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11Alexandr11 [23.1K]
3 years ago
14

How are the helium atoms in this model different from real helium atoms?

Physics
2 answers:
Taya2010 [7]3 years ago
5 0

Answer;

Real helium atoms are too small to see

Explanation;

-A helium atom is an atom of the chemical element helium.

-Helium is composed of two electrons bound by the electromagnetic force to a nucleus containing two protons along with either one or two neutrons, depending on the isotope, held together by the strong force.

KonstantinChe [14]3 years ago
3 0

With the information given in the question, we can only observe
that they are quite SIMILAR in at least one respect ... both the real
ones and those in 'this model' are quite invisible and intangible, and
we must take your word for it that they exist at all.

Like so many others in math and science, this question raises another
question:    "Which model are you talking about ?"

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In the equilibrium position, the 30-kg cylinder causes a static defl ection of 50 mm in the coiled spring. If the cylinder is de
MArishka [77]

Answer:

2.23 Hz

Explanation:

From the attached diagram below; there exists a diagrammatic representation of the equilibrium position of the cylinder.

The equilibrium position of the spring is expressed as:

mg = K\delta _{st}

where;

m = mass of the object

g = acceleration due to gravity

K = spring constant

\delta _{st} = static deflection of the string

Given that:

m = 30 kg

g = 9.81 m/s²

\delta _{st} = 50 mm = 50 × \frac{1 \ m}{1000 \ m}

= 0.05 m

Then;

30 * 9.81= k * 0.05\\k = \frac{30*9.81}{0.05} \\k = 5886 N/m

From here; let us find the angular velocity which will be needed to determine the natural frequency aftewards.

The angular velocity of the cylinder can be expressed by the formula:

\omega_{n} = \sqrt{\frac{k}{m}}

\omega_{n} = \sqrt{\frac{5886}{30}}

\omega_{n} = \sqrt{196.2}

\omega_{n} = 14.007141 \ \ rad/s

Finally; the natural frequency f_n can be calculated by using the equation

f_n = \frac{\omega_n}{2 \ \pi }

f_n = \frac{14.007141}{2 \ \pi }

f_n= 2.229305729

f_n ≅ 2.23 Hz

Thus; the resulting natural frequency of the vertical vibration of the cylinder = 2.23 Hz

3 0
3 years ago
A 10.0 L balloon contains helium gas at a pressure of 660 mmHg . What is the final pressure, in millimeters of mercury, of the h
Ksivusya [100]

Answer:

(a)= 264mmHg

(b)= 2000mmHg

(c)474.82mmHg

(d)= 511.63mmHg

Explanation:

the question deals with boyles law, which states that the volume of a given mass of gas at constant temperature is inversely proportional to its pressure

V ∝ 1/P

P₁V₁ = P₂V₂

making V₂ as the subject of formular

P₂ = P₁V₁/ V₂

with a volume of  25.0L

P₂ = 660×10 / 25

= 264mmHg

with a volume of 3.30 L

P₂  = 660 × 10 / 3.30

= 2000mmHg

with a volume of 13900 mL

= 13.9L

P₂  =660× 10 / 13.9

474.82mmHg

with a volume of 12900 mL

P₂ =660×10 / 12.9

= 511.63mmHg

6 0
3 years ago
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The answer would be Power

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Hiya!

The answer to your question is B.

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4 0
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