Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 7.0-kg projectile from rest to a speed of 2.7 x 103 m/s. The net force accelerating the projectile is 8.2 x 105 N. How much time is required for the projectile to come up to speed?
Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 7.0-kg projectile from rest to a speed of 2.7 x 103 m/s. The net force accelerating the projectile is 8.2 x 105 N. How much time is required for the projectile to come up to speed?
College Physics
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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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![### Projectile Motion Experiment
**Scenario:**
Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 7.0-kg projectile from rest to a speed of \( 2.7 \times 10^3 \) m/s. The net force accelerating the projectile is \(8.2 \times 10^5\) N. The problem asks to determine the amount of time required for the projectile to come up to speed.
**Given Data:**
- Mass of the projectile: \( 7.0 \) kg
- Final speed of the projectile: \( 2.7 \times 10^3 \) m/s
- Net force applied to the projectile: \( 8.2 \times 10^5 \) N
**Formula to Use:**
The equation \( F = ma \) relates force, mass, and acceleration. We can rearrange this equation to solve for acceleration (\(a\)):
\[ a = \frac{F}{m} \]
Once we have the acceleration, we can use kinematic equations to find the time (\(t\)). Since the projectile starts from rest, we use:
\[ v = at \]
where \( v \) is the final velocity and \( t \) is the time.
Rearranging to solve for time:
\[ t = \frac{v}{a} \]
**Solution Steps:**
1. Calculate the acceleration:
\[ a = \frac{F}{m} = \frac{8.2 \times 10^5 \text{ N}}{7.0 \text{ kg}} = 1.17 \times 10^5 \, \text{m/s}^2 \]
2. Calculate the time:
\[ t = \frac{v}{a} = \frac{2.7 \times 10^3 \text{ m/s}}{1.17 \times 10^5 \, \text{m/s}^2} \approx 0.023 \text{ s} \]
**User UI:**
- The input text box shows the answer: **0.023**.
- The dropdown menu below the input box selects the unit of measurement, in this case, **seconds (s)**.
- The system provides a hint section, although this hint is not visible in the current image](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e20d9b6-0c20-477e-99bf-00fa906d88b6%2F1bfb4fa8-a382-4099-84f2-92511a65f40d%2Fa4jv5t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Projectile Motion Experiment
**Scenario:**
Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 7.0-kg projectile from rest to a speed of \( 2.7 \times 10^3 \) m/s. The net force accelerating the projectile is \(8.2 \times 10^5\) N. The problem asks to determine the amount of time required for the projectile to come up to speed.
**Given Data:**
- Mass of the projectile: \( 7.0 \) kg
- Final speed of the projectile: \( 2.7 \times 10^3 \) m/s
- Net force applied to the projectile: \( 8.2 \times 10^5 \) N
**Formula to Use:**
The equation \( F = ma \) relates force, mass, and acceleration. We can rearrange this equation to solve for acceleration (\(a\)):
\[ a = \frac{F}{m} \]
Once we have the acceleration, we can use kinematic equations to find the time (\(t\)). Since the projectile starts from rest, we use:
\[ v = at \]
where \( v \) is the final velocity and \( t \) is the time.
Rearranging to solve for time:
\[ t = \frac{v}{a} \]
**Solution Steps:**
1. Calculate the acceleration:
\[ a = \frac{F}{m} = \frac{8.2 \times 10^5 \text{ N}}{7.0 \text{ kg}} = 1.17 \times 10^5 \, \text{m/s}^2 \]
2. Calculate the time:
\[ t = \frac{v}{a} = \frac{2.7 \times 10^3 \text{ m/s}}{1.17 \times 10^5 \, \text{m/s}^2} \approx 0.023 \text{ s} \]
**User UI:**
- The input text box shows the answer: **0.023**.
- The dropdown menu below the input box selects the unit of measurement, in this case, **seconds (s)**.
- The system provides a hint section, although this hint is not visible in the current image
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