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)
FMRI creates the images or brain maps of brain functioning by setting up and utilizing an advanced MRI scanner in such a way that increased blood flow to the activated areas of the brain shows up on the MRI scan. The MRI scanners do not actually detect blood flow or other metabolic processes.
The average act on her during the deceleration is 4.47 meters per second.
<u>Explanation</u>:
<u>Given</u>:
youngster mass m = 50.0 kg
She steps off a 1.00 m high platform that is s = 1 meter
She comes to rest in the 10-meter second
<u>To Find</u>:
The average force and momentum
<u>Formulas</u>:
p = m * v
F * Δ t = Δ p
vf^2= vi^2+2as
<u>Solution</u>:
a = 9.8 m/s
vi = 0
vf^2= 0+2(9.8)(1)
vf^2 = 19.6
vf = 4.47 m/s .
Therefore the average force is 4.47 m/s.
<u>Answer:</u> Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
<u>Explanation:</u>
To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.
This wave base is equal to half of the wavelength. The equation becomes:
Wave base = ![\frac{\text{Wavelength}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BWavelength%7D%7D%7B2%7D)
We are given:
Wavelength = 24 m
Putting values in above equation, we get:
Wave base = ![\frac{24m}{2}=12m](https://tex.z-dn.net/?f=%5Cfrac%7B24m%7D%7B2%7D%3D12m)
Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves