- #1
chom
- 3
- 0
Hello everyone, I'm new here and a physics noob. I'm doing Physics 024 correspondence via web and my tutor has a (if I'm lucky) a once a week schedule to talk with him. Thought I might have better luck here since I'm on a strict time frame to get this done.
Here goes...
I measured the displacement for each interval of a ball bearing free falling (in a picture in my textbook) next to a meter stick with a strobe light flashing in 0.050 s intervals. From this measurement I calculated the following data:
100mm (length of ruler on page 77)
I then factored the scale by using: 1000mm/100mm (actual length of meter stick)/(length of meter stick on print) = 10mm scale factor
Time(s) 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600
∆d(mm)displacement
Measured for
interval 0.5 4.0 10.0 18.0 29.0 42.5 59.0 77.5 99.0 123.0 150.0 178.5
∆d (cm)actual
Displacement for
interval .50 4.0 10.0 18.0 29.0 43.0 59.0 78.0 99.0 123.0 150.0 180.0
V (cm/s)actual average velocity
For interval 10.0 80.0 200.0 360.0 580.0 860.0 1200.0 1600.0 2000.0 2500.0 3000.0 3600.0
Sorry, the forum wouldn't let me upload my word doc for word 2007 for a proper data table.
To find actual displacement the calculation I used was ∆d(mm) x 10mm(scale factor) = ∆d(mm), then I converted to cm to get the actual displacement for ∆d(cm).
To find actual average velocity I used the calculation: v=(∆d(cm))/0.050s for each interval in question.
Next, I drew a Velocity-Time graph for the motion, I calculated average acceleration using: a=∆v/∆t,
a=3600/0.600 = 6000 = 6.0x10^3cm/s^2
converted to m/s is 6.0x10^1m/s^2 which is very far from the value of g = 9.8m/s^2.
I am fairly new to all of this and any advice to where my error is would be greatly appreciated.
Chom
Here goes...
I measured the displacement for each interval of a ball bearing free falling (in a picture in my textbook) next to a meter stick with a strobe light flashing in 0.050 s intervals. From this measurement I calculated the following data:
100mm (length of ruler on page 77)
I then factored the scale by using: 1000mm/100mm (actual length of meter stick)/(length of meter stick on print) = 10mm scale factor
Time(s) 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600
∆d(mm)displacement
Measured for
interval 0.5 4.0 10.0 18.0 29.0 42.5 59.0 77.5 99.0 123.0 150.0 178.5
∆d (cm)actual
Displacement for
interval .50 4.0 10.0 18.0 29.0 43.0 59.0 78.0 99.0 123.0 150.0 180.0
V (cm/s)actual average velocity
For interval 10.0 80.0 200.0 360.0 580.0 860.0 1200.0 1600.0 2000.0 2500.0 3000.0 3600.0
Sorry, the forum wouldn't let me upload my word doc for word 2007 for a proper data table.
To find actual displacement the calculation I used was ∆d(mm) x 10mm(scale factor) = ∆d(mm), then I converted to cm to get the actual displacement for ∆d(cm).
To find actual average velocity I used the calculation: v=(∆d(cm))/0.050s for each interval in question.
Next, I drew a Velocity-Time graph for the motion, I calculated average acceleration using: a=∆v/∆t,
a=3600/0.600 = 6000 = 6.0x10^3cm/s^2
converted to m/s is 6.0x10^1m/s^2 which is very far from the value of g = 9.8m/s^2.
I am fairly new to all of this and any advice to where my error is would be greatly appreciated.
Chom