lab 10

Average) (constant , record value as Average) [m] (constant , record value as Average) Ranges (using averages) 1 .295 .24 .04 .155 .196 7.8% 2 .145 3 .13 Average .295 .24 .04 .143 .196 7.8% 4 (with cap on) Observation s With the cap on the bottle I have found that the water does not flow. Part 2: In part two I did the same thing that I did in the first trials except instead of having the hole at the top of the bottle it was closer to the middle of the bottle. In trial four again I filled the bottle to the top and released my finger off the hole. The same thing that happened in the first trial also happened in this trial and the water did not flow. To calculate the range in these trials I used the formula R=2√hy. The value of h in this formula represents how far the hole is from the bottom of the table which is .15 m. In this formula the value of y represents how far the top of the water is from the hole which in this case was .13 m. To solve this you plug in h and y R=2√.15x.13 and once you simplify you get the answer .279 m. Data Table 2 Part II Measurements using Hole near the Middle of a Bottle Trial Total Height of Bottle, H [m] Height of hole above Vertical Distance of water Measure d Range, R [m] Calculated Range, R = 2 √ hy % Difference between

for this was that I did not have the correct value for “y” which was the distance of the water from the hole. 2. The speed of water in a hose increases from 1.96 m/s to 25.5 m/s going from the hose to the nozzle. Using Equation 3, calculate the pressure in the hose, given that the absolute pressure in the nozzle is 1.01×10 5 N/m 2 (atmospheric, as it must be) and assuming level, frictionless flow. Show your calculations. P 1 + 1 2 ρv 1 2 + ρg h 1 = P 2 + 1 2 ρv 2 2 + ρg h 2 Equation 3 The first thing you do to find the pressure of the hose it to move p1 to one side of the equation with the rest on the other side of the equal sign. When you do this you get: p 1 = p 2 + 1 2 ρv 2 2 − 1 2 ρ v 1 2 = p 2 + 1 2 ρ ( v 2 2 − v 1 2 ) After you move p1 to the other side of the equal sign you can plug in the values for pressure which is 1.01X10^5 v2 which is 25.5 and v1 which is 1.96. p 1 = 1.01 × 10 5 + 1 2 ( 10 3 ) ¿ ] p 1 = 4.24 × 10 5 Once you have plugged in the values you can simplify the answer to be 4.24x10^5. 3. A Siphon is used to drain fluid from a higher level reservoir to a lower level reservoir. If the drain is 6.75 m from the top of the higher level reservoir, what is the velocity of the fluid as it exits the drain? Show your calculation. V = √ 2 gh Equation 4 Finding the velocity of the fluid is very simple with this formula we know the height and we know the value of g to be 9.8. using these values we can plug them in to the formula and find the answer. V = √ 2 × 9.8 × 6.75 V = 11.5 Once you have plugged in the values to the formula you can simplify to find the answer that the velocity of the fluid out of the drain comes out to be 11.5 meters per second. 4. Explain your observation of the fluid flow when the cap was placed on the bottle, and you removed your finger from the hole. Was this the same for the holes in all 3 bottles?