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4 years ago

Why would scientists call solids and liquids a condensed phase of matter?

Physics

1 answer:

IceJOKER [234]4 years ago

6 0

Explanation:

Liquids and solids are often referred to as condensed phases because the particles are very close together. The following table summarizes properties of gases, liquids, and solids and identifies the microscopic behavior responsible for each property.

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Which ancient idea of the nature of light did Isaac Newton build upon? Light is a wave or disturbance that travels through space

JulijaS [17]

Answer:

The answer is C

Explanation:

Light is a substance through which particles flow from a light source

Just took the quiz on edge.

6 0

3 years ago

Earth's moon orbits Earth at a mean distance of 382,508,576 m. What is the moon's orbital speed? Earth's mass is 5.972x1024 kg.

Snezhnost [94]

Answer:

I think 1.022 or 1

Explanation:

6 0

2 years ago

An object weighs 2.6 N in air and 2.2N when completely immersed in water. Determine the relative density of the object (2mks

frez [133]

Answer:

4.8n

Explanation:

you add 2.6n+2.2n

7 0

3 years ago

The place you get your hair cut has two nearly parallel mirrors 7.00 apart. As you sit between these mirrors in the chair, your

frez [133]

Answer:

$a''_b=18\ m$

Explanation:

Given:

distance between the parallel mirrors, $l=7\ m$

distance of the head from the front mirror, $a=2.5\ m$

By the laws of reflection we know that the image of an object appears at the same distance from the surface of a plane mirror as that of the object from the surface of the mirror.

Now the distance of the head from the rear mirror:

$b=7-2.5$

$b=4.5\ m$

Now the distance of image of the back of head in the rear mirror from the object head:

$b'=2\times 4.5$

$b'=9\ m$

Now this image is reflected into the front mirror from the image of the back of the heat in the rear mirror.

So, this is the second reflection of the actual object, which will form at a distance of:

$a''_b=2\times 9$

$a''_b=18\ m$

6 0

3 years ago

A ball is rolled of the edge of a table with a

tatiyna

Answer:

Approximately $10\; \rm m \cdot s^{-1}$ at $5.6^\circ$ below the horizon.

Horizontal component of velocity: $10\; \rm m \cdot s^{-1}$ .
Vertical component of velocity: $0.981\; \rm m\cdot s^{-1}$ (downwards.)

(Assumption: air resistance on the ball is negligible; $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the air resistance on the ball is negligible. The horizontal component of the velocity of the ball would stay the same at $10\; \rm m \cdot s^{-1}$ until the ball reaches the ground.

On the other hand, the vertical component of the ball would increase (downwards) at a rate of $g = 9.81\; \rm m\cdot s^{-2}$ (where $g$ is the acceleration due to gravity.) In $0.1\; \rm s$ , the vertical component of the velocity of this ball would have increased by $9.81\; \rm m \cdot s^{-2} \times 0.10\; \rm s = 0.981\; \rm m \cdot s^{-1}$ .

However, right after the ball rolled off the edge of the table, the vertical component of the velocity of this ball was $0\; \rm m\cdot s^{-1}$ . Hence, $0.10\; \rm s$ after the ball rolled off the table, the vertical component of the velocity of this ball would be $0\; \rm m \cdot s^{-1} + 0.981\; \rm m\cdot s^{-1} = 0.981\; \rm m \cdot s^{-1}$ .

Calculate the magnitude of the velocity of this ball. Let $v_{x}$ and $v_{y}$ and denote the horizontal and vertical component of the velocity of this ball, respectively. The magnitude of the velocity of this ball would be $\displaystyle \sqrt{{v_x}^{2} + {v_y}^{2}}$ .

At $0.10\; \rm s$ after the ball rolled off the table, $v_x = 10\; \rm m \cdot s^{-1}$ while $v_y = 0.981\; \rm m \cdot s^{-1}$ . Calculate the magnitude of the velocity of the ball at this moment:

$\begin{aligned} \| v \| &= \sqrt{{v_x}^{2} + {v_y}^{2}} \\ &= \sqrt{\left(10\; \rm m \cdot s^{-1}\right)^{2} + \left(0.981\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 10.0\; \rm m\cdot s^{-1}\end{aligned}$ .

Calculate the angle between the horizon and the velocity of the ball (a vector) at that moment. Let $\theta$ denote that angle.

$\displaystyle \tan \theta = \frac{\text{rise}}{\text{run}}$ .

For the vector representing the velocity of this ball:

$\displaystyle \tan \theta = \frac{0.981\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-1}} = 0.0981$ .

Calculate the size of this angle:

$\theta = \arctan 0.0981 \approx 5.62^\circ$ .

Notice that the vertical component of the velocity of this ball at that moment points downwards (towards the ground.) Hence, the corresponding velocity should point below the horizon.

5 0

3 years ago