This cannot be hand-written, please type out the solutionUse the average time to drop and calculate the distance dropped using the equation given. Then compare this distance to the measured distance. Use Y = 1/2 g t^2, Thus Y= 4.9 t^2See my attempt attached with the data - teacher indicated my work needed correction, please helpThank you! You will drop one of the hex nuts from a height of about 6 ft. Stand near awall, door jam, post of some kind and place a tape or have some otherway of marking the height you will drop the hex nut from. Record thisheight.With a stopwatch time the drop of the hex nut and record in the data tablebelow. Repeat this for a total of 10 trials.Height of Drop: 6FTData table for hex nut drop experiment:Drop Number1Calculations:2345JWH678910AverageStandard DeviationTime (sec)0.630.610.660.670.590.630.640.690.620.680.6420.0322Using the average time from the table calculate the distance the hex nutdropped and compare to your recorded distance. y = 1/2* g * t^2where y is the distance dropped, g is the gravitational acceleration (about 9.8m/s2), and t is the time taken. Let's start by converting 6 feet to meters. Because1-foot equals 0.3048 meters, the height in meters is:6 feet tall * 0.3048 meters/foot = 1.8288 metersWe can now determine the distance dropped:y = 1/2* (9.8 m/s^2) * (0.642 s) ^2y≈ 1/2 * 9.8* 0.412164 y≈2.02074 meters2Physics I

The object is dropped from height 6 ft Now, y= 6 ft=6×0.3048 m= 1.8288 m Time required to drop =…

Answered: Use the average time to drop and…

Use the average time to drop and calculate the distance dropped using the equation given. Then compare this distance to the measured distance. Use Y = 1/2 g t^2, Thus Y= 4.9 t^2

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter5: Newton's Law Of Motion

Section: Chapter Questions

Problem 89AP: In building a house, carpenters use nails from a large box. The box is suspended from a spring twice...

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This cannot be hand-written, please type out the solution

Use the average time to drop and calculate the distance dropped using the equation given. Then compare this distance to the measured distance. Use Y = 1/2 g t^2, Thus Y= 4.9 t^2

See my attempt attached with the data - teacher indicated my work needed correction, please help

Thank you!

You will drop one of the hex nuts from a height of about 6 ft. Stand near a
wall, door jam, post of some kind and place a tape or have some other
way of marking the height you will drop the hex nut from. Record this
height.
With a stopwatch time the drop of the hex nut and record in the data table
below. Repeat this for a total of 10 trials.
Height of Drop: 6FT
Data table for hex nut drop experiment:
Drop Number
1
Calculations:
2
3
4
5
JWH
6
7
8
9
10
Average
Standard Deviation
Time (sec)
0.63
0.61
0.66
0.67
0.59
0.63
0.64
0.69
0.62
0.68
0.642
0.0322
Using the average time from the table calculate the distance the hex nut
dropped and compare to your recorded distance. y = 1/2* g * t^2
where y is the distance dropped, g is the gravitational acceleration (about 9.8
m/s2), and t is the time taken. Let's start by converting 6 feet to meters. Because
1-foot equals 0.3048 meters, the height in meters is:
6 feet tall * 0.3048 meters/foot = 1.8288 meters
We can now determine the distance dropped:
y = 1/2* (9.8 m/s^2) * (0.642 s) ^2
y≈ 1/2 * 9.8* 0.412164 y≈
2.02074 meters
2
Physics I

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