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3 years ago

Which characteristics can be used to differentiate star systems? Check all that apply.

Physics

2 answers:

hjlf3 years ago

7 0

The answer is:

the number of stars in the system

the age of the stars in the system

the way stars are organized in the system

I just finished this assignment, hope it helped.

Dovator [93]3 years ago

6 0

♡ hello there ヾ(°∇° ) ♡

Your answers are...

★ ... <u>A, C, and D</u> !!!! ★

×║hope this helps,║×

➶ have a nice day! ➷

↬ <u>ɢᴏᴛʜɪx</u> ♡

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Decribe what must happen inside of an aluminum can in order for it to be attracted to a positively-charged and to a negatively-c

horrorfan [7]

Answer:

The PROTONS in the can attract to the negatively charged object, so then the can becomes polarized and the ELECTRONS in the can attract the positively charged object.

Explanation:

5 0

3 years ago

What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic th

asambeis [7]

Answer:

$\lambda = 2.57 \times 10^{-11} m$

Explanation:

Average velocity of oxygen molecule at given temperature is

$v_{rms} = \sqrt{\frac{3RT}{M}}$

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

$v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}$

$v_{rms} = 483.4 m/s$

now for de Broglie wavelength we know that

$\lambda = \frac{h}{mv}$

$\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}$

$\lambda = 2.57 \times 10^{-11} m$

7 0

3 years ago

A starship travels to a planet that is 20 light years away. The astronauts stay on the planet for 2.0 years before returning at

ad-work [718]

Answer:

astronauts age is 32 years

correct option is e 32 years

Explanation:

given data

travels = 20 light year

stay = 2 year

return = 52 years

to find out

astronauts aged

solution

we know here they stay 2 year so time taken in traveling is

time in traveling = ( 52 -2 ) = 50 year

so it mean 25 year in going and 25 years in return

and distance is given 20 light year

so speed will be

speed = distance / time

speed = 20 / 25 = 0.8 light year

so time is

time = $\frac{t}{\sqrt{1-v^2} }$

time = $\frac{25}{\sqrt{1-0.8^2} }$

time = 15 year

so age is 15 + 2 + 15

so astronauts age is 32 years

so correct option is e 32 years

4 0

3 years ago

which changes of state should she place in the column titled "lose energy"?freezing, boiling, and meltingmelting, vaporization,

Blababa [14]

Freezing (liquid to solid)
Deposition (gas to solid)
Condensation (gas to liquid)

All three of these state changes are a result of a energy loss. When considering energy loss it is best to think of situations where temperature has dropped. Less energy in the system results in less energy the substance is exposed to or has available.

7 0

2 years ago

Read 2 more answers

The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.

Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

$\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})$

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

$R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}$

Here, $R_{\infty}$ is the "general" Rydberg constant, $m_e$ is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, $M=1.67*10^{-27}kg$ :

$R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}$

Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$ :

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$

5 0

3 years ago