Physics Solution Manual for 1100 and 2101

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Unformatted text preview: SONING AND SOLUTION We wish to convert 2.0% of the heat Q into gravitational potential energy, i.e., (0.020)Q = mgh. Thus, mg = ( 0.020 ) Q 4186 J 1 Calorie = 4.4 × 103 N ( 0.020 ) (110 Calories ) = h 2.1 m ______________________________________________________________________________ 98. REASONING Young’s modulus Y can be obtained from F = Y ( ∆L/L0 ) A (Equation 10.17), where F is the magnitude of the stretching force applied to the ruler, ∆L and L0 are the change in length and original length, respectively, and A is the cross-sectional area. Solving for Y gives F L0 Y= A ∆L The change in the length of the ruler is given by ∆L = α L0∆T (Equation 12.2), where α is the coefficient of linear expansion and ∆T is the amount by which the temperature changes. Substituting this expression for ∆L into the equation above for Y gives the desired result. SOLUTION Substituting ∆L = α L0∆T into Y = F L0 / ( A ∆L ) and using the fact that ∆T = 39 C°, we find that Y= = F L0 A ∆L = ( F L0 A α L0 ∆T ) F 1.2 ×103 N = = 7.7 ×1010 N/m 2 −1 −5 2) −5 ( Aα ∆T (1.6 ×10 m 2.5 × 10 C° ) ( 39 C° ) Chapter 12 Problems 99. 683 SSM REASONING AND SOLUTION As the rock falls through a distance h, its initial potential energy mrock gh is converted into kinetic energy. This kinetic energy is then converted into heat when the rock is brought to rest in the pail. If we ignore the heat absorbed by the pail, the principle of conservation of energy indicates that mrock gh = c rockmrock ∆T + cwater mwater ∆T where we have used Equation 12.4 to express the heat absorbed by the rock and the water. Table 12.2 gives the specific heat capacity of the water. Solving for ∆T yields ∆T = m rock gh crock m rock + cwater m water Substituting values yields (0.20 kg)(9.80 m/s2 )(15 m) = 0.016 C° [1840 J/(kg ⋅ C°)](0.20 kg) + [4186 J/(kg ⋅ C°)](0.35 kg) ______________________________________________________________________________ ∆T = 100. REASONING The change in length of the wire is the sum of the change in length of each of the two segments: ∆ L = ∆ Lal + ∆Lst . Using Equation 12.2 to express the changes in length, we have αL0 ∆T = α al L0al ∆T + α st L0st ∆T Dividing both sides by L0 and algebraically canceling ∆T gives L0al L + αst 0st L0 L0 α = α al The length of the steel segment of the wire is given by L0st = L0 − L0al . Making this substitution leads to L α = α al 0al L0 L −L 0 0al + α st L0 L L L = α al 0al + α st 0 − α st 0al L0 L0 L0 This expression can be solved for the desired quantity, L0al / L0 . 684 TEMPERATURE AND HEAT SOLUTION Solving for the ratio ( L0al / L0 ) and taking values for the coefficients of thermal expansion for aluminum and steel from Table 12.1 gives L0al α − α st 19 × 10–6 (C°) –1 − 12 × 10 –6 (C°) –1 = = = 0.6 L0 α al − α st 23 × 10 –6 (C°) –1 − 12 × 10 –6 (C°)–1 101. REASONING AND SOLUTION The heat lost by the steel rod is Q = cm∆T = cρ V0 ∆T. Table 12.2 gives the specific heat capacity c of steel. The rod contracts according to the equation ∆L = α L0∆T. Table 12.1 gives the coefficient of thermal expansion of steel. We also know that F = AY(∆L/ L0), Equation 10.17, so that F = AYα ∆T. Table 10.1 gives Young’s modulus Y for steel. Combining expressions yields, F= −1 α QY 12 × 10−6 ( C° ) (3300 J)(2.0 × 1011 Pa) = = 1.1 × 103 N 3 c ρ L0 [452 J/(kg ⋅ C°)(7860 kg/m )(2.0 m) ______________________________________________________________________________ 102. REASONING AND SOLUTION The figure below (at the left) shows the forces that act on the middle of the aluminum wire for any value of the angle θ. The figure below (at the right) shows the same forces after they have been resolved into x and y components. T T sin θ T T sin θ θ θ T cos θ W T cos θ W Applying Newton's second law to the vertical forces in figure on the right gives 2T (sin θ ) − W = 0 . Solving for T gives: T= W 2 ( sin θ ) (1) The following figures show how the angle is related to the initial length L0 of the wire and to the final length L after the temperature drops. Chapter 12 Problems L0 685 L θ0 θ x x If the distance between the supports does not change, then the distance x is the same in both figures. Thus, at the original temperature, x = L0 (cos θ0) (2) x = L (cos θ ) (3) while at the lower temperature Equating the right hand sides of Equations (2) and (3) leads to cos θ = L0 cos θ0 (4) L Now, L = L0 + ∆L, and from Equation 12.2, it follows that ∆L = α ∆T L0 Thus, Equation (4) becomes cos θ = L0 cos θ 0 L0 + ∆L = cos θ 0 1 + (∆L / L0 ) = cos θ0 1 + α ∆T From the figure in the text θ0 = 3.00°. Noting that the temperature of the wire drops by 20.0 C° (∆T = −20.0 C°) and taking the coefficient of thermal expansion of aluminum from Table 12.1, we find that the wire makes an angle θ with the horizontal that is cos θ0 1 + α ∆T θ = cos −1 cos 3.00° −1 = cos = 2.446° −1 −6 ( 1 + 23 × 10 C° ) ( −20.0...
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