That isn"t the right answer the correct answer is B.
Answer:
If we use the equation for the transformation of velocities for moving frames:
v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'
v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)
or v' = 1.2 c / (1 + .36) = .88 c
v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u (-.6 c) and we measure the speed of v as seen in the frame moving to the left
To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.
By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,
Where,
Velocity in each state
g= Gravity
h = Height
Our values are given as,
Replacing at the kinetic equation to find 
we have,
Applying the concepts of continuity,
We need to find A_2 then,
So the cross sectional area of the water stream at a point 0.11 m below the faucet is
Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 
Answer: 5 gm/cc
Explanation:
200 gm/40 cc
= 5 gm/cc
Answer:
a) P =392.4[Pa]; b) F = 706.32[N]
Explanation:
With the input data of the problem we can calculate the area of the tank base
L = length = 10[m]
W = width = 18[cm] = 0.18[m]
A = W * L = 0.18*10
A = 1.8[m^2]
a)
Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:
P = density * g *h
where:
density = 1000[kg/m^3]
g = gravity = 9.81[m/s^2]
h = heigth = 4[cm] = 0.04[m]
P = 1000*9.81*0.04
P = 392.4[Pa]
The force can be easily calculated knowing the relationship between pressure and force:
P = F/A
F = P*A
F = 392.4*1.8
F = 706.32[N]
