Answer:
Force per unit plate area is 0.1344
Solution:
As per the question:
The spacing between each wall and the plate, d = 10 mm = 0.01 m
Absolute viscosity of the liquid,
Speed, v = 35 mm/s = 0.035 m/s
Now,
Suppose the drag force that exist between each wall and plate is F and F' respectively:
Net Drag Force = F' + F''
where
= shear stress
A = Cross - sectional Area
Therefore,
Net Drag Force, F =
Also
F =
where
= dynamic coefficient of viscosity
Pressure, P =
Therefore,
Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:
where;
m = mass flow rate = 120 kg/min
the pressure at the inlet = 96 kPa
the pressure at the exit = ???
the pressure = 1000 kg/m³
∴
400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
The load is placed at distance 0.4 L from the end of area.
<h3>What is meant by torque?</h3>
The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.
Let the beam is of length L
Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero
also, we know that stress at both ends are same
Now from two equations we have
solving the above equation we have
so the load is placed at distance 0.4 L from the end of area.
The complete question is:
47. the beam is supported by two rods ab and cd that have cross-sectional areas of and
, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.
To learn more about torque refer to:
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