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4 years ago

What are noble gases?

Physics

1 answer:

Bas_tet [7]4 years ago

6 0

Answer:

noble gases are basically a group of gases that are similar in their chemical compounds, theres six of them : helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).

~batmans wife dun dun dun.....

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Which statement is correct about the car?

JulijaS [17]

Answer:

Explanation:

OOf we are doing this stuff atm

So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B

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3 years ago

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What is the momentum of a 65kg ball rolling at 5 m/s?

mojhsa [17]

Explanation:

Momentum = mass × velocity

p = (65 kg) (5 m/s)

p = 325 kg m/s

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3 years ago

A photoelectric experiment is performed where green light with a wavelength of 546.1 nm is shined on a metal plate, creating a p

PtichkaEL [24]

Answer:

$\phi=1.55 [eV]$

Explanation:

We can use the work function equation for a photoelectric experiment:

$\phi=\frac{hc}{\lambda}-K_{max}$

h is the plank constant
c is the speed of light
λ is the wave length
K is the kinetic energy (or K=eΔV)

So we will have:

$\phi=\frac{hc}{\lambda}-e\Delta V$

$\phi=\frac{6.63*10^{-34}*3*10^{8}}{546.1*10^{-9}}-0.728eV$

$\phi=3.64*10^{-19}[J]-0.728 [eV]$

$\phi=(3.64*10^{-19}[J]*\frac{1eV}{1.6*10^{-19}[J]})-0.728 [eV]$

$\phi=2.28 [eV] - 0.728 [eV]$

$\phi=1.55 [eV]$

I hope it helps you!

8 0

4 years ago

Read 2 more answers

Certain x-rays have a frequency of 1.0×1019hz. calculate their wavelength in air.

Murljashka [212]

For any electromagnetic wave, the relationship between frequency and wavelength is given by
$\lambda = \frac{c}{f}$
where
$\lambda$ is the wavelength
c is the speed of light (the speed of the electromagnetic wave)
f is the frequency

For the X-rays in our problem, the frequency is $f=1.0 \cdot 10^{19}Hz$ , while the speed of light is $c=3.0 \cdot 10^8 m/s$ , so the wavelength of this radiation is
$\lambda= \frac{3 \cdot 10^8 m/s}{1.0 \cdot 10^{19} Hz} =3 \cdot 10^{-11} m$

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3 years ago

What is the expected fuel consumption for a 500-nautical mile flight under the following conditions? Pressure altitude 4,000 ft

aleksandr82 [10.1K]

Answer: 36.1 gallons

Explanation:

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3 years ago