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Igoryamba
3 years ago
13

what did taking measurements reveal about the velocity of the fireball? Was the fireball’s velocity constant or changing?

Physics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Answer:

The fireball slows down when moving upward. It speeds up when moving downward. So, its velocity is always changing.

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What is the energy in joules of a mole of photons associated with red light of wavelength 7.00 × 102 nm?
konstantin123 [22]
<span>The energy of a single photon is given by E = hc/lambda, where h is Planck's constant, c is the speed of light, and lambda is the wavelength. Plugging the values in gives E = 6.63E-34 x 3.00E8 / 700E-9 = 2.84E-19 Joules Now one mole of substance is equivalent to 6.02E23 particles, so one mole of these photons will be: 2.84E-19 x 6.02E23 = 1.71E5 Joules</span>
4 0
3 years ago
wo slightly out of tune instruments will create beats when the same note is played. This is a pattern of louder and softer sound
Lostsunrise [7]

This interaction is known as <em>constructive interference</em>.  It's a result of linear superposition.

5 0
3 years ago
Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
Taigas, tundra and tropical rainforests share what common abiotic component?
e-lub [12.9K]

Answer:

water

Explanation:

hope this helps

6 0
3 years ago
How can you be both at rest and also moving at 100,000 km/h at the same time
4vir4ik [10]

You could be lying completley still on your bed, and all though it seems you are at rest, you are moving along with the earth around the sun and hence are motion. This is why 'being at rest' is more of a relative term. Hope this helps!

7 0
3 years ago
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