A. is meaningless.
B. is perigee.
C. is apogee.
D. is perihelion.
Dependent variable is your answer.
You are given a fixed rate of 15.9 cm³/s. You are also given with the amount of volume in 237 cm³. Through the approach of dimensional analysis, you can manipulate through operations such that the end result of the units must be in seconds. The solution is as follows:
237 cm³ * (1 s/15.9 cm³) = 14.9 seconds
Answer:
A) The space time coordinate x of the collision in Earth's reference frame is
.
B) The space time coordinate t of the collision in Earth's reference frame is
![t=377,29s](https://tex.z-dn.net/?f=t%3D377%2C29s)
Explanation:
We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).
An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.
<em>Lorentz transformation</em>
The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.
The Lorentz transformation is
![x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}](https://tex.z-dn.net/?f=x%27%3D%5Cfrac%7Bx-vt%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D)
![y'=y](https://tex.z-dn.net/?f=y%27%3Dy)
![z'=z](https://tex.z-dn.net/?f=z%27%3Dz)
![t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}](https://tex.z-dn.net/?f=t%27%3D%5Cfrac%7Bt-%5Cfrac%7Bv%7D%7Bc%5E%7B2%7D%7Dx%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D)
prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations
![x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bx%27%2Bvt%27%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D)
![y=y'](https://tex.z-dn.net/?f=y%3Dy%27)
![z=z'](https://tex.z-dn.net/?f=z%3Dz%27)
![t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bt%27%2B%5Cfrac%7Bv%7D%7Bc%5E%7B2%7D%7Dx%27%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D)
First we calculate the expression in the denominator
![\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%3D%5Cfrac%7B%280%2C70%29%5E%7B2%7Dc%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%20%3D%280%2C70%29%5E%7B2%7D)
![\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714](https://tex.z-dn.net/?f=%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%20%3D0%2C714)
then we calculate t
![t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bt%27%2B%5Cfrac%7Bv%7D%7Bc%5E%7B2%7D%7Dx%27%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D)
![t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B190s%2B%5Cfrac%7B0%2C70c%7D%7Bc%5E%7B2%7D%7D.3%2C4x10%5E%7B10%7Dm%7D%7B0%2C714%7D)
![t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B190s%2B%5Cfrac%7B0%2C70c%20.3%2C4x10%5E%7B10%7Dm%7D%7B299792458%5Cfrac%7Bm%7D%7Bs%7D%7D%7D%7B0%2C714%7D)
![t=\frac{190s+79,388s}{0,714}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B190s%2B79%2C388s%7D%7B0%2C714%7D)
finally we get that
![t=377,29s](https://tex.z-dn.net/?f=t%3D377%2C29s)
then we calculate x
![x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bx%27%2Bvt%27%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D)
![x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B3%2C4x10%5E%7B10%7Dm%2B0%2C70c.190s%7D%7B0%2C714%7D%7D)
![x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B3%2C4x10%5E%7B10%7Dm%2B0%2C70.299792458%5Cfrac%7Bm%7D%7Bs%7D.190s%7D%7B0%2C714%7D%7D)
![x=\frac{3,4x10^{10}m+39872396914m}{0,714}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B3%2C4x10%5E%7B10%7Dm%2B39872396914m%7D%7B0%2C714%7D%7D)
![x=\frac{73872396914m}{0,714}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B73872396914m%7D%7B0%2C714%7D%7D)
![x=103462740775,91m](https://tex.z-dn.net/?f=x%3D103462740775%2C91m)
finally we get that
![x \approx 103,46x10^{9} m](https://tex.z-dn.net/?f=x%20%5Capprox%20103%2C46x10%5E%7B9%7D%20m)
Answer:
A. Inertial Confinement and B. Magnetic Confinement