Physics 101, need help fast. Here is the question. e e puragrupn una inen unswer
ID: 1290495 • Letter: P
Question
Physics 101, need help fast. Here is the question.
e e puragrupn una inen unswer ine joiowing esions Three balls with the same volume of 1.0 × io-mare in an open tank of water that has a density (p) equal to 1.0× 103 kg/m3. The balls are in the water at different lev- cls. Ball 1 floats in water with a part of it above the surface ball 2 is completely submerged in the water, and ball 3 rests on the bottom of the tank. Any movement of the water obeys Bernoulli's equation: where P1 and P2 are the pressures at elevations yi and ½, and vi and vy are the speeds of the water. (Note; Unless oth- erwise noted, the water and the balls are stationary.) 9. The buoyant forces B, B2, and B, exerted by water on the balls are related by which of the following? 10. The densities of the balls A, P2, and are related by which of the following? B. 11. Assume that the density of ball 3 is 7.8 × 103 kg/m3. Ignoring atmospheric pressure, what is the supporting force exerted by the bottom of the tank on ball 3? A. 1.0× 10-2 N B, 6.7 × 102 N C. 7.6x 10 N D. 8.8x 10 N 12. Assume that the density of ball 1 is 8.0 x 10 kg/m3. Ignoring atmospheric pressure, what fraction of ball 1 is above the surface of the water? A. B. C. 1 D. 1 13. Ball 2 is in the water 20 cm above ball 3. What is the approximate difference in pressure between the two balls? A. 2 x 102 N/m2 B, 5 × 102 N/m2 C. 2 x 10 N/m2 D.5x 10 N/m2 14. If ball 3 is a hollow, iron ball and atmospheric pressure can be ignored, what should be the volume of the hol- low portion of ball 3 such that the force exerted by it on the bottom of the tank is 0? Note: Density of iron is 7.8 × 103 kg/.) A, 0.13 × 106 B. 0.78 x 10" m? C, 0.87 × 106 m3 D, 1.15x106 m?Explanation / Answer
9) B (SInce the volume submerged is equal in last two cases and greater than the volume in first case)
10) A (Trivial since 1st has some part above water , 2nd is completely inside and third is at the bottom)
11)B =(7.8*103-1*103)9.8*10-6=6.7*10-2
12)D (let x be the fraction outside than 1(1-x)=0.8) x=1/5
13 C) delta P=density*g*h=1*103*9.8*0.2=19.6*102 == 2*103
14) C) 1*103*g*V=7.8*103*g*x (assume x is the solid volume)
x=0.12V hollow region=0.88V=0.88*10-6=