An electron is traveling at 2.0E5 m/s parallel to a uniform electric field of 9.11e-3 N/C strength. The electron is traveling to the right. The electric field lines point to the right. What is the speed of the electron (in m/s) after traveling 1.8 meters to the right?

v = 1.85*10^5 m/s

Explanation:

In order to calculate the speed of the electron after it has traveled 1.8m, you first take into account that the electric field generates a desceleration on the electron, because the direction of the electron and electric field are the same.

You use the Newton second law, to calculate the deceleration of the electron:

     (1)

q: charge of the electron = 1.6*10^-19C

m: mass of the electron = 9.1*10^-31kg

E: magnitude of the electric field = 9.11*10^-3N/C

a: deceleration = ?

You solve the equation (1) for a, and replace the values of the other parameters:

Next, you use the following formula to calculate the final speed of the electron:

     (2)

v: final speed of the electron = ?

vo: initial speed of the electron = 2.0*10^5 m/s

x: distance traveled by the electron = 1.8m

You solve the equation (2) for v and replace the values of the other parameters:

The speed of the electron after it has traveled 1.8m is 1.85*10^5 m/s

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever