Is the detonation of an atomic bomb an exergonic (spontaneous) reaction? Please explain why? How does the entropy change during the reaction?
Many people say that craftsmanship is good because you can easily become a boss and become self-employed. Is that really true? And then you can cash in without any hassle, as if it were guaranteed to make good money with a craft business. Is that really true? That's what many people who promote skilled trades…
I don't have df. in my notes, but is that correct? So without looking at the solution I proceeded like this: lim(2x-x^2)/2x x->o So I have factored out an x in the numerator: x(2-x)/2x I then shortened the x, so I get 2-x/2 2-0 is 2 2/2 are then 1
Why is it called that?
How do you do this task? It's for work tomorrow! Thanks in advance.
A colleague and I came up with the idea and planned to fill an entire barrel with Newtonian fluid, while each of us stood in it. If it hardens while you're standing in it, will you really get stuck and not be able to get out easily? Is it even possible to get out again?…
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Hi.
would make it easier, first with the aid of a wooden beam, explain:
When a bar is thus stressed on bending, compressive stress is produced in the upper fibers of the cross-section and tensile stress is produced in the lower fibers, while the middle fibers “swell”.
So if I want to save mass on the component, why don’t I take what away in the middle area? Of course, I can’t take everything away, otherwise the bar won’t hold more than one piece together.
In the case of a double-T-carrier (i.e. no. 2) made of steel, this principle is so to speak driven to the tip: above a flange which absorbs pressure, below a flange which receives tension and in the middle a thin web which holds the whole together.
Say: little mass, much bending load.
Halfway comprehensible?
Greetings!
both 2) and 4) can be with the same mass (of course always the same) outer Dimensions) have the same load capacity for vertical loads (only for these). The two vertical “bars” of 4) are then together as thick as that of 2)
4), however, also has a greater load capacity for horizontal loads than 2).
Of course No. 2.
Number 2. Has the second most compact construction material and provides sufficient support surface up and down.
is actually not a physical question but a static one that goes beyond normal school mechanics…
but it is numero 2. Because the moment of inertia here is best. The idea is that the legs at the top and bottom are at most far away from each other and only connected to a thin carrier in the middle mechanich. so you have the best weight ratio to moment of inertia. Since you save the material in the middle.
With this you have the least bending due to the weight of your own. Because balconies and bridges do not have to bear loads but themselves
Thanks for the answer. I was really very uncertain if you can use the moment of inertia here. That is why I also have the question. How do the torque differ from the moment of inertia? I mean, both of them are about how much force you have to exert on an object. Only the moment of inertia is not m, but m2?
I’d rather have assigned moments of inertia when rotating bodies?
Puh Baumechanik 2 is really long. I would just have to refer from memory and that can be wrong. Therefore, I would rather ask Dr. Google.
Is certainly a VL value. Sit here since 11:00 and try to clarify why the steel support 2 is better suited. Of course, if you run around the moment of inertia with the wrong formulas, this explains a lot of XD. But only for security. It is correct that the carrier 1 and 4 are more stable than the carrier 3 is better suited, however, because less mass and space, wherein carrier 4 with only a little more mass is apparently better suited, is it?
yes this moment of inertia (then for the supplement) is really not easy to explain in few words
You don’t have to be normal either. This is a characteristic of the strength teaching, relatively special mechanics.
Sorry to LEO2619 that I am not answering, but that would be in a lecture :-P, but unfortunately the time is missing at the moment
I don’t know.
You’re right.
That’s what I mean Areasmoment of inertia of the cross section of the carrier, which was previously swallowed.
The double-T-carrier, 2.
It has the most mass/material, most remote from the “neutral chamfer” -> Is therefore the most rigid.