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kaheart [24]
2 years ago
15

You observe a distant galaxy. You find that a spectral line, resulting from an electron transition in hydrogen, is shifted from

its normal location in the visible part of the spectrum into the infrared part of the spectrum. What can you conclude?
Physics
1 answer:
alexandr1967 [171]2 years ago
6 0

Answer:

The galaxy is moving away from the observer

Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.

So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.

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The displacement in simple harmonic motion is a maximum when the 1. velocity is a maximum. 2. kinetic energy is a maximum. 3. ve
dlinn [17]

Answer:

3. velocity is zero.

Explanation:

The velocity of a simple harmonic motion is given by

v = \omega\sqrt{A^2-x^2}

Here, <em>ω</em> is the angular velocity, <em>A</em> is the amplitude (or maximum displacement from the equilibrium point) and <em>x</em> is the displacement at any time.

At maximum displacement, <em>x </em>=<em> A</em>.<em> </em>Then

v = \omega\sqrt{A^2-A^2} = 0

Therefore, at maximum displacement, velocity is 0.

Practically, this can be observed in a simple pendulum. As it approaches the maximum displacement, its velocity reduces. It becomes zero at this point and then reverses as the pendulum changes course. Then the velocity begins to increase. It becomes maximum at the equilibrium point but once past that, the velocity begins to reduce as it approaches the other amplitude.

For acceleration,

a = -\omega^2x

It follows that at maximum displacement, the acceleration is a maximum. The negative sign indicates that it is in an opposite direction to the displacement. Both kinetic energy (\frac{1}{2}mv^2) and linear momentum (mv) are proportional to velocity; they are therefore both zero at the maximum displacement.

5 0
3 years ago
What did Rutherford’s model of the atom include that Thomson’s model did not have?
Ksivusya [100]
Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment in which he demonstrated that the atom has a tiny and heavy nucleus. Rutherford designed an experiment to use the alpha particles emitted by a radioactive element as probes to the unseen world of atomic structure.
3 0
3 years ago
Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan
Eddi Din [679]

To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

a=0.50 m/s^2

We want to increase it by a factor of 4, i.e. to

a'=2.00 m/s^2

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

r'=\frac{1}{4}r

So the new acceleration is

a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0
3 years ago
What mean by expansion effect of heat<br>​
pishuonlain [190]

Answer:

Explanation:

-Cambio de temperatura

Al calentar un cuerpo la temperatura aumenta

Es el efecto más inmediato del calor, el aumento de la temperatura. Al calentar un cuerpo, es habitual, aunque no siempre, que el cuerpo aumente de temperatura. El aumento dependerá de la cantidad de calor que se suministra, del tipo de sustancia y de su cantidad.

-Dilatación

Cuando un objeto se calienta, su volumen aumenta. Este fenómeno se llama dilatación térmica. Por el contrario, cuando un objeto se enfría, su volumen disminuye, debido a la contracción térmica.

Cuando se calienta un cuerpo, además de cambiar de estado o variar su temperatura, también cambia su tamaño, se dilata.

Por ejemplo, los puentes no se construyen de una única pieza, sino que suelen presentar uno o varios cortes longitudinales, las llamadas juntas de dilatación. Si no existieran esas juntas, los cambios de longitud del puente entre el invierno y el verano o entre el día y la noche acabarían por romperlo.

La dilatación de un cuerpo dependerá del aumento de temperatura que experimente, de su tamaño y de la sustancia de que esté hecho. Cuanto más aumente la temperatura más aumentará su tamaño, lo mismo que cuanto mayor sea, mayor se hará.

Todos los cuerpos, ya sean sólidos, líquidos o gaseosos, varían su tamaño cuando intercambian calor con otro cuerpo.

-Cambios de estado:

Si una sustancia modifica el estado de sólido, líquido o gaseoso, se produce un cambio de estado. Un cambio de estado es una modificación en la forma en que se disponen las partículas que constituyen una sustancia.

El estado en que se encuentre un cuerpo depende de la presión a la que está sometido y de su temperatura. Para cambiar su estado se debe modificar alguna de estas variables, o ambas. Al elevar la temperatura de una sustancia sólida, aumenta la agitación de sus partículas.

5 0
2 years ago
Read 2 more answers
A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed
oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

    v=30j

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

    u=475i

c. The true velocity of the jet as a vector will be represented as w:

  w=u+v

  w=475i+30j

d.  The true speed of the jet will be calculated as:

    IwI=\sqrt{(475)^2+(30)^2}

    IwI=\sqrt{225625+900}

    IwI=\sqrt{226525}

    IwI=476 mi/h

e. The direction of the jet will be:

tita=tan^{-1}\frac{30}{475}

tita=tan^{-1}(0.0632)

tita=3.62degrees,or,N86.38degreesS

7 0
2 years ago
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