The correct answer (sample response) is:
The image seems to be behind the mirror, but nothing is really there.
Include the following in your response:
The image appears to be behind the mirror.
If someone looks behind the mirror, there is no image there.
![|Huntrw6|](https://tex.z-dn.net/?f=%7CHuntrw6%7C)
Answer:
wait what do you mean? And why is this in physics?
Is this about the iron dome or something biblical?
Explanation:
Answer: Move the small car so it appears on the left side of the lens.
Explanation:
Because the lens is reflective the small car would apear on the same side as the normal car.
Hope this helps :)
The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
To find the answer, we have to know about the Lorentz transformation.
<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>
It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.
![V_x'=0.8c\\V=0.6c\\m=4*10^5kg](https://tex.z-dn.net/?f=V_x%27%3D0.8c%5C%5CV%3D0.6c%5C%5Cm%3D4%2A10%5E5kg)
- Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,
![KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}](https://tex.z-dn.net/?f=KE%3Dm_0c%5E2%3D%5Cfrac%7Bmc%5E2%7D%7B1-%28%5Cfrac%7Bv_x%5E2%7D%7Bc%5E2%7D%20%29%7D)
- So, to
find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame. - We have an expression from Lorents transformation for relativistic law of addition of velocities as,
![V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V](https://tex.z-dn.net/?f=V_x%27%3D%5Cfrac%7BV_x-V%7D%7B1-%5Cfrac%7BVV_x%7D%7Bc%5E2%7D%20%7D%20%5C%5Cthus%2C%5C%5CV_x%3DV_x%27%281-%5Cfrac%7BVV_x%7D%7Bc%5E2%7D%20%29%2BV)
- Substituting values, we get,
![V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c](https://tex.z-dn.net/?f=V_x%3D0.8c%281-%5Cfrac%7B0.8c%2A0.6c%7D%7Bc%5E2%7D%20%29%2B0.6c%3D%280.8c%2A0.52%29%2B0.6c%3D1.016c)
![KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B4%2A10%5E5%2A%283%2A10%5E8%29%5E2%7D%7B%5Csqrt%7B1-%5Cfrac%7B%281.016c%29%5E2%7D%7Bc%5E2%7D%20%7D%20%7D%20%3D%5Cfrac%7B1.2%2A10%5E%7B22%7D%7D%7B0.179%7D%3D6.704%2A10%5E%7B22%7DJ)
Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
Learn more about frame of reference here:
brainly.com/question/20897534
SPJ4