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padilas [110]
3 years ago
10

Intermolecular distance is the distance between the particles that make up matter. The graph below compares the intermolecular d

istances in two substances. One substance is gas and the other is a liquid.
Which statement best describes the two substances?

A. Particles collide in liquids so Substance Y is a liquid
B. Particles collide in gas so Substance X is a gas.
C. Particles move around in liquids so Substance X is a liquid.
D. Particles move around in gas so Substance Y is a gas.

Physics
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

y has to be a liquid and x would be a gas

Explanation:

the closer two molecules are, greater is the force of attraction which makes them stick together and give it a structure

here y has lesser intermolecular distance so it is a liquid

particles will collide in gas though as

liquid particles have constant motion

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A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi
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  • Magnitude: 12.1 N.
  • Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

  • \displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.
  • \displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.

The sum of forces on each direction will be the resultant force on that direction:

  • Resultant force parallel to the 8 N force: (8 + \dfrac{5\sqrt{2}}{2})\;\text{N}.
  • Resultant force normal to the 8 N force: \dfrac{5\sqrt{2}}{2}\;\text{N}.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N} (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491.

Find the size of the angle using inverse tangent:

\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree.

In other words, the resultant force is 17.0° relative to the 8 N force.

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