Answer:0.153 Hz
Explanation: The relation between Time Period(T) and frequency(f) is given by T=1/f
Plug in the values and u arrive at the answer
acceleration = Velocity changes ÷ time of the velocity changes
4 m/s^2 =
4 × 10^(-3) × 3600 km / h =
4 × 3.6 =
14.4 km / h
Thus :
14.4 = V(2) - V(1) / t(2) - t(1)
14.4 = V(2) - 20 / 10
Multiply both sides by 10
10 × 14.4 = 10 × ( V(2) - 20 ) / 10
144 = V(2) - 20
Add both sides 20
144 + 20 = V(2) - 20 + 20
V(2) = 164 Km/h
Thus the final velocity after 10 seconds is 164 Km/h .
Answer:
(a) The energy of the photon is 1.632 x
J.
(b) The wavelength of the photon is 1.2 x
m.
(c) The frequency of the photon is 2.47 x
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
-
= -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x
)
-
= 1.632 x
Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x
Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x
Js), c is the speed of light (3 x
m/s) and λ is the wavelength.
10.20(1.6 x
) = (6.6 x
* 3 x
)/ λ
λ = 
= 1.213 x 
Wavelength of the photon is 1.2 x
m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x
= 6.6 x
x f
f = 
= 2.47 x
Hz
Frequency of the emitted photon is 2.47 x
Hz.
Porque el internet ayuda a la rapidez del mensaje.