Why hasn't a perpetual motion machine been created?

1 answer:

5 0

Thus, machines that extract energy from finite sources will not operate indefinitely, because they are driven by the energy stored in the source, which will eventually be exhausted. A common example is devices powered by ocean currents, whose energy is ultimately derived from the Sun, which itself will eventually burn out<span>. Machines powered by more obscure sources have been proposed, but are subject to the same inescapable laws, and will eventually wind down.</span>

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What happens when a moving object bumps into a stationary object?.

Sonbull [250]

It depends on the mass of the moving object versus the mass of the stationary object. if the mass of the moving object is larger the stationary object will get sent into motion. if the mass of the stationary object is larger than the moving object, the stationary object will stay stationary and cause the moving object to do the same. if the two objects have the same mass, they will likely move together upon impact and then eventually come to rest.

5 0

2 years ago

Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental const

ludmilkaskok [199]

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$ , Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$ , the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$ , so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$