55 meters. If she started at 10 meters and ran 45 more, 10+45=55.
Answer:
The index of refraction of the liquid is n = 1.33 equivalent to that of water
Explanation:
Solution:-
- The index of refraction of light in a medium ( n ) determines the degree of "bending" of light in that medium.
- The index of refraction is material property and proportional to density of the material.
- The denser the material the slower the light will move through associated with considerable diffraction angles.
- The lighter the material the faster the light pass through the material without being diffracted as much.
- So, in the other words index of refraction can be expressed as how fast or slow light passes through a medium.
- The reference of comparison of how fast or slow the light is the value of c = 3.0*10^8 m/s i.e speed of light in vacuum or also assumed to be the case for air.
- so we can mathematically express the index of refraction as a ratio of light speed in the material specified and speed of light.
- The light passes through a liquid with speed v = 2.25*10^8 m/s :
![n = c / v\\\\n = \frac{ 3*10^8 }{2.25*10^8} \\\\n = 1.33](https://tex.z-dn.net/?f=n%20%3D%20c%20%2F%20v%5C%5C%5C%5Cn%20%3D%20%5Cfrac%7B%203%2A10%5E8%20%7D%7B2.25%2A10%5E8%7D%20%5C%5C%5C%5Cn%20%3D%201.33)
- The index of refraction of the liquid is n = 1.33 equivalent to that of water.
If the ball is green, then it is absorbing all the colors of light except green. This means that it is reflecting or denying the absorption of green light, that is why the ball is green. Therefore, your answer would be B.
Hope I helped :)
About 85% sure on this one.
Lets do
We know
The rate of change of velocity is acceleration .
![\\ \sf\longmapsto a=\dfrac{dv}{dt}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20a%3D%5Cdfrac%7Bdv%7D%7Bdt%7D)
![\\ \sf\longmapsto dv=adt](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20dv%3Dadt)
Integrate both sides
![\\ \sf\longmapsto \int dv=a\int dt](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cint%20dv%3Da%5Cint%20dt)
As acceleration is constant .Take it outside of integral .On velocity we can take limit u to v and time from 0 to t
![\\ \sf\longmapsto {\displaystyle{\int}}^v_u dv=a{\displaystyle{\int}}^t_0 dt](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%7B%5Cdisplaystyle%7B%5Cint%7D%7D%5Ev_u%20dv%3Da%7B%5Cdisplaystyle%7B%5Cint%7D%7D%5Et_0%20dt)
Hence
![\\ \sf\longmapsto v{\huge{|}}^v_u=at](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20v%7B%5Chuge%7B%7C%7D%7D%5Ev_u%3Dat)
![\\ \sf\longmapsto v-u=at](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20v-u%3Dat)
![\\ \sf\longmapsto v=u+at](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20v%3Du%2Bat)
![\large \mathfrak{Solution : }](https://tex.z-dn.net/?f=%20%5Clarge%20%5Cmathfrak%7BSolution%20%3A%20%7D)
let's use first equation of motion to solve this ;
Velocity after 30 seconds = 170 m/s