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- A force É = (5.90 N )î + (5.40 N )f + (6.40 N )k acts on a 1.20 kg mobile object that moves from an initial position of (4.20 m )î + (2.10 m )ŷ + (7.40 m )k to a final position of á f = (5.90 m )î + (8.20 m )ŷ + (5.50 m )k in 5.70 s. i Find (a) the work done on the object by the force in the 5.70s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors d ; and d f. i (a) Number i Units (b) Number i Units (c) Number i Unitsarrow_forwardA force F = (4.40 N) î + (3.00 N)ĵ + (1.80 N) k acts on a 5.40 kg mobile object that moves from an initial position of á ; = (4.90 m) î + (3.50 m)ĵ + (5.90 m) k to a final position of d; = (3.20 m)î + (6.60 m)ĵ + (5.10 m) k in 2.20 s. Find (a) the work done on the object by the force in the 2.20 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors d ; and d f. (a) Number Units (b) Number Units (c) Number i Units >arrow_forwardWhat is the average force exerted by a 70 kg passenger on his seatbelt when his car crashes onto a concrete wall? The car moving at 17.90 m/s comes to halt in 0.10sarrow_forward
- A high jumper with a body weight of 820 N exerts an average vertical force of 1640 N down on the floor for 0.75 s. The average vertical force exerted on the jumper by the floor is:arrow_forwardAn object with a mass of 1.58 kg is initially at rest upon a horizontal, frictionless surface. An applied force of 4.79 N i acts on the object for 2.94 s. What is the object's final speed?arrow_forwardTwo forces, F1 = (6.10î + 4.30ĵ) N and F2 = (3.40î + 7.65ĵ) N, act on a particle of mass 2.20 kg that is initially at rest at coordinates (+2.05 m, −4.50 m). (d) What are the coordinates of the particle at t = 10.7 s? x=_________m y=__________marrow_forward
- f Daniella messaged X 8 Timer for Google F X + form-timer.com/app Apps M Gmail YouTube Maps 2 h 20 min Find the magnitude of the resultant of the concurrent force system below y B (15.0 m) D (10.0 m) 25.0° M Sent Mail - loraine X ← → C 53.00 30.0° X A (8.00 m) C (12.0 m) 90.2 N (755 dea from y in the 3rd quadrant) ☐ zFVLOdcF Carrow_forwardA 8.7-kg cat moves from rest at the origin to hunk of cheese located 9.5 m along the x-axis while acted on by a net force F(x) =a − Bx+yx² with a = 5.2 N, 6 = 4.9 N/m, and y = 1.2 N/m². Find the cat's speed v as it passes the hunk of cheese. V= m/sarrow_forwardA mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be Foooooarrow_forward
- Two forces, F, = (2.70î – 3.40j) N and F, = (3.70î – 7.80j) N, act on a particle of mass 2.10 kg that is initially at rest at coordinates (-1.50 m, +4.40 m). 1 (b) In what direction is the particle moving at t = 11.5 s? ° counterclockwise from the +x-axisarrow_forwardWE-4 A 3.00kg block is initially sliding to the right on a frictionless surface at 7.00m/s. Force P is directed to the left and is applied to the block continuously starting at the initial position as shown below. The magnitude of P varies as shown in the graph to the right. V=7.00m/s y, P P (N) 20 15 10 5 0 0 1 2 X 3 4 (m) a) Find the magnitude of the work done by the force P as the block moves from x-0 to x=4m. b) Find the kinetic energy of the block at x=4m. Neglect air resistance (and the surface is frictionless). c) Find the speed of the block at x-4m.arrow_forwardAt a particular instant, a 1.0-kg particle’s position is r → = (2.0i ^ − 4.0j ^ + 6.0k ^ )m , its velocity is v → = (−1.0i ^ + 4.0j ^ + 1.0k ^ )m/s , and the force on it is F → = (10.0i ^ + 15.0j ^ )N . (a) What is the angular momentum of the particle about the origin? (b) What is the torque on the particle about the origin? (c) What is the time rate of change of the particle’s angular momentum at this instant?arrow_forward
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