What is the kinetic energy K of an electron with momentum 1.05×10−24 kilogram meters per second?Express your answer in electron volts to two significant figures.

3.7eV

The kinetic energy (K) of a particle is the half the product of the mass (m) and the square of the velocity (v) of the particle. i.e

K = x m x v²        --------------------(i)

Also, the momentum (p) of a particle is the product of the mass and the velocity of the particle. i.e

p = m x v                   [square both sides]

p² = (mv)²

p² = m²v²                   [make v² subject of the formula]

v² =      

Now substitute v² =  into equation (i) as follows;

K = x m x

K = x          --------------------------(ii)

From the question the particle is an electron with the following details;

m = mass of the electron = 9.1 x 10⁻³¹kg                        [known value]

p = momentum of the electron = 1.05 x 10⁻²⁴kgm/s    [given value]

Substitute these values into equation (ii) as follows;

K = x

K = x

K = x 0.12 x 10⁻¹⁷

K = 0.06 x 10⁻¹⁷

K = 6.00 x 10⁻¹⁹ J

Convert the result to electron volts (eV) as follows;

1 J = 6.2 x 10¹⁸ eV

6.00 x 10⁻¹⁹ J = 6.00 x 10⁻¹⁹ J x 6.2 x 10¹⁸ eV / 1 J = 3.7eV

Therefore, the kinetic energy K of an electron with that momentum is 3.7eV

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this