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Sav [38]
3 years ago
9

Which state of matter can be compressed?

Physics
2 answers:
aivan3 [116]3 years ago
6 0
D is the answer I believe
MA_775_DIABLO [31]3 years ago
3 0
Gases............<span>In gases, the atoms are much more spread out than in solids or liquids, and the atoms collide randomly with one another. A gas </span>will<span> fill any container, but if the container is not sealed, the gas </span>will<span> escape. Gas </span>can be compressed<span> much more easily than a liquid or solid</span>
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The mass of an atom is ________________________.
alexandr402 [8]
The mass of an atom comes from the protons and neutrons that is found in the nucleus. The number of protons is the atomic number of an element. To find the number of neutrons, subtract the atomic number from the mass of an atom. For example, sodium’s atomic number is 11. This will tell us that sodium has 11 protons in it. The atomic mass of sodium is 23. So subtract 23 form 11 gives us 12. Therefore, there are 12 neutrons in sodium.
7 0
3 years ago
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most
Oxana [17]

Answer:

A)  I_{total} = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

     I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

      I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

    I = I_{cm} + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

       I_{body} = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

    I_{arm} = M D²

Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

    I_{total} = 1.12 + 0.32

    I_{total} = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0
3 years ago
What are the 3 main branches of natural sciences
ICE Princess25 [194]

Answer:

Physical science, Earth science, and life science.

4 0
3 years ago
Read 2 more answers
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
1. What types of natural phenomena could serve as time standards?<br>​
Alex Ar [27]
In practice, something that follows a very predictable pattern can be used as a time standard. This include things like radioactive decay, planetary orbit, and the speed of light, among others.
8 0
2 years ago
Read 2 more answers
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