dN F(V)dV : N ‚MV². -)dv 2RT M = 47(- 3?v² exp(÷ 2 7RT' Find the Vm value by considering the following curve ax 10 20 x 10 298 I5 x 10 10 x 10 1500 K Sx 10 Vm Vm
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- why did you do times 3.2?Instrument Ruler Quantities Diameter Units m Meas #1 Meas #2 Meas #3 Meas #4 Meas #5 Meas #6 0.011 0.017 0.019 0.016 0.009 0.021 Vernier Caliper Micro-meter Diameter Diameter 0.0120 0.0166 0.0182 0.0151 0.0100 0.0225 0.01268 0.01745 0.01904 0.01587 0.00947 0.02221 Ruler Volume 7.0E-07 2.6E-06 3.6E-06 2.1E-06 4.E-07 4.8E-06 Measurements: diameter and volume. Results: mass Vernier Caliper Volume Measurements: diameter and mass. Results: volume O Measurements: diameter and mass. Results: none. 9.05E-07 2.40E-06 3.16E-06 1.80E-06 5.24E-07 5.96E-06 Among this data, what is considered raw measurement and plain result? Micro-meter Volume 1.067E-06 2.782E-06 3.614E-06 2.093E-06 4.45E-07 5.736E-06 Measurements: diameter measured with the ruler only and mass. Results: volume. Scale Mass kg 0.00850 0.02080 0.02720 0.01680 0.00389 0.04423Find the number of significant figures in each of the following. (a) 61.8 ± 0.3 (b) 3.86 ✕ 109 (c) 2.80000 ✕ 10−6 (d) 0.0045
- S Can 6 PAR к Торс K Unit K In x K Moti = Cop K Unit S Spee S Topo S Math Micr eb.kamihq.com/web/viewer.html?source-filepicker&document_identifier=137VZR5BZOVSAIMOA55WU555_CvJ9NacO + 100 P e Interpreting Graphs Answer the questions following the graphs on each side Dietance va. Time 2 7 10 11 12 13 14 Time in soconds 1. From 1 second to 2 seconds, how fast is the object traveling. (Take the difference in distance and divide it by the time in between the 2 distances) 2. Is the object going as fast between 9 and 12 seconds as it is between 1 and 4 seconds? How can you tell? 3. What is the motion of the object between 4 and 6 seconds? acerderive the second equation from the firstThe value of (jxk). (kx î) is: O √3 O-1 0 O 3 O +1
- I need help with a University Physics 1 (significant figures) - The problem is described in the image below:a) Write down the metric and the inverse metric, and use the definitionrpµv = 1/2gpσ(δµgvσ + δvgµσ - δσgµv) = rpvµto reproduce the results written above for rθ ØØ and rØ θØ.In Figure (a), string 1 has a linear density of 2.08 g/m, and string 2 has a linear density of 4.84 g/m. They are under tension due to the hanging block of mass M = 677 g. Calculate the wave speed on (a) string 1 and (b) string 2. (Hint: When a string loops halfway around a pulley, it pulls on the pulley with a net force that is twice the tension in the string.) Next the block is divided into two blocks (with M, + M2 = M) and the apparatus is rearranged as shown in Figure (b). Find (c) M1 and (d) M2 such that the wave speeds in the two strings are equal. - String 1 String 2 Knot M (а) - String 1 String 2 M1 M2 (b)