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- A total hip replacement consisting of a 28 mm diameter Co-Cr femoral head articulating against a UHMWPE cup functions in a patient for 5 years before it is removed during revision arthroplasty. You perform a detailed analysis of the explanted cup. The articular surface of the explanted UHMWPE cup shows a change in geometry such that the cup is 0.75 mm deeper (see photo). Radiograph of THR at the time of surgery Assumptions: 28 mm Patient weighs 70kg with an average load of a gate cycle of 2.5BW Neglect creep for all questions Assume 1 million steps per year ANSWER COLUMN THR at day 1 QUESTIONS 0.75 mm THR at 5 years Total angular motion per gait stride (hip flexion-extension-flexion OR heel strike to heel strike) is 80 degrees Hint: Recall details of UHMWPE materials and applications in THR from Lecture 9 Notes. UHMWPE after 5 years showing progression of worn region (dark grey) Q1B: Calculate the number of wear particles produced per year, assuming spherical UHMWPE particles of 5…The phenol concentration of wastewater was determined from three (3) parallel determinations, giving a mean of 0.513 μg L-1 and a standard deviation of 0.05 μg L-1. Decide at a risk level of 5% (or a confidence level of 95%) whether a reference value of 0.520 μg L-1 is statistically different.The organophosphate Azinphos-methyl has a96-hour LC50 value of 3 ppb for rainbow trout. Inan unfortunate incident, 200 g of this pesticidewas sprayed on a field, and a subsequent heavyrainfall washed 35% of it into a nearby lakehaving a surface area of 30,000 m2and anaverage depth of 0.5 m. Would the pesticideconcentration in the lake water have beensufficient to kill a significant fraction of therainbow trout in the lake?
- Given the extinction coefficient for a species is 650 M-1cm-1 and the Absorbance measured at lambda max is .79 in a 1cm pathlength cuvette, what is the concentration(in units of mM) of the species.Prepare 1L 20X Trace minerals stock in a media bottle: Fill out the last column of this table: Component NaCl MnSO4.H₂O H3BO3 CaSO4.5H2O NazMo04.2H₂O COSO4.7H2O KI CaCl2.2H₂O NiCl2 ZnSO4.7H2O FeSO4(NH4)2. 6H20 1X concentration 0.06 g/L 0.024 g/L 0.5 mg/L 0.5 mg/L 2 mg/L 0.0023 mg/L 0.1 mg/L 0.2 g/L 0.0025 mg/L 12 mg/L 35 mg/L 20X (Provide unit)Relative Microbial Activity 1.00 0.80 0.60 0.40 0.20 0.00 20 ⚫02 consumption • CO2 production 40 60 80 100 Water Filled Pore Space (% of total pores) Figure 1. The relationship between microbial activity and water filled pore space. The blue solid line indicates O2 consumption and the orange dashed line indicates CO2 production. 4. With your new-found appreciation for the cause(s) of the relationship between 02/CO2 and water filled pore space, I will now ask you to draw on Figure 1 to show the relative rate of N2 production from 10 to 95% water filled pore space. What is molecule that is the e acceptor when N2 is produced?
- 46. is a proximate value for particle density of most soils. 0 2.65 g/cm3 O2.66 g/cm3 O 2.55 g/cm3 O2.45 g/cm3Plants absorb several ions and water from soil. For this reason, it is necessary for the soil to have a pH value at the range of 6.0 – 8.0. Again, if the p4 is less than 3.0 or greater than 10, then the beneficial microbes of the soil die. Normally rainwater is slightly acidic, its p# value is 5.5. During thunderstorm nitrogen and oxygen of air react to form nitric oxide, which is oxidized to form nitrogen-dioxide. This nitrogen-dioxide reacts with water vapor and forms nitrous acid and nitric acid. Acid-rain occurs because of it. 2NO2+ H2O → HNO2+HNO3 Examination of soil in a region of Rajshahi revealed that its p" value was 6.43. Nitrogen-dioxide is formed in the air because of thunderstorms, and its concentration is 0.297 ppm. What will be the difference between the p" value of the soil after rain and the lowest standard pH value ?RESULTS Please give the time-absorbance values tab? Draw graph-growth curve in Excel? Find the equation of slope line? Calculate the generation time? Choose the values in log phase Abs 0.005 25 0.003 50 0.023 75 0.026 100 0.029 125 0.035 150 0.1 175 0.26 200 0.32 225 0.375 250 0.42 275 0.46 300 0.475 325 0.489 350 0.506 375 0.515 400 0.521 425 0.526 450 0.532 475 0.522 500 0.515 525 0.514 550 0.506 575 0.489 600 0.465 625 0.42 650 0.38 675 0.36 700 0,345 725 0.333 BACTERIAL GRO WTH CURVE
- The number of cells in a culture is estimated based on turbidity. If the following standards are used for comparison: Number of cells/mL Optical Density (measure of turbidity) 0.025 0.050 000'000'৮ 000 000'000' కరరక How many cells are present in a culture that has an optical density of 0.075? O A. Less than 1,000,000 O B. Between 2,000,000 and 4,000,000 O C. 16,000,000 O D. Between 4,000,000 and 8,000,000 Reset SelectionResearchers examining decomposition in a freshwater marsh estimated the tollowing decomposition rate constants: k 0.058-0.61 d1 for the soft tissues of fish and waterfowl; and k = 0.03-0.05 d for aquatic plants. Decomposition was measured under similar environmental conditions. %3D %3D (A) Based on the estimated rate constants, which type of material-animal (i.e. fish, waterfowl) or plant (i.e. aquatic plants)-decomposes faster? (B) What is a likely explanation for this pattern? E.g., how would C:N compare between the two materials? [2-3 sentences]explain the results of this experiment Table 1 Absorption Spectrum of Cobalt Chloride Using a Spectrophotometer Wavelength Absorbance 400 410 420 430 440 450 460 370 480 490 500 510 520 530 540 550 560 570 580 590 600 610 620 630 640 650 660 670 680 690 700 0.036 0.044 0.063 0.095 0.147 0.215 0.284 0.329 0.362 0.402 0.450 0.479 0.464 0.398 0.314 0.216 0.142 0.087 0.054 0.041 0.036 0.036 0.034 0.028 0.030 0.023 0.023 0.023 0.014 0.014 0.011 Test Tube Number Cobalt Chloride Concentration (mol/mL) Absorbance at 510 nm 1 0.000 0.000 2 0.009 0.042 3 0.018 0.92 4 0.027 0.139 5 6 0.036 0.045 0.190 0.236 7 (Unknown) 0.151