Answer:
The speed it reaches the bottom is

Explanation:
Given:
, 
Using the conservation of energy theorem


, 
![m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2](https://tex.z-dn.net/?f=m%2Ag%2Ah%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%28r%2Aw%29%5E2%20%2B%5Cfrac%7B1%7D%7B2%7D%2A%5B%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Ar%5E2%5D%2Aw%5E2)


Solve to w'





Answer:
71.19 C
Explanation:
25C = 25 + 273 = 298 K
Applying the ideal gas equation we have

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
<h3>De Broglie wavelength:</h3>
The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
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