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A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9° past the lowest point on its way up, its total acceleration is
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Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card for Serway/Jewett's Physics for Scientists and Engineers, 10th, Multi-Term
- A small object moves at constant speed in a horizontal circle of radius 0.425 m. If the object makes two complete revolutions in one second, what is the magnitude of the acceleration of the object?arrow_forwardA small object with mass 0.200 kg moves with constant speed in a vertical circle of radius 0.500 m. It takes the object 0.100 s to complete one revolution. As the object passes through the lowest point of its motion, what is the magnitude of the acceleration of the object?arrow_forwardA ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9° past the lowest point on its way up, its resultant acceleration is a = (-22.5i + 20.2j) m/s2 , where i is the unit vector along the horizontal direction (you can treat it as pointing to the right), and j is the unit vector along the vertical direction (you can treat it as pointing upward). For that instant, (i) determine the magnitude of its centripetal acceleration in m/s2 (ii) determine the speed of the ball in in m/s (iii) determine the velocity of the ball in m/s, i and j.arrow_forward
- A woman rides a carnival Ferris wheel at radius 22 m, completing 4.1 turns about its horizontal axis every minute. What are (a) the period of the motion, and the magnitude of her centripetal acceleration at (b) the highest point and (c) the lowest point?arrow_forwardA car drives around a circular track of diameter 156 m at a constant speed of 33.0 m/s. During the time it takes the car to travel 102 degrees around, what is the magnitude of the car s average acceleration? NOT ALLOWED TO USE CENTRIPITAL MOTION EQUATION, must be average acceleration. 13.96 m/s^2 6.98 m/s^2 12.19 m/s^2 0 m/s^2arrow_forwardA skater is gliding along the ice at 2.2 m/s, when she undergoes an acceleration of magnitude 1.2 m/s2 for 3.0 s. At the end of that time she is moving at 5.8 m/s. (a) What must be the angle between the acceleration vector and the initial velocity vector?arrow_forward
- The Ferris wheel in the figure, which rotates counterclockwise, is just starting up. At a given instant, a passenger on the rim of the wheel and passing through the lowest point of his circular motion is moving at 3.00 m/sm/s and is gaining speed at a rate of 0.500 m/s2m/s2. 1.Find the magnitude of the passenger's acceleration at this instant in m/s^2 2.Find the direction of the passenger's acceleration at this instant. θ=___∘ to the right of verticalarrow_forwardA pursue at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the pursue is (2.00 m/s^2)i + (4.00 m/s^2)j. At the instant and in unit-vector notation, what is the acceleration of the wallet?arrow_forwardA cat rides a merry-go-round turning with uniform circular motion. At time t1 = 2.20 s, the cat's velocity is (3.80 m/s) i^ + (3.40 m/s) j^, measured on a horizontal xy coordinate system. At t2 = 5.10 s, its velocity is (-3.80 m/s) i^ + (-3.40) j^. What are (a) the magnitude of the cat's centripetal acceleration and (b) the magnitude of the cat's average acceleration during the time interval t2 – t1, which is less than a period of the motion? I just need (b) solved. It's not 4.45.arrow_forward
- A cat rides a merry-go-round turning with uniform circular motion. At time t1 = 2.20 s, the cat's velocity is (3.80 m/s) i^ + (3.40 m/s) j^, measured on a horizontal xy coordinate system. At t2 = 5.10 s, its velocity is (-3.80 m/s) i^ + (-3.40) j^. What are (a) the magnitude of the cat's centripetal acceleration and (b) the magnitude of the cat's average acceleration during the time interval t2 – t1, which is less than a period of the motion?arrow_forwardIf a particle moving in a circular path of radius 8.1 m has a velocity function v =3.5t2(units: m/s), what is the magnitude of its total acceleration at t = 2.1s?arrow_forwardA cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal. Express the initial velocity as a linear combination of its unit vector components. ( m/s) + ( m/s) At the maximum height, the speed of the cannon ball is v = ? m/s and the magnitude of its acceleration is a = ? m/s2. The time needed to reach maximum height is t = ? s. The maximum height reached by the cannon ball is H = ? m.arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningAn Introduction to Physical SciencePhysicsISBN:9781305079137Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar TorresPublisher:Cengage Learning