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2 years ago

If a container of gas is at rest, the average velocity of molecules must be zero. yet theaverage speed is not zero. explain.

Physics

1 answer:

lakkis [162]2 years ago

5 0

To consider the molecule in gas at rest, suppose a container of gas at rest.

We know that velocity is a vector quantity and it includes direction but speed is a scalar quantity and it does not include a direction.

Although the molecules are moving around a lot, they are only moving around within the fixed walls of the container. As the container isn't moving (given), the net velocity of the gas molecules is zero. Think about their random movements as vectors, then think about adding up all those vectors over time. They cancel each other out over time. So, the average velocity of a molecule in gas at rest is zero.

But it's only the directions that cancel each other out, which is why the average speed (ignoring direction) is non-zero.

To learn more about average speed refer to the link:

brainly.com/question/12322912

#SPJ4

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A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

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Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is:

$F = k \frac{|q_1q_2|}{d^2}$

where $k$ is Coulomb's constant.

Taking the values:

$d = 4.6 \ 10^{-15} m$

$q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}$

and knowing the value of the Coulomb's constant:

$k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}$

Taking all this in consideration:

$F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}$

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3 years ago

What type of pollution did the Clean Water Act succeed in limiting?

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4 years ago

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A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

$P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m$

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$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m$

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If h = 35 m

From energy conservation

$65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0$

Solving the above equation we get

$x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45$

The compression of the net is 1.52 m

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Your anwser is A
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Explanation:

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