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:
I=1.48 A
Explanation:
Given that
B=3.1 x 10⁻5 T
b= 4.2 cm
l= 9.5 cm
The relationship for magnetic field and current given as
![B=\dfrac{2\mu _oI}{\pi}D](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B2%5Cmu%20_oI%7D%7B%5Cpi%7DD)
Where
![D=\dfrac{\sqrt{l^2+b^2}}{lb}](https://tex.z-dn.net/?f=D%3D%5Cdfrac%7B%5Csqrt%7Bl%5E2%2Bb%5E2%7D%7D%7Blb%7D)
By putting the values
![D=\dfrac{\sqrt{l^2+b^2}}{lb}](https://tex.z-dn.net/?f=D%3D%5Cdfrac%7B%5Csqrt%7Bl%5E2%2Bb%5E2%7D%7D%7Blb%7D)
![D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}](https://tex.z-dn.net/?f=D%3D%5Cdfrac%7B%5Csqrt%7B0.042%5E2%2B0.095%5E2%7D%7D%7B0.042%5Ctimes%200.095%7D)
D=26.03 m⁻¹
![B=\dfrac{2\mu _oI}{\pi}D](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B2%5Cmu%20_oI%7D%7B%5Cpi%7DD)
![3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03](https://tex.z-dn.net/?f=3.1%5Ctimes%2010%5E%7B-5%7D%3D%5Cdfrac%7B2%5Ctimes%204%5Ctimes%20%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%20I%7D%7B%5Cpi%7D%5Ctimes%2026.03)
![I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B3.1%5Ctimes%2010%5E%7B-5%7D%7D%7B%7B2%5Ctimes%204%5Ctimes%2010%5E%7B-7%7D%20%7D%5Ctimes%2026.03%7D)
I=1.48 A
Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.
![V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t](https://tex.z-dn.net/?f=V_x%20.t%27%20%3D%203d%5C%5C%20%5Cfrac%7B3d%7D%7Bt%7D%20.t%27%20%3D%203d%5C%5C%20t%27%3Dt)
Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.
![V_y.t = \frac{2d}{t} .t = 2d](https://tex.z-dn.net/?f=V_y.t%20%3D%20%5Cfrac%7B2d%7D%7Bt%7D%20.t%20%3D%202d)
We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
Answer:
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