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1 year ago

3. If nothing can ever be at absolute zero, why does the concept exist?

1 answer:

3 0

The absolute zero in temperature refers to the minimal possible temperature. It is the temperature at which the molecules of a system stop moving, so it is a really useful reference point.

<h3>Why absolute zero can't be reached?</h3>

It would mean that we need to remove all the energy from a system, but to do this we need to interact with the system in some way, and by interacting with it we give it "some" energy.

Actually, from a quantum mechanical point of view, the absolute zero has a residual energy (so it is not actually zero) and it is called the "zero point". This happens because it must meet <u>Heisenberg's uncertainty principle</u>.

So yes, the absolute zero can't be reached, but there are really good approximations (At the moment there is a difference of about 150 nanokelvins between the absolute zero and the smallest temperature reached). Also, there are a lot of investigations near the absolute zero, like people that try to reach it or people that just need to work with really low temperatures, like in type I superconductors.

So, concluding, why does the concept exist?

Because it is a reference point.
It is the theoretical temperature at which the molecules stop moving, defining this as the <u>minimum possible temperature.</u>

If you want to learn more about the absolute zero, you can read:

brainly.com/question/3795971

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2 years ago

Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed

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Answer:

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Explanation:

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2 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

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2 years ago

4. What are these parts commonly called?

patriot [66]

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The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.

The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.

These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.

Learn more about carburetors here:

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2 years ago

What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?

nasty-shy [4]

Answer:

$0.14\ lb/in^{3}$

Explanation:

Density is defined as mass ler unit volume, expressed as

$\rho=\frac {m}{v}$

Where m is mass, $\rho$ is density and v is the volume. For a sphere, volume is given as

$v=\frac {4\pi r^{3}}{3}$

Replacing this into the formula of density then

$\rho=\frac {m}{\frac {4\pi r^{3}}{3}}$

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

$\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}$

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2 years ago