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3 years ago

Why can't there be a number lower than absolute zero

2 answers:

5 0

Absolute zero is not about numbers. It's about temperature, and the
motion of molecules in gases.

You know that the temperature we feel with our skin is the result of the
average speed of all the tiny molecules zipping around or vibrating in
the solid, liquid, or gas.

The faster they're all moving, the warmer the substance feels to us.
The slower they're all moving, the cooler the substance feels to us.

When molecules slow down to zero and lose all of their kinetic energy,
that temperature is what we call 'absolute zero' ... if they're not moving
at all, then they can't move any slower.

Dafna1 [17]3 years ago

5 0

Temperature lower than absolute zero is not possible. Lower than absolute zero a gas contain negative pressure and volume which is not possible. temperature is the energy created by moving atoms absoulte zero is a complete stand still you can make particles move to the end of the universe but the coldest temperature will be stuck at the point where atoms are motionless

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A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

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Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

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We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

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4 years ago