You'd be surprised at how different these two machines are. Although they're called wind 'turbines', their operating principles are somewhat different from real gas or steam turbines.
A simpler way to find the net force of two or more vectors is to decompose each vector into its x and y components. Once that is done, the individual x and y components are added together algebraically, and the sum of these components gives you the component of the net force, or the resultant...
The fact remains, however, that human civilization, except in a few instances, has abandoned the H/G model in favor of agriculture in order to provide a wider variety of food sources and a greater supply of same. The proportion of the population involved in procuring food in a H/G group would...
Sure, modern groups of H/G which still exist may benefit from factors which allow for increased longevity among the human population at large, but in older times, you would be unlikely to find H/Gs who made it to 32, let alone 72. However, I think the OP is wondering about H/Gs as they existed...
This is prolly true, if you mean total work over a lifetime. H/G probably had a life expectancy at birth of about 25 years or so, less if times were lean.
No one worked for 40 years as a H/G becuz they were dead a good while by then.
Well, H/G certainly led more active and exciting lives than...
At this stage, you can only estimate the deadweight of the barge by assuming a light draft and a loaded draft and taking the difference in displacement at those two drafts. Later on, if more information about the barge becomes available, you can refine this initial estimate of the DWT.
There...
I don't think that's possible.
In the limit, the denominator ##x^2 - 4## goes to 0 as x → 2.
Cross multiplying means that you are multiplying the RHS by zero as well, leaving ##x^2 - cx + d = 0##, for which there can be an infinite number o' solutions.
Evaluating limit expressions sometimes...
Bending stresses are not distributed uniformly to counteract the bending moment, unless you mean that the stresses are distributed uniformly along the length of the beam (i.e., the stress in each cross section is the same, because the local bending moment is constant and equal to the applied...
No, what I'm saying is that your material properties are based on a tensile test, not a shear test. To use those test results for calculating the max. allowable shear stress, the figure 0.4 sy is supposed to account for going from a tensile loading to a shear loading and any other unknown...
You should draw a diagram for this problem.
The tension in the rope which is doing the pulling is counteracted by friction. Your F(being pull) ≠ (m)(a), since a = 0 yet F(being pull) must be > 0.
Have you constructed a free body diagram for the beam in this condition?
Take a segment of the beam AC as described above, put all of the forces and moments which act on AC on a diagram, and see what must happen for segment AC to remain in equilibrium.
Yes, but it doesn't move from side to side, thus staying in the x-z plane. In case you are wondering, the z-axis runs along the un-buckled length of the column.