Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure, = 2.95 atm
Atmospheric pressure, = 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;
where;
P₁ =
P₂ =
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;
where;
is the density of seawater = 1030 kg/m³
Therefore, the water is flowing at the rate of 28.04 m/s.
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ =
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ =
V = V₀
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀
0.1V₀ = V₀
0.1V₀/V₀ =
0.1 =
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
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.