i simply dont understand why now???????/

slope is tan^-1(4/10)

now the base is 2 metre.. so the height change should be x/2=4/10

thus x=.8

height is thus 3.6 and other will be 4.4

eureka i think answer is wrong..

base is 2 ..

x is the difference in height..

Proof in next post..

take the point on the top of the water surface

its acceleration is a with respect to the ground

it is very small let its mass be m

mg=Rcosθ
Rsinθ = ma

tan θ = a/g

thus if the difference in height is x then the base is 2

tan θ = x/2

so x = 2 tan θ = 2 a/g = 2.4/10 = .8 meters

nwo see the volume

initial volume is 3.A

Now if the height on one side is H then on other side is H+x

Total volume is (H+x/2)A = 3A

Thus, H=3-.4 = 2.6

and the other side has heigh of 2.6+x = 3.4 m

b)Find accelration at which liquid is just touching the base of front assuming container is open.Also fond spilled volume

if the container is open.. then the volume contained will be 1/2(4+0).A
=2A = 2.4 m³ = 8m³

In this case tan θ = a/g = 4/2= a/10

thus, a=20 m/s²

c)If accelration is 10m/s2 and container is colosed from top,find pressure at top and bottom of rear face.

now here you have to conserve volume ...!!

this is the more dirty question.. but the method is simpler than you think!! (see the acceleration in forward and force in downward direction are equal..) so tan θ = pi/4

See the right whtie portion of unfilled liquid has equal sides.. let x

volume of blue part = 4.4-1/2x² × 2 = 3x4 = 12

4= x²

x = 2

Now, can you solve it? Pressure is known at some points....