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Flura [38]
3 years ago
14

Qwertyuiop hhffhhhhf fiuefhasuhs afbauyhsbvzskhbdv

Physics
2 answers:
Crank3 years ago
8 0
Jaiosiqhwj w ajsishhsvwoiqoqbw s sneaks
mars1129 [50]3 years ago
4 0

Answer:

gfhvgtrtjrgvfjrrgfrfftuyrisnhdvfcgfridkjhsybvvtfvjfcgvwjfccegvghcvgrcgvrekgvrkgvkvvrvkvfgkerruuyti

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Which lists types of materials from most conductive to least conductive?
Lilit [14]
Superconductor, conductor, semiconductor, insulator
7 0
3 years ago
Read 2 more answers
How much force to move 18 kg at a rate of 3m/s​
Stolb23 [73]

Answer:

54N

Explanation:

F = ma

F = (18kg)(3m/s)

F = 54N

4 0
3 years ago
Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
As you learned in Part B, a non-burning helium core surrounded by a shell of hydrogen-burning gas characterizes the subgiant sta
Gennadij [26K]

Answer:

E- The star becomes a red giant (LATEST STAGE)

F- The surface of the star becomes brighter and cooler

C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand

A- The shell of hydrogen surrounding the star's nonburning helium core ignites.

D- The star's non burning helium core starts to contract and heat up

B- Pressure in the star's core decreases (EARLIEST STAGE)

(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.

6 0
3 years ago
Complete the sentences below using the words below, you may use each word more than once
satela [25.4K]
Decreases, stays the same, increases. The volume decreases because as air is cooled, the individual molecules collectively possess less kinetic energy and the distances between them decrease, thus leading to a decrease in the volume they occupy at a certain pressure (please note that my answer only holds under constant pressure; air, as a gas, doesn't actually have a definite volume). The mass stays the same because physical processes do not create or destroy matter. The law of conservation of mass is obeyed. You're only cooling the air, not adding more air molecules. The density decreases because as the volume decreases and mass stays the same, you have the same mass occupying a smaller volume. Density is mass divided by volume, so as mass is held constant and volume decreases, density increases.
7 0
3 years ago
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