1 mile (1.61km/mi)(1000m/km)= 1610m----S = vt-----t = S/v----t = 1610m /2.10 ------t = 766.67 secs------t = 766.67/60 --> (12.78) minutes 3) Keira starts at position = 24 along a coordinate axis. She then undergoes a displacement of -42 . What is her final position? (24+-42)=-18.0m 4) It takes Harry 36 to walk from = -14 to = -52 .Velocity? (-14--52=38/36)=-1.1m/s 5) The images of trees in the figure come from a catalog advertising fast-growing trees.
height; age; sex Results Table 2: Average Breathing Rates and Lung Volumes Breathing Rate 12.2 10.8 10.4 11.1 TV(L) 0.5 0.5 0.5 0.5 Resting Values ERV(L) IRV(L) 1.4 1.3 1.4 1.4 2.3 2.2 2.3 2.3 RV(L) 26.6 29.5 26.1 27.4 Breathing Rate 1.7 1.7 1.8 1.7 TV(L) 0.6 0.7 0.7 0.7 Exercising Values ERV(L) IRV(L) 1.9 1.7 2 1.9 1.6 1.6 1.6 1.6 RV(L) 1.6 1.6 1.6 1.6 Subject1 Subject2 Subject3 Averages Comparison of Resting and Exercising Lung Volumes and Breathing Rate 1. Does exercise increase, decrease, or does not change breathing rate? Exercise increases the breathing rate. 2. Does exercise increase, decrease, or does not change TV?
|Mass of insulated test tube, spin bar, cyclohexane and |353.25g | | |unknown | | |4. |Mass of unknown |0.15g | A. Warming Curve Data-Pure Cyclohexane |Time |Temp °C |Time |Temp °C |Time |Temp °C |Time |Temp °C | |2.56.30 |6.9 |2.59.30 |6.8 |3.02.30 |6.8 |3.05.30 |7.9 | |2.57.00 |6.9 |3.00.00 |6.9 |3.03.00 |6.9 |3.06.00 |8.2 | |2.57.30 |6.9 |3.00.30 |6.9 |3.03.30 |6.9 |3.06.30 |8.5 | |2.58.00 |6.8 |3.01.00 |6.8 |3.04.00 |7 |3.07.00 |8.9 | |2.58.30 |6.8 |3.01.30 |6.8 |3.04.30 |7.1 |3.07.30 |9.2 | |2.59.00 |6.9 |3.02.00 |6.8 |3.05.00 |7.6 |3.08.00 |9.5 | [pic] 5. Freezing point of pure cyclohexane, from warming Curve is 6.0°C B. Warming Curve Data-Solution Containing Unknown – 941 Trial 1 |Time
Data: Procedure 1 m = .0657kg | M= .2426kg | y1= .059cm | Trial | p | y2 (m) | y2-y1 (m) | V (m/s) | Xvo (m/s) | 1 | 39⁰ | .145 | .087 | 1.31 | .315 | 2 | 38.5⁰ | .144 | .086 | 1.30 | .315 | 3 | 38.5⁰ | .143 | .084 | 1.28 | .311 | 4 | 38.5⁰ | .144 | .085 | 1.29 | .313 | 5 | 38.5⁰ | .144 | .085 | 1.29 | .313 | Our m is the mass of our ball and M is the mass of the pendulum just by itself. Our y1 is the distance from the table to the free hanging pendulum. To get V, we took the square root of 2g(y2-y1) which aided in calculating the initial velocity. The equation for Xvo was the mass of the ball plus the mass of the pendulum, divided by the mass of the ball all multiplied by the number we got for V. To get our y2 we measured the height of the pendulum
Together they have 160 foreign coins. Find how many coins Mary has. 36) 37) The area of a circle is found by the equation A = r2. If the area A of a certain circle is 36 square centimeters, find its radius r. 37) 10 38) Find the volume V of a sphere of diameter 3 ft. Use 3.14 for . If necessary, round the result to the nearest tenth.
The purpose of this experiment was to measure centripetal acceleration and a centripetal force of a mass. The mass was then hung on an apparatus and attached to a spring. The mass was then rotated. The centripetal force was calculated using measurements of radius of the path, the time it takes to revolve around that path, and the mass. Procedure The group first took measurements such as the mass of the object, the radius of the rotation, the tension of the mass when we attached it to the apparatus.
Arkansas, 435 U.S. 475 (1978). ............................................ 13 Morgan v. United States, 309 F.2d 234, 237 (D.C. Cir. 1962) ...................... 8 People v. Belge, 83 Misc. 2d 186 (N.Y. County Ct. 1975) ......................... 14 People v. Collie, 30 Cal. 3d 43 (1981)....................................................... 13 Richmond Newspapers, Inc. v. Virginia, 448 U.S. 555 (1980) ................... 13 State v. Kociolek, 23 N.J. 400 (1957)........................................................ 14 United States ex.
Radioactive Decay Simulation Experiment quantitative data Test 1 Dice: Roll Number | Number of Dice left | 0 | 100 | 1 | 77 | 2 | 60 | 3 | 53 | 4 | 43 | 5 | 35 | 6 | 29 | 7 | 26 | 8 | 20 | 9 | 14 | 10 | 11 | 11 | 6 | Test 2 Dice: Roll Number | Number of Dice | 0 | 100 | 1 | 84 | 2 | 68 | 3 | 58 | 4 | 51 | 5 | 46 | 6 | 38 | 7 | 33 | 8 | 28 | 9 | 26 | 10 | 20 | 11 | 19 | 12 | 16 | 13 | 14 | 14 | 8 | Test 1 Coins: Roll Number | Number of coins | 0 | 100 | 1 | 49 | 2 | 23 | 3 | 13 | 4 | 6 | 5 | 4 | 6 | 3 | 7 | 2 | 8 | 1 | 9 | 0 | Test 2 Coins: Roll Number | Number of coins | 0 | 100 | 1 | 43 | 2 | 25 | 3 | 14 | 4 | 8 | 5 | 3 | In the graphs above show the roll number as the time period and the number of coins left as the particles remaining, therefore simulating radioactive decay on 4 instances. Qualitative data As this is only a simulation there are limits as to how accurate the results can be. The results are not consistent as the probably of decay in dice is 1 in 6 where as with the coins the probability of decay is 1 in 2. The box shaken in this experiment was not of a consistent amount of times or force due to different people shaking it. Some of the dice coins landed on top of one another and therefore had to be moved to show all the dice, this would reduce the randomness of shaking the box.
Displace the masses to cause oscillation 5. Measure the time taken for 10 oscillations, therefore get the time for 1 oscillation Set up the experiment as shown below: Results: length | Time | Ball size | 138.5 | 2.48 | S | 45.8 | 1.39 | S | 39.2 | 1.19 | S | 32.5 | 1.2 | M | 42.8 | 1.25 | M | 36.5 | 1.18 | M | 33.4 | 1.15 | M | 26.4 | 0.94 | L | 19.2 | 0.8 | L | 12 | 0.63 | L | Analysis: The graph of pendulum is shown on the page enclosed. The graph of the time taken for 1 oscillation and length of pendulum is a linear graph. The above table shows that time period of a pendulum varies with the length of the pendulum.The time taken for 1 oscillation is calculated by dividing the time for 10 oscillations. Question discussion: Why did we time 10 oscillations of the pendulum?
Physics 201 Section 401 ARCHIMEDES’ PRINCIPLE Jordan Nix Partners: Kelsey Carter and Taylor Pfau Date Performed: November 10, 2014 TA: Andrew Whitley Abstract Table 1 | Mass (g) | Mass Submerged (g) | Mass Apparent (m') (g) | Radius (cm) | Length (cm) | Length Submerged (cm) | Volume (cm3) | Aluminum | 58.50 | | 42.00 | 0.95 | 7.60 | | 21.32 | Wood | 30.50 | 8.85 | 20.00 | 1.15 | 9.89 | 2.90 | 12.05 | Table 2 | Theoretical Density(theo) (g/cm3) | Experimental Density(exp) (g/cm3) | Percent Difference(%) | Aluminum | 2.74 | 3.54 | 29.20 | Wood | 2.54 | 2.90 | 14.17 | Discussion In the first part, a 58.5 g aluminum cylinder was hung from a string attached to a scale and submerged in a graduated cylinder filled with water. The cylinder displaced approximately 17 mLs of water and had an “apparent mass” of 42 g while submerged. Two oppositional forces acted upon the cylinder: gravity drew it downward while buoyant force pushed it towards the water’s surface. Using Archimedes’ Principle that force on a body in a fluid is equal in magnitude to the weight displaced by the fluid, we were able to calculate the experimental density of the aluminum cylinder as 3.54 g/cm3. Measurements of the radius and length of the cylinder provided its volume and theoretical density of 2.74 g/cm^3 which confirmed Archimedes’ Principle.