Answer:
-0.64525g
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 95 km/h
v = Final velocity = 0 km/h
s = Displacement
a = Acceleration
Equation of motion

Converting to m/s²

g = Acceleration due to gravity = 9.81 m/s²
Dividing both the accelerations, we get

Hence, acceleration of the car is -0.64525g
Answer:
14 × a^3
Explanation:
Product means multiplication
Cubed means to the power of 3
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
Newton's first law of motion is that an object in motion will tend to stay in motion unless an external force acts upon it.