Respuesta :
Answer: a) 2.5 * 10^14, b) t = 1.2*10^-8 s, c) F = 2.2775 * 10^-15 N
Explanation: Since it starts from rest, initial velocity = 0, final velocity (v) = 3*10^6 m/s, distance covered (s) = 1.80cm = 1.80/100 = 0.018m
Since the force on the electron is constant, it acceleration will be constant too thus making newton's laws of motion valid.
Question a)
To get the acceleration, we use the formulae that
v² = u² + 2as
But u = 0
v² = 2as
(3*10^6)² = 2*a*(0.018)
9* 10^12 = 0.036*a
a = 9 * 10^12 / 0.036
a = 250 * 10^12
a = 2.5 * 10^14 m/s².
Question b)
To get the time, we use
v = u + at
But u = 0
v = at
3*10^6 = 2.5 * 10^14 * t
t = 3*10^6 / 2.5*10^14
t = 1.2*10^-8 s
Question c)
To get the force, we use the formulae below
F = ma
F = 9.11*10^-31 * 2.5 * 10^14
F = 22.775 * 10^-17
F = 2.2775 * 10^-15 N
a) The acceleration of the electron as it travels in a straight line to the grid is 2.5 × 10¹⁴m/s².
b) The time taken for the electron to reach the grid is 1.2 × 10⁻⁸ seconds.
c) The net force that is accelerating the electron is 2.28 × 10⁻¹⁶ Newton.
Given the data in the question
- Mass of electron; [tex]m = 9.11 * 10^{31} kg[/tex]
- Initial velocity; [tex]u = 0[/tex]'
- Final velocity; [tex]v = 3.00*10^6 m/s[/tex]
- Distance traveled; [tex]s = 1.80cm = 0.018m[/tex]
Acceleration; [tex]a = \ ?[/tex]
Time taken; [tex]t = \ ?[/tex]
Net force; [tex]F = \ ?[/tex]
a)
To determine the acceleration of the electron, we use the third equation of motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance traveled.
We substitute our given values into the equation
[tex](3.00 *10^6 m/s)^2 = 0^2 + [ 2 * a * 0.018m ]\\\\9.0 *10^{12} m^2/s^2 = a * 0.036m\\\\a = \frac{9.0 *10^{12} m^2/s^2}{0.036m} \\\\a = 2.5 * 10^{14}m/s^2[/tex]
Therefore, the acceleration of the electron as it travels in a straight line to the grid is 2.5 × 10¹⁴m/s².
b)
To the the time taken to reach the grid, we use the first equation of motion:
[tex]v = u + at[/tex]
Where v is the final velocity, u is initial velocity, a is the acceleration and t is the time taken
We substitute our values into the equation
[tex]3.00 * 10^6 m/s = 0 + [ ( 2.5*10^{14}m/s^2) * t\\\\3.00 * 10^6 m/s = ( 2.5*10^{14}m/s^2) * t\\\\t = \frac{3.00 * 10^6 m/s}{2.5*10^{14}m/s^2} \\\\t = 1.2 * 10^{-8}s[/tex]
Therefore, time taken for the electron to reach the grid is 1.2 × 10⁻⁸ seconds.
c)
To determine the net force of the electron, we use the expression from newton second law of motion:
[tex]F = m * a[/tex]
Where m is mass and a is the acceleration
We substitute our values into the equation
[tex]F = ( 9.11 * 10^{-31} kg ) * ( 2.5 * 10^{14}m/s^2)\\\\F = 2.28 * 10^{-16} kg.m/s^2\\\\F = 2.28 * 10^{-16}N[/tex]
Therefore, the net force that is accelerating the electron is 2.28 × 10⁻¹⁶ Newton.
Learn more: https://brainly.com/question/16968602
