The vectors A- and D- in the figure have the following magnitudes: A=8.05 m and
ID: 1536226 • Letter: T
Question
The vectors A- and D- in the figure have the following magnitudes: A=8.05 m and D = 9.6 m(Assume this is a right-handed coordinate system.) Find the magnitude and direction of A- times D Find the magnitude and direction of D- times A-. In a rectangular coordinate system a positive point charge q = 7.20 times 10^-9 C is placed at the point x = +0.150 m, y = 0, and an identical charge is placed at x = -0.150 m, y = 0. Find the x- and y components, the magnitude of the electric field at the following points. A proton is projected into a uniform electric field that points vertically upward and has magnitude E. The initial velocity of the proton has a magnitude v_o and is directed at an angle a below the horizontal. find the maximum distance h_max that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (Use any variable or symbol stated above as necessary.) After what horizontal distance d does the proton return to its original elevation? (Use any variable or symbol stated above as necessary.) Sketch the trajectory of the proton. Find the numerical values of h_max and d if E = 480 N/C, v_0 = 4.50 times 10^5 m/s and a = 31.0 degree.Explanation / Answer
take sine(31) and multiply by v(not) and E,
finding the acceleration when it is crossing the point at which it entered using vector calculus.
h(min) = this number-(9.8m/s^2)(sine31)[v(not)](E).
H(min) = - 1.0902 * 10^9
Acceleration when crossing the "axis" is a function of the gravity and the field which is combatting gravity, equalling the energy of the field minus the energy of constant gravity. Then take sine(59) on a calculator and multiply by [a(axis)(E)]. h(max) = this number-(9.8m/s^2)(sine59)[v(not)](E)].
H(max) = 1.814.* 10^9.
This is assuming Earth gravity.
Take it from here, finding d by subtracting h(max) from h(min).
D = 2.904 * 10^9
I hope that helps!
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