Answer:
v₂> v₃ velocity canoe is more than velocity fishing boat
Explanation:
For this exercise we must define a system consisting of the girl, Sally and the boat, in one case the canoe and in the other the fishing boat; for this system we can use moment conservation
Initial moment. Before the jump
p₀ = (M + m₂) v
Final moment. After the jump
= M v₁ - m₂ v₂
Where m and v are the masses and speed of the canoe
p₀ = p_{f}
(M + m₂) v = M v₁ - m₂ v₂
In the case of changing the canoe for the heaviest fishing boat, the final moment is
p_{f} = M v₁ - m₃ v₃
p₀ = p_{f}
(M + m₃) v = M v₁ - m₃ v₃
Since the canoe is stopped the speed v = 0, we write the speed of each boat
Canoe
0 = M v₁ - m₂ v₂
v₂ = M / m₂ v₁
Fishing boat
0 = M v₁ - m₃ v₃
v₃ = M / m₃ v₁
Since the masses of the fishing boat (m₃) is greater than the mass of the canoe (m₂) the speed of the fishing boat is less than the speed of the canoe, we can find the relationship between the two speeds
v₂ / v₃ = m₃ / m₂
Here you can see what v₂> v₃ velocity canoe is more than velocity fishing boat
Answer:
They are the same (assuming there is no air friction)
Explanation:
Take a look at the picture.
When the first ball (the one thrown upward) gets to the point marked as A, the speed will has the exact same value V but the velocity will now point downward (just like the second ball).
So if you think about it, the first ball, from point A to the ground, will behave exactly like the second ball (same initial speed, same height).
That is why the speeds will be the same when they reach the ground.
Answer:
Explanation:
28 / 70 = 0.3857142... = 0.39 hr
280 / 100 = 2.8 hrs.
(100 - 0) / 10 = 10 m/s²
(60 - 20) / 4 = 10 m/s²
Transform boundary
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