Physics exercises_solution: Chapter 14

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 14 | w mg pVg x 103 kg m3 n x 10-2 m 2 s2 N or 42 N to two places. A cart is not necessary. m m p v a x 1022 kg _ g3 x 103 n x 106 mJ kg m3. p m - 0 0158kg r 103 kg m3. You were cheated. K V mm3 01 The length L of a side of the cube is L V1 f m3 f-4 3 p x 103 kg m3 cm. m pV nr3 p Same mass means ra3 pa rf p1 a aluminum l lead i 1 3 X t 1 3 r f a f n-3 x 10 r I Pa I x 103 a D Msun x 1030 kg x 1030 kg Vsun _ n x 108m 3 x 1027 m3 x 103 kg m3 b D 1030 kg l0 kg 1 7 3 7--------- ------------ g- x 1017 kg m3 x 104m 1013m3 x 1016 kg m3 p - P0 Pgh h P-P0 Pg x 105 Pa 1030 kg m3 m s2 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 pgh 5980 Pa h 82 59 80iNm2 gh 1050 kg m3 m s2 a pgh 600 kg m3 s2 706Pa. b 706Pa 1000 kg m3 m s2 x 103 Pa. is a The pressure used to find the area is the gauge pressure and so the total area 16-5 x 103 N 805 cm - 205 x 103 Pa b With the extra weight repeating the above calculation gives 1250 cm2. a pgh x 103 kg m3 m s 250 m x 106 Pa. b The pressure difference is the gauge pressure and the net force due to the water and the air is x 106 Pa tf m 2 x 105 N. p pgh x 103 kg m3 m s 640 m x 106 Pa atm. a pa pgy2 980 x 102 Pa x 103 kg m3 m s2 x 10 2 m x 105 Pa. b Repeating the calcultion with y y2 - y1 cm instead of y2 gives x 105 Pa. c The absolute pressure is that found in part b x 105 Pa. d y2 -y1 pg x 103 Pa this is not the same as the difference between the results of parts a and b due to roundoff error . pgh x 103kg m3 m s2 m x 104 Pa. With just the mercury the gauge pressure at the bottom of the .