Question

finish all of them and show process

(g) What is the volume of the oil displaced by the block, in terms of the variables of the problem? Vo= Problem 2 - Pressure and Buoyancy with Oil and Water You have a block of a light wood that is 10 cm by 10 cm by 10 cm. You also have a bucket that you fill with a depth of 20 cm of water (has a density of 1000 kg/m'), and then another 20 cm of oil (has a density of 850 kg/m) which floats on top of the water. You drop your block of wood into the bucket, and it floats trapped at the water/oil interface such that 4 cm of the block is above the oil/water line, and 6 cm is below the oil/water line. Assume that the bucket is very large so that the water level (and oil level) do not change when the block of wood is dropped into the bucket. (a) Draw this system, and label distances. (h) What is the volume of the water that is displaced by the block, in terms of the variables of the problem? Vw= (b) The absolute pressure at the top of Mt Diablo today is exactly 100,000 Pa. The top of Mt Diablo is 980 m higher than your bucket with water, oil and a block in it. The density of air is 1.29 kg/m'. Find the absolute pressure here at DVC, at the top of your bucket (i) Substitute your expressions from (g) and (h) into your answer from (f) so that you have an expression relating the force of the fluids on the block in terms of densities and volumes only. P = (c) Find the pressure at the top of the block in your bucket. FB = P = (d) Find the pressure at the oil/water interface line. () Find the density of the block. To do this, use Newton's First Law and the relationship between density, mass and volume. Also, how is the volume of the block related to the volume of oil displaced and the volume of water displaced? P- (e) Find the pressure at the bottom of the block in your bucket. P. (1) Find the pressure at the bottom of your bucket. PE (g) Find the force acting upwards on the bottom of the block. Also find the force acting downwards on the top of your block. F and Fawn (h)What is the net force of these two pressure forces acting on the block? What direction does it point? and and it points

Answer 1

Nolm loom ofl 20 cm 114 cm Olm water interface. om 7 L 14ml 16cm water 20cm -Pc 100000 Pa P= 1.29 Kg/7 900m | golong? level of pe potegh 100000 + 129X10X 980: 100000+12642 top of bucket = 1 126 42 pa ... Pass = 112642 pa. at top of bucket Pass = 112642 pa roa Pol=85058/m3 P= Pressure at topt pigxh - 112642 + 85oxloxo. 2-12642 41700 3114342 pal At top of bucket hans = 112642 Pa I - Pon - Pabst Pou gx h=112642 +858x 1080.16 = 114002 Pa ibim a Pressure at the interface - 114342 Pa (calculated ind parl) p= Pinyt Pugh = 114342 of 1000x10x6.06 - 114942 pa 2010 | Potom - Pint & Pwxg xh a 114342 I 116342 t 1000XLOXO. 2 Pa.

1 *P, = 114002 Pa p = 114942 pa f FPLA up i P2 A 2 114942X 0.17 0.1 = 1149.42 N & Fdown - PiA = 114002 X 0.100.1=1140.02N a tret = Fup-Flow - 1149.42-1140.02 =.9.40. lend page is think thes t I think but these are the other questions i am solving it by taking data of this questo volume of oil displaced = Axhor 0-1X0.1x0.04 = 4x100 m3. 7 volume of water displaced = AX how = 0.140.180.06 = 6x100 ? (1) A vou displaced xgx fou to vioare displaced XgX Purata - Voog Pot VW g Pw mg = Fg sy v Pag = Volog + Kopong P = 4x10x 850 + 6x 10 x 1000 I (3) 0.1X01 X OIL (850x4 + 6000 = 940 kg/m3 Po= quorg/m3 (o 6x 10-4