The time passed on earth is mathematically given as
t' = 24.79 hrs
<h3>What is the time passed on earth?</h3>
Generally, the equation for is time mathematically given as
![t' = \gamma t](https://tex.z-dn.net/?f=t%27%20%3D%20%5Cgamma%20t)
Where
![\gamma = Lorentz\ factor \\\\\gamma = 1/ \sqrt {(1 - v^2/c^2)}](https://tex.z-dn.net/?f=%5Cgamma%20%3D%20Lorentz%5C%20factor%20%5C%5C%5C%5C%5Cgamma%20%3D%201%2F%20%5Csqrt%20%7B%281%20-%20v%5E2%2Fc%5E2%29%7D)
![t' = t/ \sqrt {(1 - v^2/c^2)}](https://tex.z-dn.net/?f=t%27%20%3D%20t%2F%20%5Csqrt%20%7B%281%20-%20v%5E2%2Fc%5E2%29%7D)
Therefore
![t' = 24/ \sqrt {(1 - (0.25c/c)^2) }](https://tex.z-dn.net/?f=t%27%20%3D%2024%2F%20%5Csqrt%20%7B%281%20-%20%280.25c%2Fc%29%5E2%29%20%7D)
![t'= 24/ \sqrt {(1 - 0.25^2)](https://tex.z-dn.net/?f=t%27%3D%2024%2F%20%5Csqrt%20%7B%281%20-%200.25%5E2%29)
t' = 24.79 hrs
In conclusion, the time passed on earth
t' = 24.79 hrs
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Answer:
Landed before it explodes
Explanation:
vf = vi + at,
0 = 145 - (9.8)t,
t = 14.79 s (Time to reach highest point)
14.79 x 2 = 29.59 s (Time to land on the ground)
It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.
<span>In most cases, magma differentiation (a.k.a. fractional crystallization produces magma with higher silica content than the parent magma. Fractional crystallization removes early formed minerals in magma. The liquid that does not react to the process remains in the magma. </span>
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia, ![I_1=900\ kg.m^2](https://tex.z-dn.net/?f=I_1%3D900%5C%20kg.m%5E2)
Initial angular velocity of the platform, ![\omega=0.95\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D0.95%5C%20rad%2Fs)
Part A,
Let
is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :
![I_1\omega_1=I_2\omega_2](https://tex.z-dn.net/?f=I_1%5Comega_1%3DI_2%5Comega_2)
Here, ![I_2=I_1+mr^2](https://tex.z-dn.net/?f=I_2%3DI_1%2Bmr%5E2)
![I_1\omega_1=(I_1+mr^2)\omega_2](https://tex.z-dn.net/?f=I_1%5Comega_1%3D%28I_1%2Bmr%5E2%29%5Comega_2)
![900\times 0.95=(900+70\times (2.9)^2)\omega_2](https://tex.z-dn.net/?f=900%5Ctimes%200.95%3D%28900%2B70%5Ctimes%20%282.9%29%5E2%29%5Comega_2)
Solving the above equation, we get the value as :
![\omega_2=0.574\ rad/s](https://tex.z-dn.net/?f=%5Comega_2%3D0.574%5C%20rad%2Fs)
Part B,
The initial rotational kinetic energy is given by :
![k_i=\dfrac{1}{2}I_1\omega_1^2](https://tex.z-dn.net/?f=k_i%3D%5Cdfrac%7B1%7D%7B2%7DI_1%5Comega_1%5E2)
![k_i=\dfrac{1}{2}\times 900\times (0.95)^2](https://tex.z-dn.net/?f=k_i%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20900%5Ctimes%20%280.95%29%5E2)
![k_i=406.12\ rad/s](https://tex.z-dn.net/?f=k_i%3D406.12%5C%20rad%2Fs)
The final rotational kinetic energy is given by :
![k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2](https://tex.z-dn.net/?f=k_f%3D%5Cdfrac%7B1%7D%7B2%7D%28I_1%2Bmr%5E2%29%5Comega_1%5E2)
![k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2](https://tex.z-dn.net/?f=k_f%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%28900%2B70%5Ctimes%20%282.9%29%5E2%29%280.574%29%5E2)
![k_f=245.24\ rad/s](https://tex.z-dn.net/?f=k_f%3D245.24%5C%20rad%2Fs)
Hence, this is the required solution.
It is a solid when is frozen and a liquid when it melts