Near Earth's surface, gravitational acceleration is approximately 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.81 metres per second every second.
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Answer:
4.64m/s
Explanation:
We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).
v = √2(9.8)(1.1)
v ≈ 4.64m/s
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Answer:
Explanation:
1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is
100kg/h*0.5h = 50kg

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity
3. If we assume that the force of the boat before the raining is

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts
And if we take the net force as

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.
I hope this is useful for you
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