<h2>
The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 3 s
Substituting
s = ut + 0.5 at²
s = 0 x 3 + 0.5 x 9.81 x 3²
s = 44.145 m
The seagull's approximate height above the ground at the time the clam was dropped is 4 m
Answer:
I will answer in English.
Here we will use the relation
Velocity*time = distance
So:
a) velocity = 3m/s
time = 2s
Distance = 3m/s*2s = 6m
b) velocity = 2m/s
time = 3.5s
Distance = 2m/s*3.5s = 7m
c) velocity = 10m/s
time = 0.5s
Distance = 10m/s*0.5s = 5m
d) velocity = 4m/s
time = 2.5s
Distance = 4m/s*2.5s = 9m
e) velocity = 1.5m/s
time = 5s
Distance = 1.5m/s*5s = 7.5m
Answer:
I would strongly recommend Exploring Quantum Physics through Hands-on Projects for physics practicals.
Explanation:
- Though it is not about books, but it is solely on you how you want to get knowledge. If you are truly passionate about learning physics in a practical way "Exploring Quantum Physics through Hands-on Projects" will be the best one out there.
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