Explanation:
Given that,
Initial speed of the electron, 
Distance, s = 5 cm = 0.05 cm
Acceleration of the electron,
(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :


v = 871779.788 m/s
or

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:



Hence, this is the required solution.
Answer:
0.60 m/s
Explanation:
The average velocity from t = a to t = b is:
v_avg = (x(b) − x(a)) / (b − a)
Given that x(t) = 0.36t² − 1.20t, and the time is from 1.0 to 4.0:
v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)
v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0
v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0
v_avg = [0.96 − (-0.84)] / 3.0
v_avg = 0.60
The average speed is 0.60 m/s.
Answer:
The most correct option is;
B. 10 km
Explanation:

Where:
y = Distance between the two headlights
d = Aperture of observers eye
λ = Wavelength of light
L = Distance between the observer and the headlight
Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.
Also we have
sinθ = y/L = 1.22 (λ/d)
sinθ = 1.22×10⁻⁴ rad
consider the motion of con from top to bottom
Y = vertical displacement = 1000 m
a = acceleration due to gravity = 9.8 m/s²
v₀ = initial velocity at the top = 0 m/s
v = final velocity at the bottom = ?
using the kinematics equation
v² = v²₀ + 2 aY
v² = 0² + 2 (9.8) (1000)
v = 140 m/s
t = time taken to hit the ground
Using the equation
v = v₀ + at
140 = 0 + 9.8 t
t = 14.3 sec