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1 year 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]1 year 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 charge in motion is called a (n) ______.

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3 years ago

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What is the increase in pressure required to decrease volume of mercury by 0.001%

REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0

3 years ago

For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$

$y-2.25=-\frac{1}{16}x+\frac{1}{4}$

$y=-\frac{1}{16}x+\frac{1}{4}+2.25$

$y=-\frac{1}{16}x+\frac{5}{2}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{16}x+\frac{5}{2}=0$

and solve for x

$x=40$

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

8 0

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Answer:

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explanation:

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