Answer:
true b and c
Explanation:
n the electromechanical transitions of the atoms the relationship must be fulfilled
= R (1 / nf - 1 / no²)
where for the final state nf = 1 giving in the case of hydrogen the Lymma series whose smallest wavelength is lam = 122 nm with nf = 1 and there are a series of spectral lines for each value of n of the final state
in the case of sodium so well it has a transition from an excited state to the kiss state (bad)
Now let's review the different proposals
a) False. The electronic potential for sodium is much lower than for hydrognosia
b) True
c) True
d) true
The density of the nugget is
and is made of gold
Explanation:
The density of an object can be calculated as

where
d is the density
m is the mass
V is the volume of the object
We have to note that density of an object actually depends on the material the object is made of (therefore, two objects made of the same material can have different mass and different volume, but they have same density).
For the nugget in this problem, we have:
mass: m = 38 g
volume: 
So, its density is

And by looking at the table, we see that this value corresponds approximately to the density of gold, so the nugget is made of gold.
Learn more about density:
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Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
Responder:
20.3 ° C
Explicación:
<u>Según la ley de Charles</u>: <em>cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa.
</em>
Paso uno:
datos dados
Temperatura T1 = 20 ° C
Temperatura T2 =?
Volumen V1 = 12.2 cm ^ 3
Volumen V2 = 12.4 cm ^ 3
Aplicar la relación temperatura y volumen

sustituyendo tenemos

Cruz multiplicar tenemos

Temperatura delle braci 20.3°C