College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- Two unequal masses m1 and m2 are suspended from different points on a meterstick in rotational equilibrium. Which of the following equations is correct for MEASUREMENT 1? mi MEASUREMENT 1 MEASUREMENT 2 O 30 m1 = 45 m2 O 30 m1 = 95 m2 O 20 m1 = 95 m2 O 20 m1 = 5 m2 >0 cm >20 cm >50 cm >95 cm >100 cm >0 cm →30 cm wɔ os aɔ 08 > 100 cmarrow_forwardA hollow cylinder of mass M1 and radius R1 rolls without slipping on the inside surface of another hollow cylinder of mass M2 and radius R2. Assume R1<<RZ. Both axes are horizontal, and the larger cylinder is free to rotate about its axis What is the angular velocity or frequency of small oscillationsarrow_forward1. A 97 g ball bobbing up and down on the ocean as the waves roll by has a vertical position that can be described as a function of time by x(t) = cos(10t). What is the force on the ball at a time of 4.6 seconds? Note: The argument of the cosine function is in radians. Hint: Don't forget the sign of the force. F = 6.74 X√ Narrow_forward
- Item 5 In this version of the "Giant Swing", the seat is connected to two cables, one of which is horizontal (Figure 1). The seat swings in a horizontal circle at a rate of 26.0 rpm (rev/min). For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of A conical pendulum. Figure 40.0⁰ -7.50 m- 1 of 1 > 20 ▼ 999 OOD F4 If the seat weighs 305 N and a 875 N person is sitting in it, find the tension Thorizontal in the horizontal cable. Express your answer with the appropriate units. ► View Available Hint(s) Submit Thorizontal Value Part B Tinclined = O Submit Previous Answers X Incorrect; Try Again; 9 attempts remaining HA 0 If the seat weighs 305 N and a 875 N person is sitting in it, find the tension Tinclined in the inclined cable. Express your answer with the appropriate units. ▸ View Available Hint(s) F5 Value μÅ 3 Previous Answers Units F6 BR Units www F7 ? ? FB F9arrow_forwardSand moves without slipping at 6.0 m/sm/s down a conveyer that is tilted at 15∘∘. The sand enters a pipe hhh = 5.0 mm below the end of the conveyer belt, as shown in the figure (Figure 1).arrow_forwardIn Fig. the upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both 50.0 cm long. The upper ball has mass 2.00 kg, and it is initially 10.0 cm higher than the lower ball, which has mass 3.00 kg. Find the frequency and maximum angular displacement of the motion after the collision.arrow_forward
- Can you please explain how to find the angle of theta in this problem? The answer states that it is 88.8 degrees but does not explain how. A propeller blade at rest starts to rotate from t = 0 s to t = 5.0 s with a tangential acceleration of the tip of the blade at 3.00m/s^2. The tip of the blade is 1.5 m from the axis of rotation. At t = 5.0 s, what is the total acceleration of the tip of the blade?arrow_forwardNOTE: The answer of 85.715 rpm was given to me and was incorrect. What is the correct answer?arrow_forwardA pendulum's angular position is given by ?= 0.0340 cos (?t), where ? is in radians and ? = 6.23 rad/s. Find the period (in s) and length (in m) of the pendulum.arrow_forward
- The smooth hollow tube assembly rotates about a vertical axis with angular velocity ω=θ˙=3.8ω=θ˙=3.8 rad/s and ω˙=θ¨=−2.4ω˙=θ¨=-2.4 rad/s2. A small 0.23-kg slider P moves inside the horizontal tube portion under the control of the string which passes out the bottom of the assembly. If r=0.88r=0.88 m, r˙=−1.9r˙=-1.9 m/s, and r¨=3.8r¨=3.8 m/s2, determine the tension T in the string and the horizontal force Fθ exerted on the slider by the tube. T is positive in tension, and Fθ is positive if in the positive θ direction.arrow_forwardWrite the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates for the instant when 0 = 54° if 0 = 2.06 rad/sec and 0= 4.395 rad/sec². Answers: a = (i a = ( i 2.7' C i+ i Me e) ft/sec² j) ft/sec² 3arrow_forward4:48 O ll B/s docs.google.com/forms 11 225 J Consider a place where the gravity is 4 times the gravity on Earth (g' = 4g), then the frequency of oscillation of a simple pendulum in that place, f', as compared to its frequency on earth is: O f' = f/2 f' = 9f f' = f/3 O f' = 4f O f' = 2f A simple pendulum consists of a 0.8- kg bob connected to a massless inextensible cord with a length L = 1.1 m. The bob is set into motion and its angular displacement is given by e(t) = 0.11cos(wt), where e is in radiansarrow_forward
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