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ludmilkaskok [199]
3 years ago
12

What is the relationship between matter and energy as it changes states of matter (phase changes?)

Physics
1 answer:
AlekseyPX3 years ago
7 0

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.

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A sample of helium gas has a volume of 900. milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the t
Sladkaya [172]

Answer:

<em>The new pressure = 5.64 atm.</em>

Explanation:

Using The general gas equation,

P₁V₁/T₁ = P₂V₂/T₂..................... Equation 1

making P₂ the subject of the equation

P₂ = P₁V₁T₂/V₂T₁ ...................... Equation 2

Where P₁ = initial pressure, V₁ = initial Volume, T₁ = Initial temperature, P₂ = final pressure, V₂ = final volume, T₂ = final Temperature.

<em>Given: P ₁= 2.5 atm, T₁ = 298 K, V₁= 900 milliliters, T ₂= 336 K, V₂ = 450 milliliters</em>

<em>Substituting these values into equation 2,</em>

<em>P₂ = (2.5×900×336)/(298×450)</em>

P₂ = 756000/134100

P₂ = 5.64 atm.

<em>Thus the new pressure = 5.64 atm.</em>

5 0
3 years ago
Is the color spectrum simply a small segment of the electromagnetic spectrum?
ch4aika [34]

Answer:

yup, u r correct

Explanation:

3 0
3 years ago
Taylor mixes two liquids together in a beaker. A solid forms at the bottom of the beaker, and the liquid changes from pink to bl
Dimas [21]
It is a chemical change and a physical change
3 0
3 years ago
A racecar driver circles a racetrack with a constant speed of 110 km/hr. Which BEST describes the speed and velocity of the car?
tangare [24]
The answer is B, The speed is constant and the velocity is changing.
8 0
3 years ago
Read 2 more answers
Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

8 0
3 years ago
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