Answer:
A) s=1/2at^2
t=√(2s/a)=√(2x400)/10.0)=9.0s
B) v=at
v=10.0x9=90m/s
Given that,
Mass of trackler, m₁ = 100 kg
Speed of trackler, u₁ = 2.6 m/s
Mass of halfback, m₂ = 92 kg
Speed of halfback, u₂ = -5 m/s (direction is opposite)
To find,
Mutual speed immediately after the collision.
Solution,
The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.
Answer:
Option C and D only
Explanation:
Option A is incorrect because refractive index of a material is the ratio of speed of light in vacuum to the speed of light in a any given medium
Option B is correct as the speed of light in vacuum is always greater than the speed of light in any given medium.
Option C is correct
Option D is incorrect
Option E is incorrect because the denser the medium the more is the refractive index. Water is denser than air, hence it should have more refractive index as compared to that of air.
Explanation:
It is given that,
Force, 
Position vector, 
(a) The torque on the particle about the origin is given by :

(b) To find the angle between r and F use dot product formula as :

Hence, this is the required solution.
Answer. Second Option: .85p_o=p_o e^-.00012h
Solution:
P(h)=Po e^(-0.00012h)
Air pressure: P(h)
Height above the surface of the Earth (in meters): h
Air pressure at the sea level: Po
Height at which air pressure is 85% of the air pressure at sea level:
h=?, P(h)=85% Po
P(h)=(85/100) Po
P(h)=0.85 Po
Replacing P(h) by 0.85 Po in the formula above:
P(h)=Po e^(-0.00012h)
0.85 Po = Po e^(-0.00012h)