do a very fast walk through the corridor from start to finish. Record the time in seconds. End Start Total Displacement Total time Time (s) Avg Speed Avg Velocity Distance (m) (m) (s/u) (s/m) Move slowly Move medium Move very fast % ణ 4. 5. 9-
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where does the 0.75 come from isint it supposed to be 0.5?
A tailroad flatcar of mass 2,000 kilograms nlls to the ight at 10 me ters per Ccond and ollides with a platcar of mass frat S meters per second the flatoars couple fogcther. thei speedafterffe collisn a.)Im/s 3,600 kilagrams that o nling to th b.2.5mls d.7mls C. Sm/s A 40 kilogram girl climbs a vertical dis tancc of 5meters in twenty Second at a constant. How much wor has the girl dme ? 9.245 J 6.490J d.)1940S on a horizontal frichiontess surface. A horizanta A SSou Wo al 48 force of 5 Nowfns is applied to the mass. After the force has been for Iseoond the velocity of the mass is . 9.0meter per second 10.5 meters persecond C.S meters persecond d.50 meters per seand 4. An impulse of 10 kilogram-meter per second acting an an objed G kilogram ml cause a change in the objcte velocity of: 9.0.5moters per secmd b.2 meters per second C. 1o meters per second d. so meters per secand 5. The fime needed for a net firce of 10 Newtons to change fre velscd of G 5 kilogroms mass by 3 meters/ second…