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2. Knight P11.8 A 2.00 kg object is moving to the right with a speed of 1.00 m/s when it encounters the force shown below. (a) What impulse does the force exert? (b) What is the object's speed and direction after the force ends? F. (N) 2 0 -2 1.0 s t(s) Answers to one significant figure. These are given so you

2. Knight P11.8 A 2.00 kg object is moving to the right with a speed of 1.00 m/s when it encounters the force shown below. (a) What impulse does the force exert? (b) What is the object's speed and direction after the force ends? F. (N) 2 0 -2 1.0 s t(s) Answers to one significant figure. These are given so you

College Physics
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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**Problem Statement:** 2. Knight P11.8: A 2.00 kg object is moving to the right with a speed of 1.00 m/s when it encounters the force shown below. **Questions:** (a) What impulse does the force exert? (b) What is the object's speed and direction after the force ends? **Description of the Graph:** The graph displays Force (F, in Newtons) as a function of time (t, in seconds). The x-axis represents time, ranging from -2 to 2 seconds, with a specific point marked at 0 s and 1.0 s. The y-axis represents force, ranging from -2 to 2 Newtons, with specific points marked at 0 N and 2 N. **Graph Explanation:** - The force is represented by a red rectangle. - The force exists over a 1.0-second interval from 0 s to 1.0 s. - The force exerted is 2.0 N during this interval. The force acts in a constant manner over the 1.0-second duration.
Transcribed Image Text:**Problem Statement:** 2. Knight P11.8: A 2.00 kg object is moving to the right with a speed of 1.00 m/s when it encounters the force shown below. **Questions:** (a) What impulse does the force exert? (b) What is the object's speed and direction after the force ends? **Description of the Graph:** The graph displays Force (F, in Newtons) as a function of time (t, in seconds). The x-axis represents time, ranging from -2 to 2 seconds, with a specific point marked at 0 s and 1.0 s. The y-axis represents force, ranging from -2 to 2 Newtons, with specific points marked at 0 N and 2 N. **Graph Explanation:** - The force is represented by a red rectangle. - The force exists over a 1.0-second interval from 0 s to 1.0 s. - The force exerted is 2.0 N during this interval. The force acts in a constant manner over the 1.0-second duration.
### Useful Information **Gravitational Acceleration** - \( g = 9.8 \, \text{m/s}^2 \) **Gravity** - \( F_g = -mg \) - \( U_g = mgy \) **Power** - \( P = \frac{\Delta E_{\text{sys}}}{\Delta t} \rightarrow \frac{dE_{\text{sys}}}{dt} \) - \( P = F \cos \theta \, v_x \) **Momentum** - \( \vec{p} = m\vec{v} \) **Newton's 2nd Law** - \( \vec{F} = \frac{\Delta \vec{p}}{\Delta t} \rightarrow \frac{d\vec{p}}{dt} \) **Impulse** - \( \Delta p = \int_{t_0}^{t_f} F(t) \, dt \) - This represents the area under the Force-Time (F-t) graph.
Transcribed Image Text:### Useful Information **Gravitational Acceleration** - \( g = 9.8 \, \text{m/s}^2 \) **Gravity** - \( F_g = -mg \) - \( U_g = mgy \) **Power** - \( P = \frac{\Delta E_{\text{sys}}}{\Delta t} \rightarrow \frac{dE_{\text{sys}}}{dt} \) - \( P = F \cos \theta \, v_x \) **Momentum** - \( \vec{p} = m\vec{v} \) **Newton's 2nd Law** - \( \vec{F} = \frac{\Delta \vec{p}}{\Delta t} \rightarrow \frac{d\vec{p}}{dt} \) **Impulse** - \( \Delta p = \int_{t_0}^{t_f} F(t) \, dt \) - This represents the area under the Force-Time (F-t) graph.
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