Answer:
1. Changing Beam Material
2. Corrugation
3. Changing Beam form
4. Steel Reinforcing Bars
Explanation:
Changing Beam Material
Some materials are stronger when used in beams than others. Beams made of steel for instance are stronger than beams made of wood. Therefore changing material can improve the strength of the beam. It is quite important to take into account the weights of the material though as different structures have different requirements.
Corrugation.
You can fold the beam into triangular shapes to increase strength. If you look at roofs you will notice that they are folded and this increased their strength. The same logic can be applied to beams.
Changing Beam Form
Another way to make Beams stronger is to change their form or rather their shape. Straight beams are not as strong as I-beams for instance. I-beams look like the capital letter I with the lines at both ends. I-beams are usually used in construction which shows that they are quite strong.
Steel Reinforcing Bars
When placed in concrete beams, Steel Reinforcing Bars which are also called Rebar can help strengthen a beam by helping it withstand the forces of tension. A concrete beam with Rebar inside it is known as Reinforced Concrete.
Answer:
Hematology is the study of blood
Explanation:
The cellular component consists of three major cell types, all of which arise from hematopoietic stem cells located primarily in the bone marrow: red blood cells (RBCs), white blood cells (WBCs), and platelets.
It does not consist of a single element
Hope I helped! ( Smiles )
If you rewite the following word equation as a balanced chemical equation, what will the coefficient and symbol for iodine be?
bromine + potassium iodide → potassium bromide + iodine
A 2I
B I
C 2I
D I2
E 2I2
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:
Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:
Finally, the Gibbs free energy for the reaction at 298.15K is:
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