Answer:
I dont know lol sorry ddhd
Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
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
0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.
The thing could be a glass of water, a sheet of paper,
a pound of ground beef, a gallon of gas, or a snowball.
In this problem, it just happens to be a dm.
Since 1m/s=3.6 km/h, we can conclude that 10.0m/s = 36 km/h