Problem_Set_1_Solutions

Problem_Set_1_Solutions - 30. To solve the problem, we note...

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0.2 2.00 s 3 g 1kg 60 s 4.00(5.00) 3.00 g/s 0.101g/s 0.101 s 1000 g 1 min 6.05 10 kg/min. t dm dt = ªº = =− ¬¼ × 30. To solve the problem, we note that the first derivative of the function with respect to time gives the rate. Setting the rate to zero gives the time at which an extreme value of the variable mass occurs; here that extreme value is a maximum. (a) Differentiating 0.8 ( ) 5.00 3.00 20.00 mt t t + with respect to t gives 0.2 4.00 3.00. dm t dt The water mass is the greatest when / 0, dm dt = or at 1/0.2 (4.00/3.00) 4.21s. t == (b) At 4.21s, t = the water mass is 0.8 ( 4.21s) 5.00(4.21) 3.00(4.21) 20.00 23.2 g. + = (c) The rate of mass change at 2.00 s t = is 0.2 2.00 s 2 g 60 s 4.00(2.00) 3.00 g/s 0.48 g/s 0.48 s 1000 g 1 min 2.89 10 kg/min. t dm dt = = = (d) Similarly, the rate of mass change at 5.00 s t = is
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31. The mass density of the candy is 43 4 3 3 0.0200 g 4.00 10 g/mm 4.00 10 kg/cm . 50.0 mm m V ρ −− = = If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in the container when filled to height h is , M Ah = where 2 (14.0 cm)(17.0 cm) 238 cm A == is the base area of the container that remains unchanged. Thus, the rate of mass change is given by 2 () (4.00 10 kg/cm )(238 cm )(0.250 cm/s) 0.0238 kg/s 1.43 kg/min. dM d Ah dh A dt dt dt = ×
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3. The speed (assumed constant) is v = (90 km/h)(1000 m/km) (3600 s/h) = 25 m/s. Thus, in 0.50 s, the car travels (0.50 s)(25 m/s) 13 m.
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6. (a) Using the fact that time = distance/velocity while the velocity is constant, we find avg 73.2 m 73.2 m 3.05 m 1.22 m/s 73.2 m 73.2 m 1.74 m/s. v + == + (b) Using the fact that distance = vt while the velocity v is constant, we find v avg m / s)(60 s) m / s)(60 s) s m/s. = + = (. . 122 305 120 214 (c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer.
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10. The amount of time it takes for each person to move a distance L with speed s v is / s tL v Δ= . With each additional person, the depth increases by one body depth d (a) The rate of increase of the layer of people is (0.25 m)(3.50 m/s) 0.50 m/s /1 . 7 5 m s s dv dd R v L == = = = Δ (b) The amount of time required to reach a depth of 5.0 m D = is 5.0 m 10 s 0.50 m/s D t R =
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12. (a) Let the fast and the slow cars be separated by a distance
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This note was uploaded on 11/16/2009 for the course PHYSICS 11a taught by Professor Silvera during the Fall '07 term at Harvard.

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Problem_Set_1_Solutions - 30. To solve the problem, we note...

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