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2 years ago

Most of the universe is believed to be ______

Physics

2 answers:

vivado [14]2 years ago

6 0

Answer:

B. dark matter

Explanation:

The<u> current theory</u> is that the dark matter composes about 85% of the matter in the whole universe, possibly in large majority made out of unknown material. What we found so far is that the majority of the dark matter is NOT composed of baryonic material (the stuff that makes stars and planet).

We know dark matter exists and we suppose there's a HUGE amount of it that fills the gaps between stars and planets.

Minchanka [31]2 years ago

6 0

The answer would be B.dark matter

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What is the work done if you push a tree with<br> 50 N of force but the tree does not move?

Soloha48 [4]

Answer:

The work done is 0.

Explanation:

The reason no work is done is because the equation W = Fs.

W = work

F= force

s= displacement

In this scenario F = 50 and s= 0

Therefore.

W = 50(0)

W = 0

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3 years ago

Changes of state

Mrrafil [7]

Answer:

Option A

Explanation:

At segment T-U, the substance changes from a liquid to a gas and does not change temperature.

The reason is because latent heat of vaporisation allows for the absorption of heat in the change of state and temperature remains constant until it has fully changed state.

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1 year ago

Read 2 more answers

What would scientists using classical, Newtonian physics expect to observe

Sav [38]

Answer:

A. That enough light of any frequency would cause electrons to flow.

Explanation:

A P E X

5 0

2 years ago

Read 2 more answers

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

When Mendeleev organized elements in his periodic table in order of increasing mass, similar elements with similar properties we

Temka [501]

Similar elements with similar properties were in the same groups and periods for instance lithium(Li) and sodium(Na) are alkaline metals and so belong to the same group (that is group 1).Also Hydrogen(H) and Helium(He) both have only one shell or energy level and so belong to the same period.

8 0

3 years ago