Answer:
Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.
Explanation:
Solution
Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating pressure shortens the hose life.
In systems where pressure peaks are severe, select or pick a hose with higher maximum operating pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.
Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.
Answer:

Explanation:
given data:
height of tank = 60cm
diameter of tank =40cm
accelration = 4 m/s2
suppose x- axis - direction of motion
z -axis - vertical direction
= water surface angle with horizontal surface
accelration in x direction
accelration in z direction
slope in xz plane is



the maximum height of water surface at mid of inclination is



the maximu height of wwater to avoid spilling is

= 60 - 8.2

the height requird if no spill water is 
Explanation:
Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)
The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)
Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current
I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches
R1 = (8 V)/(3 A) = 8/3 Ω
R2 = (8 V)/(4 A) = 2 Ω
R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω
V1 = V2 = V3 = 8 V
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)

Now we can use the equation that defines the percentage of increase, in this case for speed

Now we use the equation obtained in the previous step, and replace values

the percent increase in the velocity of air is 25.65%