An electron is to be accelerated from a velocity of 5.00×106 m/s to a velocity of 9.90×106 m/s . through what potential difference must the electron pass to accomplish this

Potential difference needed= 207.9 V

Explanation=

using work energy principle

Work done on the electron= change in kinetic energy

W= 1/2m Vf²- 1/2 mVi²

m= mass of electron=9.11 x 10⁻³¹ Kg

Vf= final velocity= 9.9 x 10⁶ m/s

Vi= initial velocity= 5 x 10⁶ m/s

so W = 1/2 (9.11 x 10⁻³¹ ) (9.9 x 10⁶ )²- 1/2 (9.11 x 10⁻³¹ ) (5 x 10⁶ )²

W=3.33 x 10⁻¹⁷ J

Now work done= q V

q= charge of electron= 1.6 x 10⁻¹⁹J

V= potential difference

3.33 x 10⁻¹⁷= (1.6 x 10⁻¹⁹) V

v= 207.9 V

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle