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yanalaym [24]
2 years ago
9

Do any individual air molecules travel the length of the tube? how do you know?

Physics
1 answer:
Ivanshal [37]2 years ago
7 0

When a gas is inside the tube, the individual air molecules travel the length of the tube.

<h3>What is molecules?</h3>

The individual particle of a gas element moving randomly inside the element structure.

When the gas contained in a container, the molecules move randomly inside it. According to the kinetic theory of gases, molecules travel the mean distance. The air molecules vibrates horizontally parallel to the length of the tube.

In an open tube, air molecules are distant from each other. So, they can travel the length of the tube.

Thus, individual air molecules travel the length of the tube.

Learn more about molecules.

brainly.com/question/19922822

#SPJ4

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Help !!
Vaselesa [24]
The answer is ; 6cm

Hope this helps!
Please give Brainliest!

This is because of the diagram below:

6 0
2 years ago
Many years ago, scientists believed that an atom was the smallest unit of matter. Eventually, evidence was discovered that indic
dybincka [34]
Advances in technology used to study and observe atoms lead to the discovery of electrons, protons, nuetrons, and the quarq
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3 years ago
When you turn on the hot water to wash dishes, the water pipes have to heat up. How much heat is absorbed by a copper water pipe
bearhunter [10]

Answer:

53,130 J

Explanation:

When a certain substance absorbs heat, the temperature of the substance increases according to the equation:

Q=mC\Delta T

where

Q is the amount of heat absorbed

m is the mass of the substance

C the specific heat capacity

\Delta T the change in temperature of the substance

In this problem:

m = 2.3 kg is the mass of copper

\Delta T=80.0C-20.0C=60.0^{\circ}C is the increase in temperature

C=385J/kgC is the specific heat of copper

So, the amount of heat absorbed is:

Q=(2.3)(385)(60)=53,130 J

7 0
3 years ago
What is the correct answer
posledela

Answer:

D and compound

Explanation:

because N2 is = to a compound

6 0
3 years ago
A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 40
11111nata11111 [884]

Quasi frequency = 4√6

Quasi period = π√6/12

t ≈ 0.4045

<u>Explanation:</u>

Given:

Mass, m = 20g

τ = 400 dyn.s/cm

k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

Since, a = -10 and b = 4√6

u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t

u(0) = 2

u'(0) = 0

(b)

Quasi frequency:

μ = \frac{\sqrt{4km - y^2} }{2m}

= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}

(c)

Quasi period:

T = 2π / μ

T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6}  }{12}

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

8 0
3 years ago
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