7.8 kg object undergoes an acceleration of 2.9 m/s^2 A. What is the magnitude of the resultant force acting on it? N B. If this same force is applied to a 4.7kg object, what acceleration is produced? m/s^2
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- A 1.0 kg object has a velocity of 5.0f m/s at t = 0. A constant resultant force of (3.0€ + 4.0j ) N then acts on the object for 3.0 s. What is the magnitude of object's velocity at the end of the 3.0 s interval? A B D E 9.8m/s 17.7m/s 9.2 m/s 15.0 m/s 16.3 m/s Select one: O A O B O D O E1. A force F=100 N is acting on an object of mass m=10 kg. The force makes an angle of 10° with the x-axis. Calculate the acceleration of the 9.85 m/s 8.69 m/s object. A в 7.52 m/s D 6.34 m/s 2. A 1000 kg object is rising. Its speed is increasing at 3 m/s. The force on the object is: A 1000 N B 3000 N 9800 N D 12800 N 3. A 16 kg object is initially at rest. A constant force of 8.0 N is exerted for 4 s on that object. The change in speed of this object will be 2.0 m/s 6.0 m/s 0.5 m/s 4.0 m/s A B D45. Review. Two constant forces act on an object of mass m = QC 5.00 kg moving in the xy plane as shown in Figure P7.45. Force F, is 25.0 N at 35.0°, and force F, is 42.0 N at 150°. At time t = 0, the object is at the origin and has velocity (4.00i2.50j m/s. (a) Express the two forces in unit-vector other answers notation. Use unit-vector notation for your (b) Find the total force exerted on the object. (c) Find the object's acceleration. Now, considering the instant t = 3.00 s find (d) velocity, (e) its position (f) its kinetic energy from m and (g) its the object's kinetic from 1500 energy m .T. (h) What conclusion can you 35.00 draw x m by comparing the answers to parts (f) and (g)? Figure P7.45
- A 1.0 kg object has a velocity of 5.0f m/s at t = 0. A constant resultant force of (3.0î + 4.0f ) N then acts on the object for 3.0 s. What is the magnitude of object's velocity at the end of the 3.0 s interval? A B D E 9.8m/s 17.7m/s 9.2 m/s 15.0 m/s 16.3 m/s Lütfen birini seçin: O A O B O C O D O EAn 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? Enter a number rounded to 2 decimal places and assume the answer has proper SI Units.An object with a mass of 1.53 kg is initially at rest upon a horizontal, frictionless surface. An applied force of 4.94 N i acts on the object for 2.2 s. What is the object's final speed?
- A 9.50 kg block is at rest on a horizontal floor, and then you pull it with a cord that exerts a tension force of 12.0 N on the block at an angle of 24.0o above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.055. What is the speed of the block 0.85 s after it starts moving? Nothing I do is working! I only have three more tries to get it right and what should be working is not.The figure below is for a 6 kg box on a horizontal floor. Initially, you are pushing horizontally on the box and friction is opposing you. Then, you stop pushing the box. Use g = 10 m/s?. v(m/s) 8.0 6.0 4.0 2.0 1.0 2.0 3.0 t(s) Find the normal force acting on the box.A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are us 0.40 and uk = 0.30, respectively. The puck's initial speed is 4.9 m/s. What speed does it have when it slides back down to its starting point? 2.8 m/s 4.9 m/s 4.2 m/s 3.5 m/s
- Istap. A constant force F = 5.0 [N] is applied at a certain angle from the horizontal on a 0.50 [kg] block that is initially moving to the right at v; = 3.0 [m/s]. If the block stops after traveling a distance s = 0.80 [m], what is the angle in the figure?? F Vi Vf = 0 0. 34° O 79° 124° 56°DETAILS PREVIOUS ANSWERS SERPSE10 7.4.P.016. MY NOTES ASK YOUR TEACHER The force acting on a particle is F, = (10x – 17) N, where x is in meters. (a) Make a plot of this force versus x from x = 0 to x = 3.00 m. F (N) F (N) 50 20 40 10 3어 20 0,5 1.0 15 20 2,5 x (m) 30 10 -10 x (m) 3.0 0.5 1.0 1.5 2.0 2.5 F (N) F (N) 0,5 1.0 + x (m) 2,5 30 1아 1.5 2,0 -1어 2,5 x (m) 30 -20 0,5 1,0 15 -30 -10 - 40 -20 - 50 (b) From your graph, find the net work done by this force on the particle as x = 0 to x = 2.55 m. (Include the correct sign.) it moves from Need Help? Read It1. A 45.0 kg object is pushed up an incline as shown with a horizontal force of 300.0 N. The object is initially at rest but it starts moving up the incline. Find the speed of the object after 3.50 s if the coefficient of kinetic friction between the object and the incline is 0.25 and the object is still on the incline after 3.50s. 240 N I. 20.00