Answer:
There is a loss of fluid in the container of 0.475L
Explanation:
To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.
The formula that describes this thermal expansion process is given by:

Where,
Change in volume
Initial Volume
Change in temperature
coefficient of volume expansion (Coefficient of copper and of the liquid for this case)
There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

Where,
= Change in the volume of liquid
= Change in the volume of copper
Then replacing with the previous equation we have:


Our values are given as,
Thermal expansion coefficient for copper and the liquid to 20°C is




Replacing we have that,



Therefore there is a loss of fluid in the container of 0.475L
Question:
<em>What happens to electrical energy that is used by objects in our homes? (1 point)</em>
<em>a It is absorbed by batteries. </em>
<em>b It is destroyed. </em>
<em>c It is stored in solar panels. </em>
<em>d It is transformed into other forms of energy.</em>
<em />
Answer:
D
It's very hard to see the self-portrait, so I can't identify him.
Answer:
The last statement is false.
Explanation:
Photons (Electromagnetic radiation) are released when electrons drop from a higher energy lever to a lower energy level. Therefore the opposite insinuated by the last statement is wrong.
At the player's maximum height, their velocity is 0. Recall that

which tells us the player's initial velocity
is

The player's height at time
is given by

so we find their airtime to be
