The two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.
<h3 /><h3 /><h3>What is a covalent Bond?</h3>
- A covalent bond is a type of chemical bond that involves the sharing of pairs of electron between atoms.
Examples of compounds with covalent bond include the following;
- Distilled water
- Sucrose
- Ethanol
Olive oil is a mixture not a compound
Sodium Chloride & Potassium lodide are examples of ionic bond.
Thus, the two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.
Learn more about covalent bonds here: brainly.com/question/12732708
Answer:
No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.
Explanation:
A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.
Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:
* When the diameter and length are comparable (i.e have the same measurement)
When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.
Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.
Answer:
Continental polar (cP) or continental arctic (cA) air masses are cold, dry, and stable. These air masses originate over northern Canada and Alaska as a result of radiational cooling. Maritime polar (mP) air masses are cool, moist, and unstable.
Explanation:
Answer:
8.27°
Explanation:
To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.
![d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m](https://tex.z-dn.net/?f=d%3Dy_1-y_2%3D%5Cfrac%7BF_1%7D%7Bk_1%7D-%5Cfrac%7BF_2%7D%7Bk_2%7D%3D%5Cfrac%7B0.5mg%7D%7B31N%2Fm%7D-%5Cfrac%7B0.5mg%7D%7B63N%2Fm%7D%5C%5C%5C%5Cd%3D0.5%281.6kg%29%289.8m%2Fs%5E2%29%5B%5Cfrac%7B1%7D%7B31N%2Fm%7D-%5Cfrac%7B1%7D%7B63N%2Fm%7D%5D%3D0.128m)
by a simple trigonometric relation you obtain that the angle:
hence, the angle between the rod and the horizontal is 8.27°
