A velocity vector has a magnitude of 19.0 m/s. If its y component is -12.0 m/s, what are the possible values of its x component? Express your answer in meters per second. ΑΣφ 土 m/s
A velocity vector has a magnitude of 19.0 m/s. If its y component is -12.0 m/s, what are the possible values of its x component? Express your answer in meters per second. ΑΣφ 土 m/s
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![**Vector Magnitude Problem**
A velocity vector has a magnitude of 19.0 m/s. If its \( y \) component is -12.0 m/s, what are the possible values of its \( x \) component?
*Express your answer in meters per second.*
**Input Box for Answer:**
- There is a space provided for entering the answer, with a button to input special symbols or characters often used in mathematical expressions.
**Explanation for Solution:**
To find the possible values of the \( x \) component (\( v_x \)) of the vector, use the Pythagorean theorem in the context of vector magnitudes:
\[
v = \sqrt{v_x^2 + v_y^2}
\]
where:
- \( v \) is the magnitude of the vector (19.0 m/s),
- \( v_y \) is the \( y \) component (-12.0 m/s).
1. Substitute the known values into the formula:
\[
19.0 = \sqrt{v_x^2 + (-12.0)^2}
\]
2. Square both sides to solve for \( v_x^2 \):
\[
361 = v_x^2 + 144
\]
3. Rearrange to find \( v_x^2 \):
\[
v_x^2 = 361 - 144 = 217
\]
4. Calculate \( v_x \) by taking the square root:
\[
v_x = \pm \sqrt{217}
\]
5. Solve for \( v_x \):
\[
v_x \approx \pm 14.73 \, \text{m/s}
\]
Therefore, the possible values for \( x \) component are approximately \( 14.73 \, \text{m/s} \) and \( -14.73 \, \text{m/s} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca2f9b55-192d-4d11-af1a-09dd4cdb3489%2F01fc737a-bbce-42dc-88cd-0a5a0d2e4012%2Frcdkbi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Vector Magnitude Problem**
A velocity vector has a magnitude of 19.0 m/s. If its \( y \) component is -12.0 m/s, what are the possible values of its \( x \) component?
*Express your answer in meters per second.*
**Input Box for Answer:**
- There is a space provided for entering the answer, with a button to input special symbols or characters often used in mathematical expressions.
**Explanation for Solution:**
To find the possible values of the \( x \) component (\( v_x \)) of the vector, use the Pythagorean theorem in the context of vector magnitudes:
\[
v = \sqrt{v_x^2 + v_y^2}
\]
where:
- \( v \) is the magnitude of the vector (19.0 m/s),
- \( v_y \) is the \( y \) component (-12.0 m/s).
1. Substitute the known values into the formula:
\[
19.0 = \sqrt{v_x^2 + (-12.0)^2}
\]
2. Square both sides to solve for \( v_x^2 \):
\[
361 = v_x^2 + 144
\]
3. Rearrange to find \( v_x^2 \):
\[
v_x^2 = 361 - 144 = 217
\]
4. Calculate \( v_x \) by taking the square root:
\[
v_x = \pm \sqrt{217}
\]
5. Solve for \( v_x \):
\[
v_x \approx \pm 14.73 \, \text{m/s}
\]
Therefore, the possible values for \( x \) component are approximately \( 14.73 \, \text{m/s} \) and \( -14.73 \, \text{m/s} \).
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