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3 years ago

Looking for some help with the left side of the table and the questions followed :)

Physics

1 answer:

Inessa [10]3 years ago

6 0

Answer:

(1). Going in the order: Gas, Liquid, Solid.

Particle separation: Very large; large; small.

Strength of forces between particles: very weak; weak; strong.

(2). a. The change of state is condensation.

b. The density of the substance is increasing as particles move closer together.

(3). a. The liquid's temperature decreases until it reaches solidification (freezing) point, and then the temperature will remain constant until all of the liquid has solidified.

b. The particles move with less and less energy until they have low enough energy to come together and solidify, then they will release more energy when they come together to form bonds, and vibrate about their equilibrium positions once they are part of a molecular bond.

Additional explanation:

(1). The particle separation in a gas is very large because the gas has a very low density, and since particles are not that close together,<em> the strength of forces between particles is very weak</em>.

The liquid is somewhat more denser than the gas and therefore the particle separation is large but less than that in a gas,<em> The strength of forces between particles is still weak but greater than in a gas. </em>

Finally, particle separation in a solid is small because the solid is still more denser than the liquid, and the strength of forces between particles is strong because it's these forces that give a solid its rigid shape.

(2). a.The change of state of a substance in which particles slow down and move close together is called condensation.

b. As the particles move close together, they are occupying less and less volume; therefore, the density is increasing.

(3). The explanation is already given in the answer.

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2 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

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Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$