How does increasing the distance between charged objects affect the electric force between them?
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B) initial speed (u)=3
final speed (v)=8
time (t)= 10
acceleration (a)=?
v=u+at
8=3+a(10)
5=10a
a=1/2= 0.5 m/s²
c)displacement (s)=?
1st section of journey
v ²=u ²+2as
8 ²=3 ²+2(0.5)s
64=9+1s
s=55meters
2nd section of journey
a=0
t=7
u=8
s=?
s=ut+1/2at²
s=8(7)+1/2(0)(7 ²)
s=56 meters
56+55= 111 meters
(For graph x axis =time (sec) & y axis= velocity (m/s))
final speed (v)=8
time (t)= 10
acceleration (a)=?
v=u+at
8=3+a(10)
5=10a
a=1/2= 0.5 m/s²
c)displacement (s)=?
1st section of journey
v ²=u ²+2as
8 ²=3 ²+2(0.5)s
64=9+1s
s=55meters
2nd section of journey
a=0
t=7
u=8
s=?
s=ut+1/2at²
s=8(7)+1/2(0)(7 ²)
s=56 meters
56+55= 111 meters
(For graph x axis =time (sec) & y axis= velocity (m/s))
The speed of light in that medium is .
<u>Explanation:</u>
It is known that the light's speed is constant when it travels in vacuum and the value is . When the light enters another medium other than vacuum, its speed get decreased as the light gets refracted by an angle.
The amount of refraction can be determined by the index of refraction or refractive index of the medium. The refraction index is measured as the ratios of speed of light in vacuum to that in the medium. It is represented as η =
So, here η is the index of refraction of a medium which is given as 1.4, c is the light's speed in vacuum () and v is the light's speed in that medium which we need to find.
Thus the speed of light in that medium is