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Scorpion4ik [409]

3 years ago

8

The primary gases of the atmosphere are _____. nitrogen and carbon dioxide nitrogen, oxygen, and argon nitrogen and oxygen nitro

gen, oxygen, and carbon dioxide

2 answers:

Paladinen [302]3 years ago

6 0

<span>The primary gases of the atmosphere are _____:
</span>
nitrogen, oxygen, and carbondioxide

inna [77]3 years ago

3 0

Answer:

nitrogen and oxygen

Explanation:

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What kind of light would be the best to use to look inside a cold dark cloud and see the warm stars forming inside?

lys-0071 [83]

<h2>Answer: Infrared light</h2>

A dark nebula is a cloud of dust and cold gas, which does not emit visible light and hides the stars it contains.

These types of nebulae are composed mainly of the hydrogen they obtain from nearby stars, which is their fuel.

It is using infrared light that we can "observe" and analyze in detail what happens in the inner parts of these nebulae.

7 0

2 years ago

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

$m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg$

After that, we compute the potential energy 1.00 m above the reference point:

$U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J$

Then, we compute the average kinetic energy at the specified temperature:

$K=\frac{3}{2}\frac{R}{Na}T$

Whereas $N_A$ stands for the Avogadro's number for which we have:

$K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J$

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0

3 years ago

Lifting a piano into a truck is an example of doing negative work. A. True B. False

Elis [28]

Answer:

False

Explanation:

5 0

3 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

6 0

3 years ago

Read 2 more answers

Mice21 [21]

Answer:

A and A

Explanation:

see paper for work! :)

8 0

2 years ago