EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 14, Problem 14P
To determine
The effect on clock when the length of string is increased.
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Chapter 14 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
Ch. 14 - Prob. 1PCh. 14 - Prob. 2PCh. 14 - Prob. 3PCh. 14 - Prob. 4PCh. 14 - Prob. 5PCh. 14 - Prob. 6PCh. 14 - Prob. 7PCh. 14 - Prob. 8PCh. 14 - Prob. 9PCh. 14 - Prob. 10P
Ch. 14 - Prob. 11PCh. 14 - Prob. 12PCh. 14 - Prob. 13PCh. 14 - Prob. 14PCh. 14 - Prob. 15PCh. 14 - Prob. 16PCh. 14 - Prob. 17PCh. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Prob. 21PCh. 14 - Prob. 22PCh. 14 - Prob. 23PCh. 14 - Prob. 24PCh. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - Prob. 27PCh. 14 - Prob. 28PCh. 14 - Prob. 29PCh. 14 - Prob. 30PCh. 14 - Prob. 31PCh. 14 - Prob. 32PCh. 14 - Prob. 33PCh. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - Prob. 36PCh. 14 - Prob. 37PCh. 14 - Prob. 38PCh. 14 - Prob. 39PCh. 14 - Prob. 40PCh. 14 - Prob. 41PCh. 14 - Prob. 42PCh. 14 - Prob. 43PCh. 14 - Prob. 44PCh. 14 - Prob. 45PCh. 14 - Prob. 46PCh. 14 - Prob. 47PCh. 14 - Prob. 48PCh. 14 - Prob. 49PCh. 14 - Prob. 50PCh. 14 - Prob. 51PCh. 14 - Prob. 52PCh. 14 - Prob. 53PCh. 14 - Prob. 54PCh. 14 - Prob. 55PCh. 14 - Prob. 56PCh. 14 - Prob. 57PCh. 14 - Prob. 58PCh. 14 - Prob. 59PCh. 14 - Prob. 60PCh. 14 - Prob. 61PCh. 14 - Prob. 62PCh. 14 - Prob. 63PCh. 14 - Prob. 64PCh. 14 - Prob. 65PCh. 14 - Prob. 66PCh. 14 - Prob. 67PCh. 14 - Prob. 68PCh. 14 - Prob. 69PCh. 14 - Prob. 70PCh. 14 - Prob. 71PCh. 14 - Prob. 72PCh. 14 - Prob. 73PCh. 14 - Prob. 74PCh. 14 - Prob. 75PCh. 14 - Prob. 76PCh. 14 - Prob. 77PCh. 14 - Prob. 78PCh. 14 - Prob. 79PCh. 14 - Prob. 80PCh. 14 - Prob. 81PCh. 14 - Prob. 82PCh. 14 - Prob. 83PCh. 14 - Prob. 84PCh. 14 - Prob. 85PCh. 14 - Prob. 86PCh. 14 - Prob. 87PCh. 14 - Prob. 88PCh. 14 - Prob. 89PCh. 14 - Prob. 90PCh. 14 - Prob. 91PCh. 14 - Prob. 92PCh. 14 - Prob. 93PCh. 14 - Prob. 94PCh. 14 - Prob. 95PCh. 14 - Prob. 96PCh. 14 - Prob. 97PCh. 14 - Prob. 98PCh. 14 - Prob. 99PCh. 14 - Prob. 100PCh. 14 - Prob. 101PCh. 14 - Prob. 103PCh. 14 - Prob. 104PCh. 14 - Prob. 105PCh. 14 - Prob. 106P
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- The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?arrow_forwardOne type of toy car contains a spring that is compressed as the wheels are rolled backward along a surface. The spring remains compressed until the wheels are freed and the car is allowed to roll forward. Jose learns that if he rolls the car backward for a greater distance (up to a certain point), the car will go faster when he releases it. The spring compresses 1.00 cm for every 10.0 cm the car is rolled backward. a. Assuming the spring constant is 150.0 N/m, what is the elastic potential energy stored in the spring when Jose rolls the car backward 20.0 cm? b. What is the elastic potential energy stored in the spring when he rolls the car backward 30.0 cm? c. Explain the correlation between the results for parts (a) and (b) and Joses observations of different speeds.arrow_forwardA particle of mass m moving in one dimension has potential energy U(x) = U0[2(x/a)2 (x/a)4], where U0 and a are positive constants. (a) Find the force F(x), which acts on the particle. (b) Sketch U(x). Find the positions of stable and unstable equilibrium. (c) What is the angular frequency of oscillations about the point of stable equilibrium? (d) What is the minimum speed the particle must have at the origin to escape to infinity? (e) At t = 0 the particle is at the origin and its velocity is positive and equal in magnitude to the escape speed of part (d). Find x(t) and sketch the result.arrow_forward
- A pendulum, comprising a light string of length L and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. P8.40). (a) Show that if the sphere is released from a height below that of the peg, it will return to this height after the string strikes the peg. (b) Show that if the pendulum is released from rest at the horizontal position ( = 90) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5. Figure P8.40arrow_forwardIf the speed of the observer is increased by 5.0%, what is the period of the pendulum when measured by this observer?arrow_forwardA block of mass M rests on a table. It is fastened to the lower end of a light, vertical spring. The upper end of the spring is fastened to a block of mass m. The upper block is pushed down by an additional force 3mg, so the spring compression is 4mg/k. In this configuration, the upper block is released from rest. The spring lifts the lower block off the table. In terms of m, what is the greatest possible value for m?arrow_forward
- Check Your Understanding An engineer builds two simple pendulums. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendulums will differ if the bobs are both displaced by 12°.arrow_forwardA lightweight spring with spring constant k = 225 N/m is attached to a block of mass m1 = 4.50 kg on a frictionless, horizontal table. The blockspring system is initially in the equilibrium configuration. A second block of mass m2 = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m2 loses contact with the springblock 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B). a. What is the subsequent speed of block 2? b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a). 77. (a) The energy of the system initially is entirely potential energy. E0=U0=12kymax2=12(225N/m)(0.150m)2=2.53J At the equilibrium position, the total energy is the total kinetic energy of both blocks: 12(m1+m2)v2=12(4.50kg+3.00kg)v2=(3.75kg)v2=2.53J Therefore, the speed of each block is v=2.53J3.75kg=0.822m/s (b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or v=0.822m/s. FIGURE P16.77arrow_forwardA block of mass m = 2.00 kg is attached to a spring of force constant k = 500 N/m as shown in Figure P7.15. The block is pulled to a position xi = 5.00 cm to the right of equilibrium and released from rest. Find the speed the block has as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is k = 0.350. Figure P7.15arrow_forward
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