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
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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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!
![---
## Hex Nut Drop Experiment
**Objective:**
Conduct an experiment to measure and calculate the time it takes for a hex nut to fall from a height of 6 feet.
### Procedure:
1. Drop one of the hex nuts from a height of about 6 feet. Stand near a wall, door jam, post, or similar fixture, and mark the height from which you will drop the hex nut. Record this height.
2. Use a stopwatch to time the drop of the hex nut and record the time in the data table below. Repeat this process for a total of 10 trials.
### Parameters:
- **Height of Drop:** 6 feet (1.8288 meters)
### Data Table:
| Drop Number | Time (sec) |
|-------------|------------|
| 1 | 0.63 |
| 2 | 0.61 |
| 3 | 0.66 |
| 4 | 0.67 |
| 5 | 0.59 |
| 6 | 0.63 |
| 7 | 0.64 |
| 8 | 0.69 |
| 9 | 0.62 |
| 10 | 0.68 |
| **Average** | **0.642** |
| **Standard Deviation** | **0.0322** |
### Calculations:
**Calculate the distance the hex nut dropped and compare it to the recorded distance.**
**Formula:**
\[ y = \frac{1}{2} \cdot g \cdot t^2 \]
Where:
- \( y \) is the distance dropped
- \( g \) is the gravitational acceleration (approximately 9.8 m/s²)
- \( t \) is the time taken
### Steps:
1. Convert 6 feet to meters:
\[ 6 \text{ feet} \times 0.3048 \text{ meters/foot} = 1.8288 \text{ meters} \]
2. Determine the distance dropped using the average time from the experiment:
\[ y = \frac{1}{2} \cdot (9.8 \text{ m/s}^2) \cdot (0.642 \text{ s})^2 \]
\[ y = \frac{1}{2} \cdot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd02a090e-cd7e-49b1-8a18-39d4656afeb3%2F4af536bd-2cdb-4db0-a082-831955dba96d%2Fsy39964_processed.png&w=3840&q=75)
Transcribed Image Text:---
## Hex Nut Drop Experiment
**Objective:**
Conduct an experiment to measure and calculate the time it takes for a hex nut to fall from a height of 6 feet.
### Procedure:
1. Drop one of the hex nuts from a height of about 6 feet. Stand near a wall, door jam, post, or similar fixture, and mark the height from which you will drop the hex nut. Record this height.
2. Use a stopwatch to time the drop of the hex nut and record the time in the data table below. Repeat this process for a total of 10 trials.
### Parameters:
- **Height of Drop:** 6 feet (1.8288 meters)
### Data Table:
| Drop Number | Time (sec) |
|-------------|------------|
| 1 | 0.63 |
| 2 | 0.61 |
| 3 | 0.66 |
| 4 | 0.67 |
| 5 | 0.59 |
| 6 | 0.63 |
| 7 | 0.64 |
| 8 | 0.69 |
| 9 | 0.62 |
| 10 | 0.68 |
| **Average** | **0.642** |
| **Standard Deviation** | **0.0322** |
### Calculations:
**Calculate the distance the hex nut dropped and compare it to the recorded distance.**
**Formula:**
\[ y = \frac{1}{2} \cdot g \cdot t^2 \]
Where:
- \( y \) is the distance dropped
- \( g \) is the gravitational acceleration (approximately 9.8 m/s²)
- \( t \) is the time taken
### Steps:
1. Convert 6 feet to meters:
\[ 6 \text{ feet} \times 0.3048 \text{ meters/foot} = 1.8288 \text{ meters} \]
2. Determine the distance dropped using the average time from the experiment:
\[ y = \frac{1}{2} \cdot (9.8 \text{ m/s}^2) \cdot (0.642 \text{ s})^2 \]
\[ y = \frac{1}{2} \cdot
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