Answer: The 6 kg rock sitting on a 3.2 m cliff.
Explanation:
The potential energy of an object of mass M that is at a height H above the ground us:
U = M*H*g
where g is the gravitational acceleration:
g = 9.8m/s^2
Then:
"An 8 kg rock sitting on a 2.2 m cliff"
M = 8kg
H = 2.2m
U = 8kg*2.2m*9.8 m/s^2 = 172.48 J
"a 6 kg rock sitting on a 3.2 m cliff"
M = 6kg
H = 3.2m
U = 6kg*3.2m*9.8m/s^2 = 188.16 J
You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.
The speed of tsunami is a.0.32 km.
Steps involved :
The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?
Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32
As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.
To learn more about tsunami refer : brainly.com/question/11687903
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For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero.
