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Unformatted text preview: 14.1: ρVg mg w = = ( 29 ( 29 ( 29 ( 29 N 8 . 41 s m 80 . 9 m 10 43 . 1 m 858 . m kg 10 8 . 7 2 2 2 3 3 = × × = π or 42 N to two places. A cart is not necessary. 14.2: ( 29 ( 29 . m kg 10 33 . 3 m 10 74 . 1 kg 10 35 . 7 3 3 3 6 3 4 22 3 3 4 × = × × = = = π πr m V m ρ 14.3: ( 29 ( 29 . m kg 10 02 . 7 3 3 mm . 30 . 15 . 5 kg 0158 . 3 × = = = × × V m ρ You were cheated. 14.4: The length L of a side of the cube is cm. 3 . 12 m kg 10 4 . 21 kg . 40 3 1 3 1 3 1 3 3 = × = = = ρ m V L 14.5: ρ πr ρV m 3 3 4 = = Same mass means ( 29 lead l aluminum, a 1 3 1 a 3 a = = = ρ r ρ r 6 . 1 10 7 . 2 10 3 . 11 3 1 3 3 3 1 a 1 1 a = × × = = ρ ρ r r 14.6: a) ( 29 3 27 30 3 8 3 4 30 sun sun m 10 412 . 1 kg 10 99 . 1 m 10 96 . 6 kg 10 99 . 1 × × = × × = = π V M D 3 3 m kg 10 409 . 1 × = b) ( 29 3 17 3 13 30 3 4 3 4 30 m kg 10 594 . m 10 351 . 3 kg 10 99 . 1 m 10 00 . 2 kg 10 99 . 1 × = × × = × × = π D 3 16 m kg 10 94 . 5 × = 14.7: ρgh p p = m 91 . 9 ) s m 80 . 9 ( ) m kg 1030 ( Pa 10 00 . 1 2 3 5 = × = = ρg p p h 14.8: The pressure difference between the top and bottom of the tube must be at least 5980 Pa in order to force fluid into the vein: Pa 5980 = ρgh m 581 . ) s m 80 . 9 ( ) m kg 1050 ( m N 5980 Pa 5980 2 3 2 = = = gh h 14.9: a) ( 29 ( 29 ( 29 Pa. 706 m 12 . s m 80 . 9 m kg 600 2 3 = = ρgh b) ( 29 ( 29 ( 29 Pa. 10 16 . 3 m 250 . s m 80 . 9 m kg 1000 Pa 706 3 2 3 × = + 14.10: a) The pressure used to find the area is the gauge pressure, and so the total area is ⋅ = × × 2 3 3 cm 805 ) Pa 10 205 ( ) N 10 5 . 16 ( b) With the extra weight, repeating the above calculation gives 2 cm 1250 . 14.11: a) Pa. 10 52 . 2 ) m 250 )( s m 80 . 9 )( m kg 10 03 . 1 ( 6 2 3 3 × = × = ρgh b) The pressure difference is the gauge pressure, and the net force due to the water and the air is N. 10 78 . 1 ) ) m 15 . ( )( Pa 10 52 . 2 ( 5 2 6 × = × π 14.12: atm. 61.9 Pa 10 27 . 6 ) m 640 )( s m 80 . 9 )( m kg 10 00 . 1 ( 6 2 3 3 = × = × = = ρgh p 14.13: a) = × × + × = + ) m 10 00 . 7 )( s m 80 . 9 )( m kg 10 6 . 13 ( Pa 10 980 2 2 3 3 2 2 a ρgy p Pa. 10 07 . 1 5 × b) Repeating the calcultion with 00 . 4 1 2 = = y y y cm instead of Pa. 10 1.03 gives 5 2 × y c) The absolute pressure is that found in part (b), 1.03 Pa. 10 5 × d) Pa 10 33 . 5 ) ( 3 1 2 × = ρg y y (this is not the same as the difference between the results of parts (a) and (b) due to roundoff error). 14.14: Pa. 10 . 6 ) m 1 . 6 )( s m 80 . 9 )( m kg 10 00 . 1 ( 4 2 3 3 × = × = ρgh 14.15: With just the mercury, the gauge pressure at the bottom of the cylinder is ⋅ + = m m gh p p p With the water to a depth w h , the gauge pressure at the bottom of the cylinder is ....
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This note was uploaded on 08/23/2008 for the course PHYS 2048 taught by Professor Hiebig during the Spring '08 term at Nova Southeastern University.
 Spring '08
 HIEBIG
 Physics

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