Question
In: Physics
A proton moves through a uniform magnetic field given by B→=(5.52i^−10.0j^+20.9k^) mT. At time t1, the proton...
A proton moves through a uniform magnetic field given by B→=(5.52i^−10.0j^+20.9k^) mT. At time t1, the proton has a velocity given by v→=vxi^+vyj^+(2.19 km/s)k^ and the magnetic force on the proton is F→B=(3.69×10−17 N)i^+(2.03×10−17 N)j^ . (a) At that instant, what is vx? (b) At that instant, what is vy?
(a) _ m/s
(b) _ m/s
Solutions
genius_generous answered 8 months agoRelated Solutions
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