Astrology is a sub-discipline of astronomy.<br> ОА. A<br> True<br> ОВ.<br> False

The orientation of which of the following does not influence the phases of the moon? a. Earth c. Sun b. the moon d. Stars Please

beks73 [17]

I'm pretty sure it's D. The stars don't influence the moon's phases.

6 0

3 years ago

A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstrat

Likurg_2 [28]

Answer:

Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

a = 2 lines 1/600

a = 2/600

a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

tan θ = y / L

as the measured angles are small

tan θ = sin θ / cos θ sin θ

sin θ = y / L

we substitute

a y/L = λ

y = λ L / a

for λ = 400 10-9 m

I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

i = 346.89 10⁻⁶ m

f

or λ = 700 nm

y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

y_f = 609.609 10⁻⁶ m

the separation of this spectrum

Δr = v_f - i

Dr = (609.609 - 346) 10 ⁻⁶

Dr = 263 10⁻⁶ m

4 0

3 years ago

Which statement best defines an electric field?

Shkiper50 [21]

A region around a charged particle or object within which a force would be exerted on other charged particles or objects

5 0

4 years ago

A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

salantis [7]

Answer:

130m

Explanation:

You just have to multiply velocity by the time traveled:

100m/s * 1.3s = 130m!

5 0

3 years ago

Read 2 more answers

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

Astrology is a sub-discipline of astronomy. ОА. A True ОВ. False