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3 years ago

How does approaching a temperature of absolute zero affect kinetic energy.?

Physics

1 answer:

Alexus [3.1K]3 years ago

7 0

First of all the kinetic energy is when the particles move in continuous random motion.

If the temperature is high the colliding particles will collide more. and if the temperature is low the colliding particles will collide less.

Low temperature result in low kinetic energy
High temperature result in high kinetic energy

Absolute zero is the point where where all molecules have no kinetic energy. It is a theoretical value (it has never been reached).

The Kelvin temperature scale is based on absolute zero being the lowest possible temperature that could theoretically be reached. That is why there is no such thing as a negative Kelvin temperature value.

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We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one

Len [333]

Answer:

A=1

B=-2

Explanation:

Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.

From part B as attached, it shows that the right option is C which is

2A+3B=-4

Substituting B with 3A-5 then we form the second equation as shown

2A+3(3A-5)=-4

By simplifying the above equation, we obtain

2A+9A-15=-4

Re-arranging, then

11A=-4+15

Finally

11A=11

A=1

To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain

B=3(1)-5=-2

Therefore, required values are 1 and -2

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3 years ago

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2 years ago

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3 years ago

Which represents a deletion of the section of DNA shown here, a-c-t-g-g-a-t?

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3 years ago

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

4 0

2 years ago

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