The process of flask becoming cold is due to endothermic reaction.
Answer: Option B
<u>Explanation:</u>
So two kinds of heat transfer can be possible in any chemical reaction. If the sample is considered as system and the sample container is considered as the surrounding, then heat transfer can occur between them.
If the heat is transferred from the surrounding to the system , then it is an endothermic reaction. And in those cases, the sample holder will be becoming colder. This is because the heat from the surrounding that is the container will be utilized to complete the reaction.
While when there is transfer of heat from the system to surrounding , it will be exothermic reaction and the beaker will be getting hot in this process. So in the present case, the container is becoming cold because of occurrence of endothermic process.
Answer: 5.5
Explanation:
Let's start by explaining that hardness is a property that materials have related to the opposition or resistance they offer to alterations such as penetration, abrasion, scratching, cutting, and permanent deformations, among others.
In this context, several hardness scales have been developed to catalog the materials (specifically minerals), being the Mohs scale the best known. This scale, proposed by the German geologist Friedrich Mohs in 1825, consists of a ratio of ten minerals numbered in increasing order by hardness, from least to greatest.
This is how the scale starts at 1 with the talc (considered the softest material) and ends at 10 with the diamond as the hardest.
Now, if we are told that the glass is in the middle of the hardness scale that goes from 1 to 10, logically its value will be 5 (volcanic glass). However, according to the scale, the glass is at 5.5.
a yoyo in someones hand is an example of potential energy
A 500 g ball swings in a vertical circle at the end of a 1.4-m-long string. when the ball is at the bottom of the circle, the tension in the string is 18 n.
Explanation:
Formula to determine the critical crack is as follows.
= 1, 
= 24.1
[/tex]\sigma_{y}[/tex] = 570
and, 
= 427.5
Hence, we will calculate the critical crack length as follows.
a = 
= 
= 
Therefore, largest size is as follows.
Largest size = 2a
= 
= 
Thus, we can conclude that the critical crack length for a through crack contained within the given plate is .
