University Physics with Modern Physics with Mastering Physics (11th Edition)

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14.78: (Note that increasing x corresponds to moving toward the back of the car.) a) The mass of air in the volume element is ρAdx ρdV = , and the net force on the element in the forward direction is ( 29 . Adp pA A dp p = - + From Newton’s second law, , ) ( a dx ρA Adp = from which . a dx ρ dp = b) With ρ given to be constant, and with . , 0 0 0 ρax p p x at p p + = = = c) Using 3 kg/m 1.2 = ρ in the result of part (b) gives ( 29 ( 29 ( 29 atm p -5 2 3 10 15 ~ Pa 0 . 15 m 5 . 2 s m 0 . 5 m kg 2 . 1 × = , so the fractional pressure difference is negligble. d) Following the argument in Section 14-4, the force on the balloon must be the same as the force on the same volume of air; this force is the product of the mass ρV and the acceleration, or . ρVa e) The acceleration of the balloon is the
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Unformatted text preview: force found in part (d) divided by the mass ( 29 . , bal bal a ρ ρ or V ρ The acceleration relative to the car is the difference between this acceleration and the car’s acceleration, ( 29 [ ] . 1 bal rel a ρ ρ a-= f) For a balloon filled with air, ( 29 1 bal < ρ ρ (air balloons tend to sink in still air), and so the quantity in square brackets in the result of part (e) is negative; the balloon moves to the back of the car. For a helium balloon, the quantity in square brackets is positive, and the balloon moves to the front of the car....
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