Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 42AP
Review. A wire of length L, Young’s modulus Y, and cross-sectional area A is stretched elastically by an amount ∆L. By Hooke’s law, the restoring force is −k∆L. (a) Show that k = YA/L. (b) Show that the work done in stretching the wire by an amount ∆L is
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Chapter 12 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 12.1 - Consider the object subject to the two forces of...Ch. 12.1 - Consider the object subject to the three forces in...Ch. 12.2 - A meterstick of uniform density is hung from a...Ch. 12.4 - For the three parts of this Quick Quiz, choose...Ch. 12 - You are building additional storage space in your...Ch. 12 - Why is the following situation impossible? A...Ch. 12 - Prob. 3PCh. 12 - A circular pizza of radius R has a circular piece...Ch. 12 - Your brother is opening a skateboard shop. He has...Ch. 12 - A uniform beam of length 7.60 m and weight 4.50 ...
Ch. 12 - Prob. 7PCh. 12 - A uniform beam of length L and mass m shown in...Ch. 12 - A flexible chain weighing 40.0 N hangs between two...Ch. 12 - A 20.0-kg floodlight in a park is supported at the...Ch. 12 - Prob. 11PCh. 12 - Review. While Lost-a-Lot ponders his next move in...Ch. 12 - Figure P12.13 shows a claw hammer being used to...Ch. 12 - A 10.0-kg monkey climbs a uniform ladder with...Ch. 12 - John is pushing his daughter Rachel in a...Ch. 12 - Prob. 16PCh. 12 - The deepest point in the ocean is in the Mariana...Ch. 12 - A steel wire of diameter 1 mm can support a...Ch. 12 - A child slides across a floor in a pair of...Ch. 12 - Evaluate Youngs modulus for the material whose...Ch. 12 - Prob. 21PCh. 12 - When water freezes, it expands by about 9.00%....Ch. 12 - Review. A 30.0-kg hammer, moving with speed 20.0...Ch. 12 - A uniform beam resting on two pivots has a length...Ch. 12 - A bridge of length 50.0 m and mass 8.00 104 kg is...Ch. 12 - Prob. 26APCh. 12 - The lintel of prestressed reinforced concrete in...Ch. 12 - Prob. 28APCh. 12 - A hungry bear weighing 700 N walks out on a beam...Ch. 12 - Prob. 30APCh. 12 - A uniform sign of weight Fg and width 2L hangs...Ch. 12 - When a person stands on tiptoe on one foot (a...Ch. 12 - A 10 000-N shark is supported by a rope attached...Ch. 12 - Assume a person bends forward to lift a load with...Ch. 12 - A uniform beam of mass m is inclined at an angle ...Ch. 12 - Prob. 36APCh. 12 - When a circus performer performing on the rings...Ch. 12 - Figure P12.38 shows a light truss formed from...Ch. 12 - Prob. 39APCh. 12 - A stepladder of negligible weight is constructed...Ch. 12 - A stepladder of negligible weight is constructed...Ch. 12 - Review. A wire of length L, Youngs modulus Y, and...Ch. 12 - Two racquetballs, each having a mass of 170 g, are...Ch. 12 - Prob. 44APCh. 12 - Review. An aluminum wire is 0.850 m long and has a...Ch. 12 - You have been hired as an expert witness in a case...Ch. 12 - A 500-N uniform rectangular sign 4.00 m wide and...Ch. 12 - A steel cable 3.00 cm2 in cross-sectional area has...Ch. 12 - A uniform rod of weight Fg and length L is...Ch. 12 - In the What If? section of Example 12.2, let d...
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- A uniform wire (Y = 2.0 1011 N/m2) is subjected to a longitudinal tensile stress of 4.0 107 N/m2. What is the fractional change in the length of the wire?arrow_forwardA horizontal, rigid bar of negligible weight is fixed against a vertical wall at one end and supported by a vertical string at the other end. The bar has a length of 50.0 cm and is used to support a hanging block of weight 400.0 N from a point 30.0 cm from the wall as shown in Figure P14.81. The string is made from a material with a tensile strength of 1.2 108 N/m2. Determine the largest diameter of the string for which it would still break. FIGURE P14.81arrow_forwardA spring 1.50 m long with force constant 475 N/m is hung from the ceiling of an elevator, and a block of mass 10.0 kg is attached to the bottom of the spring. (a) By how much is the spring stretched when the block is slowly lowered to its equilibrium point? (b) If the elevator subsequently accelerates upward at 2.00 m/s2, what is the position of the block, taking the equilibrium position found in part (a) as y = 0 and upwards as the positive y-direction. (c) If the elevator cable snaps during the acceleration, describe the subsequent motion of the block relative to the freely falling elevator. What is the amplitude of its motion?arrow_forward
- A 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_forwardUse the data in Table P16.59 for a block of mass m = 0.250 kg and assume friction is negligible. a. Write an expression for the force FH exerted by the spring on the block. b. Sketch FH versus t.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_forward
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