A = 10.0 m at 30 degrees above the +x-axis and B = 12.0 m at 60 degrees above the +x-axis. What is the magnitude of A+B ? 22.0 21.3 15.4 12.2 10.0
Added by Jose Antonio H.
Step 1
For vector A: Ax = A*cos(30) = 10.0 m * cos(30) = 8.66 m Ay = A*sin(30) = 10.0 m * sin(30) = 5.0 m For vector B: Bx = B*cos(60) = 12.0 m * cos(60) = 6.0 m By = B*sin(60) = 12.0 m * sin(60) = 10.39 m Now, we can add the x and y components of vectors A and B to Show more…
Show all steps
Your feedback will help us improve your experience
Suman K and 66 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
(a) Two vectors A and B have precisely equal magnitudes. For the magnitude of A + B to be larger than the magnitude of A - B by the factor n, what must be the angle between them? (b) What must be the value of the angle between the vectors if the value of n is 100?
Andrew S.
In the sum $\vec{A}+\vec{B}=\vec{C},$ vector $\vec{A}$ has a magnitude of 12.0 $\mathrm{m}$ and is angled 40. $.0^{\circ}$ counterclockwise from the $+x$ direction, and vector $\vec{C}$ has a magnitude of 15.0 $\mathrm{m}$ and is angled $20.0^{\circ}$ counterclockwise from the $-x$ direction. What are (a) the magnitude and (b) the angle (relative to $+x )$ of $\vec{B}$ ?
In the sum $\vec{A}+\vec{B}=\vec{C}$, vector $\vec{A}$ has a magnitude of $12.0 \mathrm{~m}$ and is angled $40.0^{\circ}$ counterclockwise from the $+x$ direction, and vector $\vec{C}$ has a magnitude of $15.0 \mathrm{~m}$ and is angled $20.0^{\circ}$ counterclockwise from the $-x$ direction. What are (a) the magnitude and (b) the angle (relative to $+x$ ) of $\vec{B}$ ?
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD