Describe how the number of electrons changes across a period of elements

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

$\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}$

Put the value into the formula

$525=\sqrt{\dfrac{c+v}{c-v}}\times950$

$\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2$

$\dfrac{c+v}{c-v}=0.305$

$c+v=0.305\times(c-v)$

$v(1+0.305)=c(0.305-1)$

$v=\dfrac{0.305-1}{1+0.305}c$

$v=−0.532c$

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0

3 years ago

Which of the following is true statement about isotopes?

mel-nik [20]

Notice for the Carbon question they were the same element and the shared the same number of protons. so i think d. is the answer

7 0

3 years ago

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I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, the orbital speed of this material when it was 40,000 AU from the sun is 4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

3 0

3 years ago

Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a

xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, $h_b$ .

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, $h_d$ . Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be $h_d$ and the distance travelled by the ball measured from the top be $h_b$ .

It follows that $h_d+h_b=9.8$ .

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

$h_b=0\times t + 0.5\times 9.8t^2$ (since $v_{b0} =0$ .)

$h_b=4.9t^2$

Dart:

$h_d=17.8\times t - 0.5\times9.8t^2$ (the acceleration is opposite to the displacement, hence the negative sign)

$h_d=17.8\times t - 4.9t^2$

But

$h_b+h_d =9.8$

$17.8\times t - 4.9t^2+4.9t^2 =9.8$

$17.8\times t = 9.8$

$t = 0.55$

The height of the collision is the height of the dart above the ground, $h_d$ .

$h_d=17.8\times t - 4.9t^2$

$h_d=17.8\times 0.55 - 4.9\times(0.55)^2$

$h_d=9.79 - 1.48225$

$h_d=8.3$

8 0

3 years ago

A student hears a police siren. What would change the frequency that the student hears? Check all that apply.

ser-zykov [4K]

A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source.

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.

4 0

3 years ago

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