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2 years ago

The air also contained a small amount of argon

Physics

1 answer:

AysviL [449]2 years ago

3 0

The argon molecules become solid when the temperature decrease from 20C to -190 C.

<h3>What changes occur in argon?</h3>

The changes in arrangement and movement of the particles of the argon occurs as the temperature of the air decreased because the argon gas begins to freeze. The freezing point of argon is -189 Celsius so when the temperature decrease from 20C to -190 C, the argon molecules becomes solid.

So we can conclude that the argon molecules become solid when the temperature decrease from 20C to -190 C.

Learn more about argon here: brainly.com/question/23949935

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You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

4 years ago

Help with 2 Physics questions, WILL CHOOSE BRAINLIEST

Tju [1.3M]

1) D

2) D.) Greater than $\theta_c$

Explanation:

The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.

This phenomenon occurs only if two conditions are satisfied:

The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
The angle of incidence is greater than a certain angle called critical angle

In picture 1, we have 4 different diagrams. In the diagrams:

The red arrow represents the incident ray
The green arrow represents the refracted ray
The blue arrow represents the reflected ray

Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)

As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be

D.) Greater than $\theta_c$

3 0

3 years ago

Can somebody explain it to me please?

Volgvan

Answer:

45°.

It is a property of the parabolas.. When the angle between a parabola and the x-axis is 45° the range is maximum.

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3 years ago

Read 2 more answers

The picture is attached.