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- Match the equivalent quantities: 160.2 µC A. 1.602 x 10-4c 160.2 nC B. 2.56 m v k C. 25600 m v 256 cm D. 9 x 10° N-m2/c2 E. 1.602 x 10- cS 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? acerDOWN 1. This is the term for molecules which have a slightly negative or positive side
- Question 7 of 15 A light ray refracts at the interface between two transparent materials as shown below. In material 1 the light ray is traveling at v₁ = 2.5 x 108 m/s and in material 2 it is traveling at a speed v2. What is v2? Material 1 Material 2 60° 30°D-E please!Some possibly useful numbers = 8.99 x 10° N-m²/C² €0 = 8.85 × 10-12C²/Nm² e = 1.6 × 10–19 C = 9.11 × 10-31kg mp = 1.67 x 10–27kg leV 1.6 x 10-19J %3D