Ask Question
30 April, 17:47

A proton moves at 4.50 3 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 3 103 N/C. Ignoring any gravitational effects, find (a) the time interval required for the proton to travel 5.00 cm horizontally, (b) its vertical displacement during the time interval in which it travels 5.00 cm horizontally, and (c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally. 26. Protons are projected with an initial

+4
Answers (1)
  1. 30 April, 21:33
    0
    Horizontal velocity V_x = 4.5 x 10⁵ m / s

    vertical electric field E_y = 9.6 x 10³ N/C

    acceleration in vertical direction a_y = force on proton / mass

    = 9.6 x 10³ x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷

    = 9.2 x 10¹¹ m / s²

    a) In horizontal direction it will move with uniform velocity

    time required = distance / velocity

    = 5 x 10⁻² / 4.5 x 10⁵

    = 1.11 x 10⁻⁷ s

    b)

    vertical displacement in time 1.11 x 10⁻⁷ s

    h = 1/2 x at², initial vertical velocity is zero.

    =.5 x 9.2 x 10¹¹ x (1.11 x 10⁻⁷) ²

    = 5.66 x 10⁻³ m

    5.66 mm

    c)

    Horizontal velocity will be unchanged ie 4.5 x 10⁵ m / s

    vertical velocity will change due to acceleration

    = u + at

    0 + 9.2 x 10¹¹ x1.11 x 10⁻⁷

    = 10.21 x 10⁴ m / s

    = 1.021 x 10⁵ m / s
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A proton moves at 4.50 3 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 3 103 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers