Answer:
The correct option is;
a- sea surface temperature anomaly, in degrees Celsius
Explanation:
From the diagram related to the question we have two graphs super imposed of Sea surface temperature anomaly, in degrees Celsius and cholera incidence anomaly (%) both plotted against time in years.
On the left the y-axis represents the sea surface temperature anomaly while on the right, the y-axis represents the cholera incidence anomaly (%).
The display of the graph shows the sea surface temperature anomaly in blue.
Answer:
E3 = 3.03 10⁻¹⁶ kJ, E4 = 4.09 10⁻¹⁶ kJ and E5 = 4.58 10⁻¹⁶ kJ
Explanation:
They give us some spectral lines of the Balmer series, let's take the opportunity to place the values in SI units
n = 3 λ = 656.3 nm = 656.3 10⁻⁹ m
n = 4 λ = 486.1 nm = 486.1 10⁻⁹ m
n = 5 λ=434.0 nm = 434.0 10⁻⁹ m
Let's use the Planck equation
E = h f
The speed of light equation
c = λ f
replace
E = h c /λ
Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and c is the speed of light that is worth 3 10⁸ m / s
Let's calculate the energies
E = 6.63 10⁻³⁴ 3 10⁸ / λ
E = 19.89 10⁻²⁶ /λ
n = 3
E3 = 19.89 10⁻²⁶ / 656.3 10⁻⁹
E3 = 3.03 10⁻¹⁹ J
1 kJ = 10³ J
E3 = 3.03 10⁻¹⁶ kJ
n = 4
E4 = 19.89 10⁻²⁶ /486.1 10⁻⁹
E4 = 4.09 10⁻¹⁹ J
E4 = 4.09 10⁻¹⁶ kJ
n = 5
E5 = 19.89 10⁻²⁶ /434.0 10⁻⁹
E5 = 4.58 10⁻¹⁹ J
E5 = 4.58 10⁻¹⁶ kJ
Answer:
F = (mg - kx) i ^
Explanation:
Force and potential are related
F = - (
i ^ + \frac{dU}{dy} j ^ + \frac{dU}{dz} k ^)
in this exercise we are told that the potential is
U = - m g x + ½ kx²
let's make the drive
\frac{dU}{dx} = -mg + kx
\frac{dU}{dy}= 0
\frac{dU}{dz} = 0
we substitute
F = (mg - kx) i ^
Answer:
Distance per unit time
Explanation:
Speed is the rate at which someone or something is able to move or operate.
