PLEASE HELP ASAP!!!!

Pro-lab 1a Name: kcad Lab 1a. You can nd how to solve problem 1 by reading through Lab la, and problem 2 in the \"Statistics for Experimental Physics" document (See Section 4: Error Propagation, then Section 3 for how to properly round the result). Then complete this pro-lab and tum it in at the beginning of lab class. 1. Read Appendix A in Lab 1a to see how to read a Vernier caliper. Then give the above reading, including units, to the best of your ability. (don't worry, there are a range of answers I'll accept - I know how difcult it is to see exactly which line lines up). If you need additional help, do a Google search for \"how to read Vernier calipers\". The calipers above have the same scales as the one in Appendix A. (1 point) If the mass ofan object is 32.5 :t 0.5 grams, and its volume is 11.0 :I: .1 cm3, find the density :t error. Show your work (2 points): Density = Appendix A: How to read Vernier Calipers Vernier calipers are the most reliable measuring tools compared to dial calipers and digital calipers. Unlike dial calipers and digital calipers, Vernier calipers can be exposed to coolant, oil, water and dust without worrying about damaging any internal working parts. It will definitely test your eyesight to read the Vernier, however. Take a look at the pictures below, showing measurement of a paperclip. The correct answer is 26.18 mm; read the explanation on the next page, and use it to make sense of the pictures. 50 1 2 60 metric 26 mm 8 0.029 mm O +0.18 mm KUCRAFT 10 20 30 40. 50 O 26 mm +0.18 mm 1wk1-3Metric Reading The lower scale reads in metric, and is the only scale we'll use. Each bar graduation on the main scale is 1 mm. Every 10'11 graduation is numbered (1 cm = 10 mm). Each vernier graduation on the sliding scale is 0.02 mm. Every 5'h graduation (0.1 mm) is numbered. In the above example, the measurement of 26. l 8 millimeters is determined by reading 26 mm (2.6 cm) on the main scale, and then adding 0.18 mm as determined by the vernier. The 26 is determined by looking for the closest tick mark on the main scale that the 0 on the sliding scale is to the right of. The additional 0.18 mm comes from lining up the sliding (vemier) scale and the main scale. Find the point where the lines on the two scales match up best. Look at the associated number on the sliding scale (not the main scale for this one). It's one tick mark left of the 2. Note that there are only ve tick marks between the l and the 2, which is why the far end of the sliding scale tells you that these little ticks are 0.02 mm. Five such ticks adds up to 0.1 mm, so the numbers on the sliding scale are tenths of a m. If the 2 lined up the best with the main scale, we'd need to add 0.2 mm to the main reading, but it doesn't. One left of the 2 is 0.2 mm 0.02 mm = 0.18 mm. So that's the additional piece we need to add onto the main scale reading to achieve our very accurate 26.18 mm reading. Many times it is difcult to determine exactly which line lines up the best with the main scale usually there are 3 that are very close. Pick the middle one, and then estimate the uncertainty. Remember for analog devices, your uncertainty is the smallest division you're sure of divided by 2. So if you are sure which tick marks line up best, your uncertainty will be smaller than if there are three possible tick marks that line up well. Take your reading in millimeters rst, since that is what the calipers read in, then convert to centimeters. 1) In your own words, explain the difference between systematic and random error. 2) Let's say you are using a ruler to measure the sides of a rectangular piece of metal in order to calculate its volume. If you had the choice of a larger piece of metal or a smaller piece of metal, which would you choose if you wanted to minimize the error on the volume calculation? 3) lfyou wanted to state a result with propagated error you would have to use an appropriate amount of digits based on the rules outlined in the "Statistics for Experimental Physics" document. Let's say you wanted to measure a rectangular strip of paper using inches on a ruler with tick marks spaced out 1/16 of an inch. You do the best job you can measuring the paper and you get dimensions of 3 and 5/16 inches for one side and 1 and 1/4 inches on the other. What would be the area of the paper with propagated error in inches