Then you place another 200 gram mass on the 210 degree mark. Then we have to replace the mass at the 30 degree mark with two masses, one at the 0 degree mark and one at the 80 degree mark. Essentially we are trying to calculate the x and y component vectors of a 200g mass at 30 degrees. So what we did was we just guess and checked the variables of weight at the 0 degree mark and at the 90 degree mark’s pullys until the ring was centered. You can us+e the weight on each pully to calculate the magnitude and the direction of the component vectors at 90 and 0 degrees.
2) The larger moveable knife edge was then clamped to the pendulum, at a small distance (1cm) above the centre of mass. 3) The distance h was then determined from the centre of mass to the axis of suspension. 4) The mask was then attached to the pendulum. 5) The light gate was then connected to the digit-metre and adjust its height and position relative to the mask to allow the period T, of the pendulum to be measured repeatedly. 6) The procedure was then repeated for larger values of h, until T has passed its minimum value.
Explain the function of the following duct components: (57.2) Ell are elbows used to make turns in ductwork, Wye is used to split one large duct into two smaller ducts, Transition is used to change rectangular ducts from one size to another, Boot is used at the end of a round duct to allow connection of a rectangular register or grill. 8. What is the total available static pressure for ductwork for a system with the following specifications? (57.13) 9. Why should air velocity in branch ducts be limited to 600 fpm?
After the calibration, select one of unknown blocks in the shelf. In this lab, block No.5 was chosen. Use caliper to measure the length of the block and diameter of the hole. Use micrometer to measure the height and the width of block. Finally, use scale to measure the mass of the block.
0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.
The following equation represents this relationship where k denotes the spring constant or stiffness of the spring, F=-kx Since x symbolizes the displacement or change in the length of the spring the above equation can now be surmised in the following manner, F=mg=-k∆l This new form makes it evident that a linear proportion exists between the plot of F as function of changing in length, ∆, thus confirming the spring does in fact obey Hooke’s Law. This enabled the group to determine the spring constant k. B. Derivation of Equations Definitions To gain a better understanding of the terms used here
Then you put the remaining liquid (neutral component mixture) through the suction filtration funnel to isolate the crystals. After 10-15 minutes, the crystals will be dry and you can weigh them and find their melting point. How is the neutral component of your 3-compound mixture isolated from the final methylene chloride solution? 16 of 22 4/16/12 9:15 PM StudyBlue Flashcard Printing of Lab Final 2211L UGA http://www.studyblue.com/servlet/printFlashcardDeck?deckId=... Name the four active ingredients that we will be testing for in the TLC of Analgesics lab: Ibuprophen Caffeine Acetaminophen Acetylsalicylic acid What solvent system will you be using to dissolve the common analgesics (power) in preparation for TLC spotting? Methylene chloride: ethanol (1:1 solution) What solvent system will you be using to "develop" your TLC plates in the TLC of Analgesics lab?
Allow the filter paper to dry overnight. 11. Weigh the filter paper with the contents inside. Record the weight. Data: | Fe | CuSO4 | Cu | FeSO4 | Filter Paper | Filter Paper with contents | Weigh Paper | Before | 1.445 g | 6.185 g | -- | -- | 0.770 g | -- | 0.29 g | After | -- | -- | 1.850 g | (drained out) | -- | 2.620 g | -- | Evaluation: 1.445 g Fe x 1 mol Fe x 1 mol Cu x 63.3546 g Cu = 1.64 g Cu => Theoretical Yield 55.847 g Fe 1 mol Fe 1 mol Cu 1.850 g Cu = 112.8 % => Percent Yield 1.64 g
Make the new dilution series. Remember to start with salt concentration where the egg first floated. (If you don't have enough solution from the original serial dilution, make some more by starting from the stock solution.) 8. As before, test the egg in each cup, starting with the lowest salt concentration.
These weights hang off the sides of the wheels and pull on the string at different angles, the objective is to find the point at which all the weights pull on each other so the center of the string is in the center of the force table. This was found in the lab by slowly adding weights till the right mixture was found. The forces are recorded and then shown through vectors. Adding the Vectors up shows that you have a system of equilibrium or not, depending if there is a gap between the first and last vector. The results came out to be complete vectors with the corresponding degrees of the angles with we experimented on.