<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
The Coulombs Constant k = 9.0 x 10^9
Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
Mass of the Electron = 9.11 x 10^-31
F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
Since one orbit is the distance,
one orbit = circumference = 2 x pi x r; distance s = v x t.
v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
Area=side^2=4^2=16cm^2=0.0016m^2
Work = (force) x (distance)
Power = (work) / (time)
Knowing force, distance, and time, it looks like
you can calculate both <em>work and power</em>.
But that's not all !
-- <u>Speed</u> = (distance) / (time)
-- <u>Impulse</u> = change in momentum = (force) x (time)
Answer: 3.3·10^6 m/s
Explanation: Ionization energy of H is 13.6 eV. So the kinetic energy of electron is 31,2 eV. Kinetic Energy of Electron is E = ½mv^2
1 eV = 1.60·10^-19 J and m = 9.1·10^-31 kg. Solving speed v = √( 2E / m)
v = √(2·31.2·1.60·10^-19 J /9.1·10^-31 kg) = 3.3·10^6 m/s
Answer: 0.3m
Explanation:
Given the following :
Speed of radio wave(V) = 3 × 10^8 m/s
If 1M = 10^6
Frequency(F) = 90MHz = 90 × 10^6
Wavelength (λ) =?
Speed of wave (V) = frequency(F) × wavelength(λ)
V = fλ
(3 × 10^8) = (90 × 10^6) × λ
λ = (90 × 10^6) / (3 × 10^8)
λ = 30 × 10^(6 - 8)
λ = 30 × 10^-2
λ = 0.3m
