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I need help with this problem and an explanation for the solution described below. (University Physics 1: Newton's Laws (Dynamics))
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- Three 0.300-kg billiard balls are placed on a table at the corners of a right triangle as shown in the figure. The sides of the triangle are of lengths a = = 0.500 0.400 m, b = 0.300 m, and c = m. Calculate the magnitude of the resultant force, F. m2 F. F21 m1 b. m3 7.66 x 10^-11 N 5 x 10^-11 N 3.5 x 10^-11 N 4.35 x 10^-11 Na 50 kg person rides on an elevator while standing on a scale. the elevator is traveling downward but slowing down at a rate of 2.00m/s squared. The reading on the scale is closest to. a- 708 N B- 349 N C-589 N D-120 N E- 469 NA 1.5-kg mass has an acceleration of (4.0i - 3.0j) m/s². Only two forces act on the mass. If one of the forces is (2.0i - 1.4j) N, what is the magnitude of the other force? 4.1 N 07.1 N 5.1 N O24N 6.1 N
- e 61. At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N, you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 2.7-kg projectile from rest to a speed of 8.3 x 10³ m/s. The net force accelerating the projectile is 4.9 x 105 N. How much time is required for the projectile to come up to speed? i 189.8 Units SThe 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 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. d. What is the acceleration of the cart? Ans. ä = (2.05 i + 3.87 j) or 4.38 e. What is the cart's velocity after 5.00 s? Ans. i = (13.75 i + 21.55 j)" or 25.56" f. What is the position of the cart after 5.00 s? † = 43.125i + 59.375j or73.38 mA 10 kg block of ice is at rest on a frictionless horizontal surface. A 1597 N force is applied to the block for 10 ms. During this time interval, the change of the velocity of the block is: 1.06 m/s Q3.19 m/s 2.28 m/s O 1.60 m/s O 0.80 m/s Time left 0:21:26 Clear my choiceA horizontal force of 5.0 N pushes a 0.50-kg block against a vertical wall. The block is initially at rest. If µ. – 0.60 and 0.50, the acceleration of the block is: Ⓒ4.8 m/s² Ⓒ3.8 m/s² Ⓒ 9.8 m/s² 0.0 m/s² 08.0 m/s2
- A rocket is launched vertically from the Earth, and the thrust (pushing force) from the engines is directed upward, and has a magnitude of 5.00 x 106N. The mass of the rocket is initially 2.00 x 105 kg.What is the initial acceleration of the rocket, assuming you can neglect air resistance? a 50.0 m/s2 b 15.2 m/s2 c 7.6 m/s2 d 25.0 m/s203 Body falls vertically. Its initial speed is zero and its fall duration is 2 min. The mass of the body is m=80 kgr. The air friction force is opposite to the velocity of the body with a factor of R=60 Nt s/m. Given g=10m/s². A) Draw the vertical z-axis with any direction you want and write the equation of motion of the body. B) To find the speed of the body as a function of time. C) With what speed does the body reach the ground?A mass m1 of 3.00 kg slides on a frictionless surface. This mass is connected to another mass m2 of 1.50 kg by a massless string over a frictionless pulley. The masses are held motionless and then released. What is the acceleration of m1 in m/s^2? * mi m2 O 2.14 O 3.27 O 4.98 O 6.12
