Your answer is partially correct. Try again.The velocity of a 2 kg particle is given by v = (2t i + 5t j) m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 44 N, what arethe direction (relative to the positíve direction of the x axis) of (a) the net forće and (b) the particle's direction of travel?(a) Number T84.78Unitso (degrees)(b) Number 77.2UnitsT° (degrees)Click if you would like to Show Work for this question: Open Show WorkSHOW HINTLINK TO TEXTLINK TO SAMPLE PROBLEMVIDEO MINI-LECTUREMATH HELPMATH HELPMATH HELPМАTH HELPto search10:44 PMENG4/4/2021SUS132)16f7home4.6.7.8.ER.T.00193.

Given, m=2kg v=2ti^+5t2j^ Fnet=44N

The velocity of a 2 kg particle is given by v = (2t i +5t j) m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 44 N, what are %3D the direction (relative to the positive direction of the x axis) of (a) the net forće and (b) the particle's direction of travel?

The velocity of a 2 kg particle is given by v = (2t i +5t j) m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 44 N, what are %3D the direction (relative to the positive direction of the x axis) of (a) the net forće and (b) the particle's direction of travel?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A mug of root beer is sliding across a horizontal table with a constant velocity of 5.0[m/s], and is experiencing a friction force. If the mug has a total mass of 2.60[kg] and μk = 0.250, find the force due to friction. Express your answer in N.
A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i + 18.74 j)N, F, = (-4.45 i + 12.22 j)N b. Find the total force exerted on the cart. Ans. EF = (16.36 i + 30.96 j)N or 35.02 N Find the angle at which the total force is applied. Ans. 62.15° NEERIN
A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i+ 18.74 j)N, F, =(-4.45 i+ 12.22 j)N b. Find the total force exerted on the cart. Ans. EF = (16.36 i + 30.96 j)N or 35.02 N c. Find the angle at which the total force is applied. Ans. 62.15º d. What is the acceleration of the cart? Ans. å = (2.05 i + 3.87 j)" or 4.38 EERIN
A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i + 18.74 j)N, F, = (-4.45 i + 12.22 j)N b. Find the total force exerted on the cart. Ans. EF = (16.36 i+ 30.96 j)N or 35.02 N c. Find the angle at which the total force is applied. Ans. 62.15°
wo forces are acting on a 1.08 kg object. One force is directed completely in the negative x direction and has a magnitude of 8.3 N, The second force has a magnitude of 7.5 N and is directed 28∘28∘ to the right of the negative y axis. Assume standard Cartesian coordinates. Find the magnitude of the acceleration of the object
An athlete pulls box E using an inextensible rope P while being resisted by another inextensible rope S. Let P be the tension force on rope P and S be the tension force on rope S. Consider particle analysis involving only forces P and S. P E S 2. If P = 495N, S = 465N, 0 = 10° and a = 29° , which of the following is closest to the magnitude and direction of R? 322 N, 32.9degs 161.2 N, 81.8degs 322 N, 32.9deg 161.2N, 81.1dega
There are two forces on the 2.00 kg box in the overhead view of the figure below, but only one is shown. For F1=20.0N, a= 12.0 m/s², and 0 = 30°, find (a) the second force in unit-vector notation (b) as a magnitude and an angle relative to the positive direction of the x axis.
A particle of mass 4.00 kg, moving on a horizontal table top, has an acceleration vector (in m/s2) given by a =9.10ı^ + 1.80ȷ^, where ı^ and ȷ^ are the unit vectors along the x and y axes, respectively. This particle is subjected to two horizontal forces, expressed in newtons by F⃗ 1=2.00ı^ + 3.00ȷ^ and F2. Calculate the force F2 in newtons. Give your answer to one decimal place
After the block in Fig. P-1405 has moved 10 ft from rest, the constant force P is removed. Find the velocity of the block when it returns to its initial position. Ans. v = 21.2 ft/sec
The figure shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA blocks are released, what is the tension in the cord at the right? 6.70 kg, mB 6.20 kg, and mc = 12.0 kg. When the B C Number Units
A block and sphere are connected by a cord that passes over a pulley as shown. Neglect friction and assume the cord is massless. Take m1 = 2.00 kg, m2 = 4.20 kg, and θ = 50.5°. What are the magnitude of acceleration (in m/s2) of the block and sphere?
A 125 kg body sits at rest until 3 forces act up the body: F1 = 24.5 N @ 0o, F2 = 38.0 N @ 160o, and F3 = 44.8 N @ 235o. Determine: (a) The resultant force vector (b) The resultant acceleration vector (c) The velocity vector at 10.0s (d) The position vector at 10.0s