please answer ASAP A proton accelerates from rest in a uniform electric field of

Question

please answer ASAP

A proton accelerates from rest in a uniform electric field of 660 N/C. At one later moment, its speed is 1.10 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. m/s^2 (b) Over what time interval does the proton reach this speed? s (c) How far does it move in this time interval? m (d) What is its kinetic energy at the end of this interval? J An electron and a proton are each placed at rest in a uniform electric field of magnitude 5.70 N/C. Calculate the speed of each particle 45.2 ns after being released. electron m/s proton m/s A proton moves at 3.60 times 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 7.60 times 10^3 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.50 cm horizontally. ns (b) Find its vertical displacement during the time interval in which it travels 4.50 cm horizontally. (Indicate direction with the sign of your answer.) mm (c) Find the horizontal and vertical components of its velocity after it has traveled 4.50 cm horizontally. v vector = (i + j) km/s

Explanation / Answer

(8)

Given that,

electric field E = 660 N/C

final speed of proton v = 1.1*10^6 m/s

initial speed u = 0

(a)

From equilization of forces,

ma = qE

a = qE / m

where, q = charge of proton

m = mass of proton

a = 1.6*10^(-19) * 660 / 1.67*10^(-27)

a = 6.32*10^10 m/s^2

(b)

let, time taken to reach the speed = t

from equation of motion,

v = u + at

t = v - u / a

t = 1.1*10^6 - 0 / 6.32*10^10

t = 1.74*10^(-5) s

(c)

Let, distance travelled = d

From kinematic equation,

d = ut + (1/2)at^2

d = 0 + (1/2)*6.32*10^10*(1.74*10^(-5))^2

d = 9.56 m

(d)

Kinetic energy of proton,

KE = (1/2)mv^2

KE = (1/2)*1.67*10^(-27)*(1.1*10^6)^2

KE = 1.01*10^(-15) J

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