An electron starts from rest travels a distance of 15 cm with constant acceleration and hits a television screen at a speed of 3×10^6 m/s .Calculate the acceleration of the electron.

Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop. She drops a wooden block with a mass of 0.5 kilograms on a sensor on the floor. The sensor measures the force of the impact as 4.9 newtons. What’s the acceleration of the wooden block when it hits the sensor?

1. V^2 = Vo^2 + 2a*d.

(3*10^6)^2 = 0 + 2a*0.15, a = ?.

2. F = M*a, a = F/M = 4.9/0.5 =

yeezus

ratio

To calculate the acceleration of the electron, we can use the following formula:

v^2 = u^2 + 2as

where:
v = final velocity (3×10^6 m/s)
u = initial velocity (0 m/s, as the electron starts from rest)
a = acceleration (unknown)
s = distance traveled (15 cm = 0.15 m)

Let's plug in the values and solve for a:

(3×10^6 m/s)^2 = (0 m/s)^2 + 2a(0.15 m)

(9×10^12 m^2/s^2) = 0 + (0.3a)

Dividing both sides of the equation by 0.3 gives us:

a = (9×10^12 m^2/s^2) / 0.3

a ≈ 3×10^13 m^2/s^2

Therefore, the acceleration of the electron is approximately 3×10^13 m^2/s^2.