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Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment ![M_{l}=2.90\times10^{-28}\ kg](https://tex.z-dn.net/?f=M_%7Bl%7D%3D2.90%5Ctimes10%5E%7B-28%7D%5C%20kg)
Mass of the heavier fragment ![M_{h}=1.62\times10^{-27}\ Kg](https://tex.z-dn.net/?f=M_%7Bh%7D%3D1.62%5Ctimes10%5E%7B-27%7D%5C%20Kg)
Speed of lighter fragment = 0.893c
We need to calculate the speed of the heavier fragment
Let v is the speed of the second fragment after decay
Using conservation of relativistic momentum
![0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}](https://tex.z-dn.net/?f=0%3D%5Cdrac%7Bm_%7B1%7Dv_%7B1%7D%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv_%7B1%7D%5E2%7D%7Bc%5E2%7D%7D%7D-%5Cdrac%7Bm_%7B2%7Dv_%7B2%7D%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv_%7B1%7D%5E2%7D%7Bc%5E2%7D%7D%7D)
![\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}](https://tex.z-dn.net/?f=%5Cdrac%7Bm_%7B1%7Dv_%7B1%7D%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv_%7B1%7D%5E2%7D%7Bc%5E2%7D%7D%7D%3D%5Cdrac%7Bm_%7B2%7Dv_%7B2%7D%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv_%7B1%7D%5E2%7D%7Bc%5E2%7D%7D%7D)
![\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}](https://tex.z-dn.net/?f=%5Cdfrac%7B2.90%5Ctimes10%5E%7B-28%7D%5Ctimes0.893c%7D%7B%5Csqrt%7B1-%280.893%29%5E2%7D%7D%3D%5Cdfrac%7B1.62%5Ctimes10%5E%7B-27%7Dv_%7B2%7D%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv_%7B2%7D%5E2%7D%7Bc%5E2%7D%7D%7D)
![\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_%7B2%7D%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv_%7B2%7D%5E2%7D%7Bc%5E2%7D%7D%7D%3D%5Cdfrac%7B2.90%5Ctimes10%5E%7B-28%7D%5Ctimes0.893c%7D%7B1.62%5Ctimes10%5E%7B-27%7D%5Ctimes0.45%7D)
![\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_%7B2%7D%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv_%7B2%7D%5E2%7D%7Bc%5E2%7D%7D%7D%3D0.355c)
![\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_%7B2%7D%7D%7B1-%5Cdfrac%7Bv_%7B2%7D%5E%7B2%7D%7D%7Bc%5E2%7D%7D%3D%280.355c%29%5E2)
![\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}](https://tex.z-dn.net/?f=%5Cdfrac%7B1-%5Cdfrac%7Bv_%7B2%7D%5E2%7D%7Bc%5E2%7D%7D%7Bv_%7B2%7D%5E2%7D%3D%5Cdfrac%7B1%7D%7B%280.355c%29%7D)
![\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv_%7B2%7D%5E2%7D-%5Cdfrac%7B1%7D%7Bc%5E2%7D%3D%5Cdfrac%7B1%7D%7B%280.355c%29%5E2%7D)
![\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv_%7B2%7D%5E2%7D%3D%5Cdfrac%7B1%7D%7Bc%5E2%7D%2B%5Cdfrac%7B1%7D%7B0.126c%5E2%7D)
![\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv_%7B2%7D%5E2%7D%3D%5Cdfrac%7B1%7D%7Bc%5E2%7D%281%2B%5Cdfrac%7B1%7D%7B0.126%7D%29)
![\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv_%7B2%7D%5E2%7D%3D%5Cdfrac%7B8.93%7D%7Bc%5E2%7D)
![v_{2}^2=\dfrac{c^2}{8.93}](https://tex.z-dn.net/?f=v_%7B2%7D%5E2%3D%5Cdfrac%7Bc%5E2%7D%7B8.93%7D)
![v_{2}=0.335c](https://tex.z-dn.net/?f=v_%7B2%7D%3D0.335c)
Hence, The speed of the heavier fragment is 0.335c.
weight = mg acts
downwards <span>
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is
acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left
Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ
We let the value of µs be equal to a value such that any F
will not be able to move the crate. Then, if we increase F by an amount F',
then the force pushing the crate towards the right also increases by F' cosΘ. Additionally,
the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase
by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds
by F' sin Θ.
<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>
Answer: Chopping carrots is a chemical change because you aren't changing the property its substance, you are just chopping it up, there is no new substance being formed.
Explanation: