Part a
Answer: NO
We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.
Using the second equation of motion:

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.
It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, 

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.
Part b
Answer: 29.6 m/s
The maximum distance that car can travel is 
The acceleration is same, 
The final velocity, v=0
Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.
Answer:
3Q / 4 pi (R^3 - r^3)
Explanation:
Charge density = charge / volume
volume of a spherical shell = 
Answer:
Explanation:
We shall apply law of refraction which is as follows
sin i / sinr = μ , where i is angle of incidence , r is angle of refraction and μ is refractive index
here i = θa = 22.5°
r = θb
μ = 1.77
sin22.5 / sinθb = 1.77
.3826 / sinθb = 1.77
sinθb = .216
θb = 12.5 °.
Answer:
1184 kJ/kg
Explanation:
Given:
water pressure P= 28 bar
internal energy U= 988 kJ/kg
specific volume of water v= 0.121×10^-2 m^3/kg
Now from steam table at 28 bar pressure we can write


therefore at saturated liquid we have specific enthalpy at 55 bar pressure.
that the specific enthalpy h = h at 50 bar +(55-50)/(60-50)*( h at 50 bar - h at 60 bar)

h= 1184 kJ/kg