Question 1 of 10

In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t

grigory [225]

Answer:

3. relatively high temperature, about 10,000 K, so that significant numbers of electrons are excited from the ground state, n = 1, to the first excited state, n = 2, but not too many of them have been ejected completely from the atoms

Explanation:

If hydrogen absorption lines are very strong in the visible spectrum of a particular star that means the population of electron in n = 2 is very high so on being exited they absorb radiation in Balmer series and give rise to absorption spectrum. The average temperature required to excite electron in hydrogen atom from n=1 to n = 2 is 10000K .

4 0

2 years ago

A ball is dropped from a height of 10m. At the same time, another ball is thrown

soldi70 [24.7K]

5.1 m

Explanation:

Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by

$y_1 = 10 - \frac{1}{2}gt^2$ (1)

where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground $y_2,$ is given by

$y_2 = v_0t - \frac{1}{2}gt^2$ (2)

At the instant the two balls collide, they will have the same displacement, therefore

$y_1 = y_2 \Rightarrow 10 - \frac{1}{2}gt^2 = v_0t - \frac{1}{2}gt^2$

or

$v_0t = 10\:\text{m}$

Solving for t, we get

$t = \dfrac{10\:\text{m}}{v_0} = \dfrac{10\:\text{m}}{10\:\text{m/s}} = 1\:\text{s}$

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):

$y_1 = 10\:\text{m} - \frac{1}{2}(9.8\:\text{m/s}^2)(1\:\text{s})^2$

$\:\:\:\:\:\:\:= 5.1\:\text{m}$

8 0

2 years ago

How do I solve such problem???

labwork [276]

1). Calculate how long it takes an object to fall 4,000 m after it's dropped. (Use D = (1/2) (g) (T²) . D is 4,000 m. g = 9.8 m/s². Find T .)

2). Calculate how far the object will move HORIZONTALLY in that length of time, if it's moving at 75 m/s. (Distance = (75 m/s) x (time) . )

6 0

2 years ago

Can someone explain how they got their answer or how I get the change in number? :(

enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0

1 year ago

A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall

Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object $m=2\ kg$