A 15.9 kg concrete block is being pulled 0.90 m to the right In 2.5 s
1 answer:
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Answer:
If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast
Explanation:
When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:
Vr = velocity of the river = 6[km/h}
Vc = velocity of the canoe = -6 [km/h]
We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.
Vt = Vr + Vc = 6 - 6 = 0 [km/h]
For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.
So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).
Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:
