An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 6.0 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 35 mm horizontally.
assume you mean 10^7 and 10^-16 etc
How long to move 35*10^-3 m ?
t = 35*10^-3/1.9*10^7
so
t = 18.4 * 10^-10 seconds
a = F/m = 6*10^-16 / 9.11*10^-31
= .659 * 10^15 m/s^2
d = (1/2) a t^2
= (1/2)(.659*10^15)(339*10^-20)
= 112*10^-5
= .00112 meters or 1.12 mm
To determine the vertical distance the electron is deflected, we need to calculate the acceleration of the electron in the vertical direction.
First, let's calculate the time it takes for the electron to move 35 mm horizontally. Given that speed = distance/time, we can rearrange the equation to solve for time: time = distance / speed.
Converting the distance from millimeters to meters, we have: distance = 35 mm = 35 × 10^(-3) m.
Plugging in the given speed, we have: time = (35 × 10^(-3) m) / (1.9 × 10^7 m/s).
Let's calculate this value: time = 35 × 10^(-3) / 1.9 × 10^7 = 1.842105263 × 10^(-9) seconds.
Now, let's calculate the acceleration of the electron in the vertical direction using Newton's second law: force = mass × acceleration.
Plugging in the given force and mass of the electron, we have: acceleration = force / mass.
Calculating this value: acceleration = (6.0 × 10^(-16) N) / (9.11 × 10^(-31) kg) = 6.592989e14 m/s^2.
Now that we have the acceleration and time, we can use the kinematic equation: distance = (1/2) × acceleration × time^2.
Plugging in the values, we get: distance = (1/2) × (6.592989e14 m/s^2) × (1.842105263 × 10^(-9) seconds)^2.
Calculating this, we find the vertical distance the electron is deflected to be approximately 1.83338632 × 10^(-5) meters, or 18.33 micrometers.