Match the term to the correct description.

When blueshift occurs,the preceived frequency of the wave would be?

LiRa [457]

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency $f\; {\rm Hz}$ at the source. In other words, the source of the wave sends out a peak after every $(1/f)\; {\text{seconds}}$ .

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of $(1/f)\; {\text{seconds}}$ . It would appear to the observer that consecutive peaks arrive every $(1/f)\; {\text{seconds}}\!$ . That would correspond to a frequency of $f\; {\rm Hz}$ .

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of $v \; {\rm m \cdot s^{-1}}$ .

Again, the source of this wave would send out a peak after each period of $(1/f)\; {\text{seconds}}$ . However, by the time the source sends out the second peak, the source would have been $v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}$ closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than $(1/f)\; {\text{seconds}}$ . The observed frequency of this wave would be larger than the original $f\; {\rm Hz}$ .

6 0

3 years ago

Would you classify petroleum as a renewable or nonrenewable resource? Justify your answer in two or more complete sentences.

olchik [2.2K]

Answer:

Petroleum is a nonrenewable resource. This being because it is a natural gas that take millions of years for it to reform which means each part of it is a once in a lifetime use.

3 0

3 years ago

Which best describes a force?

sashaice [31]

Force = mass*acceleration.

Therefore, a force has a magnitude and direction.

5 0

3 years ago

Read 2 more answers

Se aplica una fuerza de 2.294N a una masa de 408 kg. ¿Cuál es la aceleración de la masa?

Zanzabum

Answer:

a=5.62[m/s^2]

Explanation:

Acorde a la segunda ley de newton la cual dice que la sumatoria de fuerzas sobre un cuerpo el cual no esta en equilibrio estatico presentara una aceleracion. Esta fuerza sera igual al producto de la masa del cuerpo por su aceleracion.

$F=m*a\\F=fuerza total aplicada [N]\\m=masa del cuerpo[kg]\\a=aceleracion del cuerpo[m/s^2]\\\\$

De esta manera despejando el valor de aceleracion de la anterior ecuacion:

$a=\frac{F}{m} \\a=2294/408\\a=5.62[m/s^2]$

8 0

3 years ago

Two metal bars experience an equal change in volume due to an equal change in temperature. The first bar has a coefficient of ex

maksim [4K]

Answer:

The original volume of the first bar is half of the original volume of the second bar.

Explanation:

The coefficient of cubic expansivity of substances is given by;

γ = ΔV ÷ ( $V_{1}$ Δθ)

Given: two metal bars with equal change in volume, equal change in temperature.

Let the volume of the first metal bar be represented by $V_{1}$ , and that of the second by $V_{2}$ .

Since they have equal change in volume,

Δ $V_{1}$ = Δ $V_{2}$ = ΔV

For the first metal bar,

2γ = ΔV ÷ ( $V_{1}$ Δθ)

⇒ Δθ = ΔV ÷ (2γ $V_{1}$ )

For the second metal bar,

γ = ΔV ÷ ( $V_{2}$ Δθ)

⇒ Δθ = ΔV ÷ ( $V_{2}$ γ)

Since they have equal change in temperature,

Δθ of first bar = Δθ of the second bar

ΔV ÷ (2γ $V_{1}$ ) = ΔV ÷ ( $V_{2}$ γ)

So that;

(1 ÷ 2 $V_{1}$ ) = (1 ÷ $V_{2}$ )

2 $V_{1}$ = $V_{2}$

$V_{1}$ = $\frac{V_{2} }{2}$

Thus, original volume of the first bar is half of the original volume of the second bar.

8 0

4 years ago