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FUNDAMENTALS OF PHYSICS - EXTENDED
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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_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_forwardf 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_forward
- A 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_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, 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_forward
- A 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_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_forwardTwo constant forces act on a 5.00-kg object moving in the xy-plane, as shown in the figure below, Force F, is 25.0 N at 0 = 37.0° from x-axis, while F = 40.0 N at o = 37.0º from y-axis. At time t = 0, the object is at the origin and has velocity v, (4.0i + 2.5j) m/s. Using the unit-vector notation (a) Find the total force on the object and its acceleration, (b) What is the object's velocity at the instant t = 3.00 s?, (e) What is its location at time t = 3.0 s?, (f) What is its kinetic energy at this moment?, and (g) How much work is done by the total force in this time? y F2 F1 marrow_forward
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