Question

ent-Responses/submit?pos-18dep 225963128tags-autosave J4 points SerPSE10 14.4.P.012 20.0 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in the figure below. The 12.0 om dimension is vertical, and the top of the bloc 35 cm below the surface of the water. GMy Notes Ask Your Scale (a) What are the magnitudes of the forces (in N) acting on the top and on the bottom of the block due to the surrounding water? (Use P,-1.0130 x 10 N/m Enter your anwers accurate te at least four significant figures.) Fop N N bottom (b) What is the reading of the spring scale (in N) (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block. (Submit a file with a maximum sice of 1 M Choose File No file chosen This answer has not been graded yet Read Need Help? Practice Another Version Submit Answer 1 Type here to search

Answer 1

Sal. Given » 20.0 kg t Pa d epth uptowy 1. mans of block, m= 20.0kg Rimansion of block; d = 0.1m h breadth, b= 0.1m I height, h= 0.12 m e depth upto which block immerded, d= 0. 35m Pressure at top: Patm + 3gd . ! = 1x3x10* + (100*9-BX0-35) P = 104730 Pa force ftop - PixArea = 104730 (0.14001 Freip = 1047.3 N ..." Pressure at bottom - Pi + 'sga = 104730 + (1000x38% 0:12) P2 = 105906 Foottun PyxArea105906x0.01 I thottum = 1059.06N/ 0 ☺ Reading of scale = ng FB Red FB- Buoyant free fp = ġ ug - 1000x(0-1X0.1X ©•12)x9.8 FB = 11.76 Roading = (20x9.8) 11.76 [Reading = 184024N © Fe = 11:26 Fbottom - Pop 3 1059.06-1047-3 z 11.76 so IFB = (Foottum - From)