Isn't it "gravity" this would makes sense because grvaity difines weight
According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:

(1)
we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:
- the horizontal component of the force exerted by the rope, which is equal to

- the frictional force, acting in the opposite direction, which is equal to

By applying Newton's law (1), we can calculate the acceleration of the box:

The answer to the question is an orgamisn
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