A proton is projected in the positive x direction into a region of uniform electric field E = (-5.30 x 105) N/C at t = 0. The proton travels 6.40 cm as it comes to rest. (a) Determine the acceleration of the proton. 0.9638e7 X How..do.you find the acceleration of an object if you know the net force that acts on it? m/s² magnitude direction --Select- (b) Determine the initial speed of the proton. 0.351e7 X The electric field is constant, so the force is constant, which means the acceleration will be constant. m/s direction -Select-- magnitude (c) Determine the time interval over which the proton comes to rest. 0.3640 X You appear to have calculated the time correctly using your incorrect results from parts (a) and (b). s
A proton is projected in the positive x direction into a region of uniform electric field E = (-5.30 x 105) N/C at t = 0. The proton travels 6.40 cm as it comes to rest. (a) Determine the acceleration of the proton. 0.9638e7 X How..do.you find the acceleration of an object if you know the net force that acts on it? m/s² magnitude direction --Select- (b) Determine the initial speed of the proton. 0.351e7 X The electric field is constant, so the force is constant, which means the acceleration will be constant. m/s direction -Select-- magnitude (c) Determine the time interval over which the proton comes to rest. 0.3640 X You appear to have calculated the time correctly using your incorrect results from parts (a) and (b). s
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![### Physics Problem: Proton in a Uniform Electric Field
A proton is projected in the positive x direction into a region of uniform electric field \(\vec{E} = (-5.30 \times 10^5 \vec{i}) \ \text{N/C}\) at \(t = 0\). The proton travels 6.40 cm as it comes to rest.
#### (a) Determine the acceleration of the proton.
- **Magnitude:**
- Attempt: \(0.9638 \times 10^7 \ \text{m/s}^2\)
- Feedback: How do you find the acceleration of an object if you know the net force that acts on it?
- **Direction:**
- Selection needed
#### (b) Determine the initial speed of the proton.
- **Magnitude:**
- Attempt: \(0.351 \times 10^7 \ \text{m/s}\)
- Feedback: The electric field is constant, so the force is constant, which means the acceleration will be constant.
- **Direction:**
- Selection needed
#### (c) Determine the time interval over which the proton comes to rest.
- **Time Interval:**
- Attempt: \(0.3640 \ \text{s}\)
- Feedback: You appear to have calculated the time correctly using your incorrect results from parts (a) and (b).
### Explanation of the Steps Involved:
1. **Acceleration Calculation:**
- To find the acceleration (\(a\)), use the equation derived from Newton's second law:
\[
F = ma
\]
Here, \(F\) is the force on the proton due to the electric field \( \vec{E} \), and \(m\) is the mass of the proton.
The force \( F \) can be calculated as:
\[
F = qE
\]
where \( q \) is the charge of the proton (\(1.602 \times 10^{-19} \ \text{C}\)).
Therefore:
\[
a = \frac{F}{m} = \frac{qE}{m}
\]
2. **Initial Speed Calculation:**
- To find the initial speed (\(v_0\)), use the kinematic equation:
\[
v_f^2 = v_0^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78db42f9-6d62-48c9-a117-b436e469106b%2Fd8f2a3e0-74c3-41dc-a569-2c39bf68ca93%2Fvlai5i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Proton in a Uniform Electric Field
A proton is projected in the positive x direction into a region of uniform electric field \(\vec{E} = (-5.30 \times 10^5 \vec{i}) \ \text{N/C}\) at \(t = 0\). The proton travels 6.40 cm as it comes to rest.
#### (a) Determine the acceleration of the proton.
- **Magnitude:**
- Attempt: \(0.9638 \times 10^7 \ \text{m/s}^2\)
- Feedback: How do you find the acceleration of an object if you know the net force that acts on it?
- **Direction:**
- Selection needed
#### (b) Determine the initial speed of the proton.
- **Magnitude:**
- Attempt: \(0.351 \times 10^7 \ \text{m/s}\)
- Feedback: The electric field is constant, so the force is constant, which means the acceleration will be constant.
- **Direction:**
- Selection needed
#### (c) Determine the time interval over which the proton comes to rest.
- **Time Interval:**
- Attempt: \(0.3640 \ \text{s}\)
- Feedback: You appear to have calculated the time correctly using your incorrect results from parts (a) and (b).
### Explanation of the Steps Involved:
1. **Acceleration Calculation:**
- To find the acceleration (\(a\)), use the equation derived from Newton's second law:
\[
F = ma
\]
Here, \(F\) is the force on the proton due to the electric field \( \vec{E} \), and \(m\) is the mass of the proton.
The force \( F \) can be calculated as:
\[
F = qE
\]
where \( q \) is the charge of the proton (\(1.602 \times 10^{-19} \ \text{C}\)).
Therefore:
\[
a = \frac{F}{m} = \frac{qE}{m}
\]
2. **Initial Speed Calculation:**
- To find the initial speed (\(v_0\)), use the kinematic equation:
\[
v_f^2 = v_0^2
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