Answer:
Yes equation is valid.
Explanation:
Given:
h = (0.04 to 0.09)(D/d)^4*V^2/2*g
Using SI units to assign dimensions to every quantity as follows:
Energy loss per unit weight h = J / N = kg m ^2 s^-2 / kg m s^-2 = [m]
Hose diameter D = [m]
Nozzle tip diameter d = [m]
Fluid velocity in the hose V = [ m s^-1 ]
Acceleration of gravity g = [ m s^-2 ]
Using the Given Equation and plug the SI units of respective quantities:
h = (0.04 to 0.09)(D/d)^4*V^2/2*g
[m] = (0.04 to 0.09)([m] / [m])^4*[ m s^-1 ]^2/2*[ m s^-2 ]
Simplify the equation above:
[m] = ( 1 )^4 * [ m^2 s^-2 ] / [ m s^-2 ]
[m] = [m]
Hence, SI units of RHS of given equation = LHS of given equation, we can say the equation has consistent dimensions.
Answer:
Effective communication helps decision makers by gathering and providing the information to the right person on right time. Communication performs as a motivator to the employees by notifying the employees about the job task, process of carrying and how it could be done better.
Answer:
These drive fittings come in four common sizes: 1⁄4 inch, 3⁄8 inch, 1⁄2 inch, and 3⁄4 inch (referred to as "drives", as in "3⁄8 drive").
Answer:
Not knowing the units the tolerance is 0.02. I would presume mm but hopefully your question has more detail.
Explanation:
The tolerance is the portion after the main dimension (+/- 0.02). In our case we have bilateral tolerance since there is tolerance in both directions (positive and negative). If you were building a part the acceptable range would be 2.98 to 3.02 based on the tolerance provided.
Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
