Hi, I need assistance on this one problem. I need all work shown to understand. The distance between windowsills I am confused about.
See attachment for diagram:
Please apply the techniques used in Lab 3 (particularly the ticker tape) to the following photo:
Distance between windowsills is 3.15 meters. Three photographs were taken each second
Present all data using excel:
- Table presenting all measured values
- Displacement vs. time
- Velocity vs time.
Comment on the values obtained for acceleration of an object as compared to the theoretical value of gravity.© BrainMass Inc. brainmass.com September 29, 2022, 2:03 pm ad1c9bdddf
SOLUTION This solution is FREE courtesy of BrainMass!
Greetings: The distance between windowsills gives you a (rather rough) measure of the distance that the falling object has fallen since being released at the starting point alongside a windowsill. The downward displacements from release point are: Windowsill #0 at 0, #1 at 3.15 m, #2 at 6.30 m, #3 at 9.45 m, #4 at 12.60 m, and so on.
Step 1. Holding a metric ruler vertically alongside adjacent windowsills gives 3.6 cm representing 3.15 m which gives a scale
of 3.15 m per 3.6 cm of measured distance, which is .875 m for every cm of vertical distance.
Step 2. With the ruler vertically alongside the falling object's positions , the distance from starting point at windowsill #0 to
successive positions is: position #0 is at 0 m, #1 is at .875 m, #2 is at 1.75 m, #3 is at 2.625 m and so on.
Step 3. Starting from rest when photographed at at windowsill #0. Time elapsed between positions is .33 second.
Step 4. The time of successive positions is: at release, 0 sec, position #1 at .33 sec, #2 at .66 sec, #3 at .99 sec, #4 at 1.32
sec and so on.
A table of the above values and also equations involving them is shown on a pdf document labeled ATTACHMENT A.
Graphs from the attachment are with time horizontal and the other parameter vertically.
I hope this is helpful for you.© BrainMass Inc. brainmass.com September 29, 2022, 2:03 pm ad1c9bdddf>