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3 years ago

Compare and contrast the way molecules behave in liquid to the way they behave in solids and gases.

Physics

2 answers:

aniked [119]3 years ago

4 0

Answer:

Atoms, molecules, ions are all present in gases, liquids, and solids, but their behavior varies depending on the process. ...gas are well separated and not arranged in any particular order. There is no regular arrangement of liquids, so they are close together. Solids are arranged in a regular pattern and are tightly packed

Explanation:

Vitek1552 [10]3 years ago

3 0

In liquids, molecules particles are close together but move in random directions.

In solids, they are closely packed together and they vibrate a little.

In gases, they love around really quickly and are farther apart (even more than liquids).

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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3 years ago

Help with Order of Magnitude question?

solong [7]

The rotation of Earth is equivalent to one day which is comprised of 24 hours. To determine the number of miles in Earth's circumference, one simply have to multiply the given rate by the appropriate conversion factor and dimensional analysis. This is shown below.

C = (1038 mi/h)(24 h/1 day)
C = 24,912 miles

From the given choices, the nearest value would have to be 20,000 mile. The answer is the second choice.

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3 years ago

A bicyclist is in a 50-km race. She says she had an average velocity of 35.

zlopas [31]

The correct answer to this is (A. Units Only).

It shows that there is a velocity of 35, but the units are missing.

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4 years ago

Assume light is traveling through an optical medium of n1 and is incident on a boundary with a different optical medium with n2.

Elza [17]

That n2 = 2*n1. That is, the index of refraction is twice as big in medium 2 since v=c/n

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3 years ago