The weight of an object is the force of gravity between Earth's
mass and the object's mass.
The forces of gravity always come in equal, opposite pairs.
The Earth's weight on the object is the same as the object's
weight on the Earth, and when the object falls to Earth, Earth
falls to the object.
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows


b) the change in the combined kinetic energy of the two-car system during this collision

substitute the value in the equation above

Hence, the change in combine kinetic energy is -2534.78J
The weight changes but the mass will stay the same.
***elements become less electropositive and more electronegative in their properties...