8.) A proton moves at 3.60 x10 5 m/s in the horizontal direction. It enters a un

Question

8.) A proton moves at 3.60 x105 m/s in the horizontal direction. It enters a uniformvertical electric field with a magnitude of 8.60 x 103 N/C. Ignore anygravitational effects. (a) Find the time interval required for theproton to travel 5.00 cm horizontally.
s

(b) Find its vertical displacement during the time interval inwhich it travels 5.00 cm horizontally.
m

(c) Find the horizontal and vertical components of its velocityafter it has traveled 5.00 cmhorizontally.
m/s (horizontal)
m/s (vertical) (a) Find the time interval required for theproton to travel 5.00 cm horizontally.
s

(b) Find its vertical displacement during the time interval inwhich it travels 5.00 cm horizontally.
m

(c) Find the horizontal and vertical components of its velocityafter it has traveled 5.00 cmhorizontally.
m/s (horizontal)
m/s (vertical)

Explanation / Answer

Proton velocity in horizontal direction v = 3.60 x 105 m/s
electric field E = 8.60 x103 N/C (a) Horizontal distance S = 5 cm = 0.05 m timetaken to travel S distance is t = S / v= 1.3888 * 10 ^ -7 s In vertical direction : -------------------- Initial velocity u = 0 Accleration a = Eq / m q = charge of proton = 1.6 * 10 ^ -19 C m = mass of proton = 1.67* 10 ^ -27 kg plug the values weget a = 8.239 * 10 ^ 11 m / s ^2 from this vertical displacement S ' = ut + ( 1/ 2) at^2                                                    = 0 + 7.946* 10 ^ -3 m                                                     =7.946* 10 ^ -3 m

(c) the horizontal component of its velocity after ithas traveled 5.00 cm horizontally is = v = 3.6*10^5 m / s
Let the horizontal component of its velocity after it hastraveled 5.00 cm horizontally be V then from the relation V ^ 2- u ^ 2 = 2aS ' V = [ 2aS ' ]       = 114.426* 10 ^ 3 m /s

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