maybe if you look it up it will come up
Answer:
The speed of the animals is 1.64m/s.
Explanation:
Let us work with variables and call the mass of the two rats
and
, and the length of the rod
.
Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,
![$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$](https://tex.z-dn.net/?f=%24m_1%5Cfrac%7BL%7D%7B2%7Dg%20-m_2%5Cfrac%7BL%7D%7B2%7Dg%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_1v%5E2%2B%5Cfrac%7B1%7D%7B2%7Dm_2v%5E2%24)
,
solving for
, we get:
![$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$](https://tex.z-dn.net/?f=%24%5Cboxed%7Bv%20%3D%20%5Csqrt%7B%5Cfrac%7B%28m_1%20-m_2%29gL%7D%7B%28m_1%2Bm_2%29%7D%7D%20%7D%24)
Putting in the values for
,
,
, and
we get:
![$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}} $](https://tex.z-dn.net/?f=%24v%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.450kg%20-0.220kg%29%289.8m%2Fs%5E2%29%280.8m%29%7D%7B%280.450kg%2B0.220kg%29%7D%7D%20%24)
![\boxed{ v= 1.64m/s}](https://tex.z-dn.net/?f=%5Cboxed%7B%20v%3D%201.64m%2Fs%7D)
Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.
Use Charles Law: V1/T1 = V2/T2
0.30 m^3/27 C = V2/127 C
27V2 = 127 * 0.3
V2= 38.1/27 = 1.4 m^3