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Tju [1.3M]
3 years ago
10

Using the Particle Theory, explain why liquids are less compressible than gases.

Physics
1 answer:
AleksandrR [38]3 years ago
5 0

Explanation:

In liquids the particles have more energy. This allows the particles in liquids to roll past each other, but they are still tightly packed. Gases can be compressed because the particles can be forced closer together. The particles in a liquid are already as close as they can be.

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Which of the following. Chemical represent the law of conservation of Matter?
vazorg [7]

Answer:

The answer is first one 1...

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______ involves organizing and breaking down information into easier groups to expand capacity. Rehearsal is the verbal repetiti
adell [148]

Hi!


The answers would be <u>chunking</u> & <u>short-term</u>


1. <u>Chunking </u>involves organizing and breaking down information into easier groups to expand capacity.

<h3>Explanation:</h3>

Chunking is a mental process that is observed to increase short-term memory by taking the information and categorizing it into small groups. For instance, a longer number taken as a single unit is harder to recall then when it is divided into smaller units. 235469350 is harder to instantly recall as compared to when it is chunked into 3 groups: 235 469 350.

This allows more information to be stored in, thereby increasing the capacity of the mind to store information.


2. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of <u>short-term</u> memory.

<h3>Explanation: </h3>

Short-term memory is lost after a couple of seconds or minutes, for instance even if you chunk the information, you might not recall it after 30 seconds. Rehearsing or repetition of information, either loudly or mentally, extends the time a particular information is retained.

So you depending on the number of times you repeat the number 235 469 350, the more your short term memory improves .


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6 0
3 years ago
The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

h_{Hg-TOP}=675mmHg=0.675m the barometric reading at the top of the building

h_{Hg-BOT}=695mmHg=0.695m the barometric reading at the bottom of the building

\rho _{air}=1.18 kg/m^{3} density of air

\rho _{Hg}=13600 kg/m^{3} density of mercury

g=9.8/m^{2} gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building P_{TOP} and the perssure at the bottom P_{BOT}:

P_{TOP}=\rho _{Hg} g h_{Hg-TOP} (1)

P_{BOT}=\rho _{Hg} g h_{Hg-BOT} (2)

From (1): P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa (3)

From (2): P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure \Delta P:

\Delta P=P_{BOT} - P_{TOP} (5)

\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa (6)

On the other hand, we have a column of air with a cross-section area A and the same height of the building, lets name it h_{air}.

As pressure is defined as the force F exerted on a specific area A, we can write:

\Delta P=\frac{F}{A} (7)

If we isolate F we have:

F= A \Delta P (8)

Also, the force gravity exerts on this column of air (its weight) is:

F=m_{air} g (9)

Knowing the density of air is: \rho_{air}=\frac{m_{air}}{V_{air}} (10)

where the volume of air can be written as: V_{air}=(A)(h_{air}) (11)

Substituting (1) in (10):

\rho_{air}=\frac{m_{air}}{(A)(h_{air}} (12)

Isolating m_{air}:

m_{air}=(\rho_{air}) (A) (h_{air}) (13)

Substituting (13) in (9):

F=(\rho_{air}) (A) (h_{air}) (g) (14)

Matching (8) and (14)

A \Delta P=(\rho_{air}) (A) (h_{air}) (g) (15)

Isolating h_{air}:

h_{air}=\frac{\Delta P}{g \rho_{air}} (16)

Substituting the known and calculated values:

h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})} (17)

Finally:

h_{air}=230.50 m This is the height of the building

8 0
3 years ago
An unknown material, m1 = 0.45 kg, at a temperature of T1 = 91 degrees C is added to a Dewer (an insulated container) which cont
goldenfox [79]

Answer:

Explanation:

Let the specific heat of material be s

heat lost by material = m₁ s (T 1 - T ) ,  (T 1 - T ) is fall in temp , m₁ is mass of material

= .45 x s x (91 - 31.4 )

= 26.82 s

Heat gained by water

= m₂ cw (T2 - T )

1.3 x 4186 x ( 31.4 - 23 )

heat lost = heat gained

m₂ cw (T2 - T ) = m₁ s (T 1 - T )

1.3 x 4186 x ( 31.4 - 23 ) =  .45 x s x (91 - 31.4 )

45711.12 = 26.82 s

s = 1704.36

7 0
3 years ago
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