College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- There are no numerical values assigned to the variables in this problem. A block of mass m1 is projected with a velocity vi at the top of the wedge. When the block reaches the horizontal surface its velocity vf to the right while the wedge moves with a velocity v2 of to the left. (a). What is the velocity of the once it reaches the horizontal surface? Your answer must be in terms of relevant variables. (b). What is the height h the wedge in terms of the relevant variables?arrow_forwardThe 150-lb man lies against the cushion for which the coefficient of static friction is u, = 0.5. He rotates about the z axis with a constant speed v = 32 ft/s. (Figure 1) Figure 1 of 1 ftarrow_forwardO Macmillan Learning An object of mass m₁ = 0.105 kg undergoes uniform circular motion. It is connected by a massless string through a hole in a frictionless table to a larger object of mass m₂ = 0.275 kg, as shown in the figure. Assuming the larger object is stationary, calculate the tension T in the string and the linear speed u of the circular motion of the smaller object on the table. Assume that the radius R of the circular path of the smaller object is equal to 1.00 m. T = V = N m/sarrow_forward
- Please don't provide handwritten solution ....arrow_forwardA 3kg block it’s on a horizontal table top connected by a rope and pulley to a 2kg mass. If the system it’s accelerating at a 2.5 m/s^2. What is the tension in the rope and coefficient of friction? Please use the equations provided.arrow_forwardcht3 8, A ball with a mass of 0.250 kg is tied to a weak string with a breaking strength of 20.0 N. It is spun in a horizontal circle on a tabletop in a circle of radius 0.950 m. At what speed will the string break?arrow_forward
- O Macmillan Learning An object of mass m₁ = 0.105 kg undergoes uniform circular motion. It is connected by a massless string through a hole in a frictionless table to a larger object of mass m₂ = 0.275 kg, as shown in the figure. Assuming the larger object is stationary, calculate the tension T in the string and the linear speed u of the circular motion of the smaller object on the table. Assume that the radius R of the circular path of the smaller object is equal to 1.00 m. T = V = N m/sarrow_forwardThe Colussus is a ferris wheel at 6 Flags amusement park with a 50.4 m diameter. It rotates at 1.5 rpm. For the following, assume the rider has a mass m = 65 kg, and that the rider sits forward with feet up so the only part of the seat that exerts a force on the rider is the horizontal seat bottom. As usual, a free body diagram is a great place to start. Find the normal force exerted by a seat on a person riding the Colossus at the highest point in the ride. Hint: the speed of the rider can be computed from the circumference of their circular path and the time it takes to make one full revolution. Find the normal force exerted by a seat on a person riding the Colossus at the lowest point in the ride.arrow_forwardASAParrow_forward
- A car of mass m is negotiating a circular turn of radius R and speed v on a banked road with an angle of banking θ. The forces on the car are shown below. Ignore friction in this problem. Draw a free body of the car. Identify and show the radial direction. Pick coordinate axes with x pointing in the radial direction towards the center of the circle and y in the vertical direction. Show the coordinate axes on the free body diagram. Set up Newton’s Laws in the radial and vertical direction. Which force provides the centripetal force? Use the equations in part C to solve for the speed of the car in terms of R, Does the speed depend on the mass of the car? If R=250.0 m, m=750.0 kg, calculate the speed of the car.arrow_forwardAnswer the followingarrow_forwardPGF A 3.60 kg mass is being rotated in the vertical plane at a constant speed of 5.00 m/s at the end of a 1.80 m - long string as shown below. Assuming the mass's speed is the same in all three positions, determine the magnitudes of string tension and total acceleration for the three positions. Draw a FBD of the mass for all three postions as part of your solution. 20 Position 1: Tension= X Acceleration = Position 2: Tension x Acceleration Position 3: Tension X Accelerationarrow_forward
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