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3 years ago

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

Physics

1 answer:

Ivanshal [37]3 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.

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The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le

Vinil7 [7]

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

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S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

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4 years ago

A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

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3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The <u>Drag Force</u> equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (<u>air in this case) </u>

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>

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3 years ago