The graph shows the variation with time t of the force F acting on an object of mass 15000 kg. The object is at rest at t=0. A. B. C. F/KN D. 7- 12 ms-¹ 180 m s-¹ 6- 5- 4- 3- 2- What is the speed of the object when t=30 s? 0.18 ms¹ 6ms-1 1 0+ 0 5 10 15 t/s 20 25 30
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- A force F = (cx – 3.00x² )î acts on a particle as the particle moves along an x axis, with F in newtons, x in meters, andca constant. At x = 0 m, the particle's kinetic energy is 20.0 J; at x = 3.00 m, it is 11.0 J. Find c. Number UnitsA child of mass 30.0 kg begins from rest and slides down a waterslide without friction, then launches into a pool. What is her speed at pt. 2 assuming the height of this point is 0.9 above the ground? pt. 1 2.0 m pt. 2 pt. 5 h pt. 3 0.8 m pt. 4 0.4 m Apt. 6An object is initially held at rest from a height of 4 m. The object is then released. A. Use conservation of energy to find the velocity right before it hits the ground B. Use the position equation to find the time it takes to reach the ground.
- On the roller-coaster shown, a car is moving 10 m/s toward the right at point 1. Assuming no friction, calculate the speed of the car at point 4. 3 40 m 26 m 15 m 29.37 m/s 22.71 m/s 20.89 m/s 26.17 m/s A. D. В. 25.07 m/s Е. С. 24.29 m/s F.A force of F=x^2 in Newtons acts on an object of mass m= 2 kg. If the object is at rest at x= 0 when the force F begins to act on it, what is the object’s speed at the position x= 3 m? A) 3 m/s B) 6 m/s C) 9 m/s D) 27 m/s4. Here we prove a version of the work-kinetic energy theorem for an object moving in 1-D subject to a constant force. Consider a mass m, traveling initially at speed v, on which a constant net force F is applied in the same direction as its motion over a distance d. After the net force has acted over this distance, the speed of the object is v2. In terms of m, F, v1, v2, and d only, V, A) What is the acceleration of the mass? t=0 m F d t>0_m B) What is the time over which the mass travels the V, distance d? [Use one equation of constant acceleration to find this.] C) Use another equation of constant acceleration to relate m, F, v,, v2, and d. Show that the product W = Fd is equal to the change in the mass’s kinetic energy.
- A kid on a bike with a combined mass of 100 kg goes 7 m/s at the top of a hill. The hill has a vertical drop of 50 m. How fast is the kid and bike going when they reach the bottom of the hill?( no friction) 1. c. d. O 27 m/s 23 m/s O 32 m/s O 980 m/sA particle of mass m = 1.50 kg moves along one dimension and is subject to a single force, whose potential energy function is given by U(x) = x4 – 2x2 + 4 (where U is measured in Joules and x is measured in meters). At the moment when the particle is located at position x = 1.50 m, its speed is v = 2.50 m/s. %3DPRINTER VERSION 1 BACK NEXT Chapter 07, Problem 001 A proton (mass m = 1.67 x 1027 kg) is being accelerated along a straight line at 4.60 x 1010 m/s2 in a machine. If the proton has an initial speed of 4.00 x 105 m/s and travels 1.10 cm, what then is (a) its speed and (b) the increase in its kinetic energy? (a) Number Units (b) Number Units Click if you would like to Show Work for this question: Open Show Work Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER 6:53 PM ENG search 3/30/2021 pause break prt sc insert delete 13) ohome 10 end mpgup 12Pgdn 45 %23 backspc 3 5. 8.
- E. 19.77 m/s 21.85 m/s C. 26.37 m/s F. 1--1 IYYVE On the roller-coaster shown, a car is moving 12 m/s toward the right at point 1. Assuming no friction, calculate the speed of the car at point 4. 3 40 m 26 m 4 15 m 21. A. 20.28 m/s D. 25.18 m/s 24.54 m/s B. 18.94 m/s E. F. C. 23.73 m/s 26.35 m/s 1--1 KTMDG A box slides from rest down a curved slope. Its initial height is 21 m above the ground, and it comes off the slope 7_m above the ground. How fast is the box moving as it leaves the slope? Assume friction is negligible. ^ Pigshow.jpeg Pig show.jpegan object is acted by a frictional force. the work done by this frictional force is a. negative because friction force is opposite its motion b. zero because fk = ukn and normal force is perpendicular to the motion c. zero because friction is not the cause of its motion d. could be positive, negative or zero because the direction of friction is unknown