Answer:
- a.
![\Delta s ^2 = 8.0888 \ 10^{17} m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%208.0888%20%5C%2010%5E%7B17%7D%20m%5E2)
- b.
![\Delta s ^2 = 3.0234 \ 10^{16} m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%203.0234%20%5C%2010%5E%7B16%7D%20m%5E2)
- c.
![\Delta s ^2 = 3.0234 \ 10^{20} m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%203.0234%20%5C%2010%5E%7B20%7D%20m%5E2)
Explanation:
The spacetime interval
is given by
![\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%20%5CDelta%20%28c%20t%29%20%5E%202%20-%20%5CDelta%20%5Cvec%7Bx%7D%5E2)
please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:
.
Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.
<h3>a.</h3>
![\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2](https://tex.z-dn.net/?f=%5CDelta%20%5Cvec%7Bx%7D%5E2%20%3D%20%287.5%20%5C%2010%20%5C%20m%29%5E2)
![\Delta \vec{x}^2 = 5,625 m^2](https://tex.z-dn.net/?f=%5CDelta%20%5Cvec%7Bx%7D%5E2%20%3D%205%2C625%20m%5E2)
![\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%20%28299%2C792%2C458%20%5Cfrac%7Bm%7D%7Bs%7D%20%5C%203%20%5C%20s%29%5E2)
![\Delta (c t) ^ 2 = (899,377,374 \ m)^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%20%28899%2C377%2C374%20%5C%20m%29%5E2)
![\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%208.0888%20%5C%2010%5E%7B17%7D%20m%5E2)
so
![\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%208.0888%20%5C%2010%5E%7B17%7D%20m%5E2%20-%205%2C625%20m%5E2)
![\Delta s ^2 = 8.0888 \ 10^{17} m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%208.0888%20%5C%2010%5E%7B17%7D%20m%5E2)
<h3>b.</h3>
![\Delta \vec{x}^2 = (5 \ 10 \ m)^2](https://tex.z-dn.net/?f=%5CDelta%20%5Cvec%7Bx%7D%5E2%20%3D%20%285%20%5C%2010%20%5C%20m%29%5E2)
![\Delta \vec{x}^2 = 2,500 m^2](https://tex.z-dn.net/?f=%5CDelta%20%5Cvec%7Bx%7D%5E2%20%3D%202%2C500%20m%5E2)
![\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%20%28299%2C792%2C458%20%5Cfrac%7Bm%7D%7Bs%7D%20%5C%200.58%20%5C%20s%29%5E2)
![\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%20%28173%2C879%2C625.6%20%5C%20m%29%5E2)
![\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%203.0234%20%5C%2010%5E%7B16%7D%20m%5E2)
so
![\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%203.0234%20%5C%2010%5E%7B16%7D%20m%5E2%20-%202%2C500%20m%5E2)
![\Delta s ^2 = 3.0234 \ 10^{16} m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%203.0234%20%5C%2010%5E%7B16%7D%20m%5E2)
<h3>c.</h3>
![\Delta \vec{x}^2 = (5 \ 10 \ m)^2](https://tex.z-dn.net/?f=%5CDelta%20%5Cvec%7Bx%7D%5E2%20%3D%20%285%20%5C%2010%20%5C%20m%29%5E2)
![\Delta \vec{x}^2 = 2,500 m^2](https://tex.z-dn.net/?f=%5CDelta%20%5Cvec%7Bx%7D%5E2%20%3D%202%2C500%20m%5E2)
![\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%20%28299%2C792%2C458%20%5Cfrac%7Bm%7D%7Bs%7D%20%5C%2058%20%5C%20s%29%5E2)
![\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%20%281.73879%20%5C%2010%5E%7B10%7D%20%5C%20m%29%5E2)
![\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2](https://tex.z-dn.net/?f=%5CDelta%20%28c%20t%29%20%5E%202%20%3D%203.0234%20%5C%2010%5E%7B20%7D%20m%5E2)
so
![\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%203.0234%20%5C%2010%5E%7B20%7D%20m%5E2%20-%202%2C500%20m%5E2)
![\Delta s ^2 = 3.0234 \ 10^{20} m^2](https://tex.z-dn.net/?f=%5CDelta%20s%20%5E2%20%3D%203.0234%20%5C%2010%5E%7B20%7D%20m%5E2)
Answer:
1.![\theta=29.84^{0}](https://tex.z-dn.net/?f=%5Ctheta%3D29.84%5E%7B0%7D)
2.![\theta=60.15^{0}](https://tex.z-dn.net/?f=%5Ctheta%3D60.15%5E%7B0%7D)
Explanation:
Polarizes axis can create two possible angles with the vertical.
first we have to find the intensity of first polarizer
which is given as
![I=\frac{I_{0} }{2}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BI_%7B0%7D%20%7D%7B2%7D)
![I= \frac{655\frac{W}{M^{2} } }{2}](https://tex.z-dn.net/?f=I%3D%20%5Cfrac%7B655%5Cfrac%7BW%7D%7BM%5E%7B2%7D%20%7D%20%7D%7B2%7D)
![I=327.5\frac{W}{m^{2} }](https://tex.z-dn.net/?f=I%3D327.5%5Cfrac%7BW%7D%7Bm%5E%7B2%7D%20%7D)
For a smaller angle for the first polarizer:
According to Malus Law
![I_{2} =I_{1} Cos^{2}(90^{0} - \theta)](https://tex.z-dn.net/?f=I_%7B2%7D%20%3DI_%7B1%7D%20Cos%5E%7B2%7D%2890%5E%7B0%7D%20-%20%5Ctheta%29)
![I_{2} =I_{1} sin^{2}\theta](https://tex.z-dn.net/?f=I_%7B2%7D%20%3DI_%7B1%7D%20sin%5E%7B2%7D%5Ctheta)
![\frac{I_{2} }{I_{1} }=Sin^{2}\theta](https://tex.z-dn.net/?f=%5Cfrac%7BI_%7B2%7D%20%7D%7BI_%7B1%7D%20%7D%3DSin%5E%7B2%7D%5Ctheta)
taking square root on both sides
![\sqrt{\frac{163}{327.5} } = sin\theta](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B163%7D%7B327.5%7D%20%7D%20%3D%20sin%5Ctheta)
![\theta=Sin^{-1}(0.4977)](https://tex.z-dn.net/?f=%5Ctheta%3DSin%5E%7B-1%7D%280.4977%29)
![\theta=29.84^{0}](https://tex.z-dn.net/?f=%5Ctheta%3D29.84%5E%7B0%7D)
For a larger angle for the first polarizer:
According to Malus Law
![I_{2} =I_{1} cos^{2}\theta](https://tex.z-dn.net/?f=I_%7B2%7D%20%3DI_%7B1%7D%20cos%5E%7B2%7D%5Ctheta)
![\frac{I_{2} }{I_{1} }=Cos^{2}\theta](https://tex.z-dn.net/?f=%5Cfrac%7BI_%7B2%7D%20%7D%7BI_%7B1%7D%20%7D%3DCos%5E%7B2%7D%5Ctheta)
taking square root on both sides
![\sqrt{\frac{163}{327.5} } = cos\theta](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B163%7D%7B327.5%7D%20%7D%20%3D%20cos%5Ctheta)
![\theta=Cos^{-1}(0.4977)](https://tex.z-dn.net/?f=%5Ctheta%3DCos%5E%7B-1%7D%280.4977%29)
![\theta=60.15^{0}](https://tex.z-dn.net/?f=%5Ctheta%3D60.15%5E%7B0%7D)
Answer: Yes.
Explanation: It is clearly stated in Newton’s first law of physics that an object will not change its motion unless a force acts on.
Answer: 585 J
Explanation:
We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:
![W=K_f -K_i](https://tex.z-dn.net/?f=W%3DK_f%20-K_i)
where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is
![K_f =\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K_f%20%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:
![W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J](https://tex.z-dn.net/?f=W%3DK_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281.3%20kg%29%2830%20m%2Fs%29%5E2%3D585%20J)
No they do not they just need to be in each other's magnetic field