Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. A certain gun has an electric field of 1.76 x 10 4 N/C. What speed would the proton obtain if the field accelerated the proton through a distance of 1.00 cm?

v = 2.75 * 10^7 m/s

Explanation: Since the electric field is a uniform one and the distance is small, the motion of the electron is of a constant acceleration, hence newton's laws of motion is applicable.

From the question

E=strength of electric field = 214N/c

q=magnitude of an electronic charge = 1.609* 10^-16c

m= mass of an electronic charge = 9.11*10^-31kg

v = velocity of electron.

S= distance covered = 1cm = 0.01m

a = acceleration of electron.

F = ma but F=Eq

Eq = ma.

a = Eq/m

a = 214 * 1.609*10^-16/ 9.11 * 10^-31

a = 344.32 * 10^-16/ 9.11 * 10 ^-31

a = 3. 779* 10^16 m/s²

Before the electron is accelerated, they are always not moving, hence initial velocity (u) = 0.

By using the equation of motion, we have

v² = u² + 2aS

But u = 0

v² = 2aS

v²= 2* 3.779*10^16 * 0.01

v² = 7.558 * 10^14

v = √7.558 * 10^14

v = 2.75 * 10^7 m/s.

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6