An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.

Explanation:

Given that  

speed u=4×10⁶ m/s

electric field E=4×10² N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates

acceleration a=qE/m=1.6×10⁻¹⁹×4×10²/9.1×10⁻³¹=0.7×10¹⁴=7×10¹³ m/s

now we find the horizantal distance travelled by electrons hit the plates

horizantal distance X=u[2y/a]^1/2

=4×10⁶[2×2×10⁻²/7×10¹³]^1/2

=9.5cm

now we find the velocity f the electron strike the plate

v²-(4×10⁶)²=2×7×10¹³×2×10⁻²

v²=16×10¹²+28×10¹¹

v²=1.88×10¹³m/s

speed after hits =>V=4.34×10⁶ m/s

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.