A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?
1 answer:
Answer:
V = - 0.5 [m/s]
Explanation:
In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:
That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left
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Answer:
W = 1250 J = 1.25 KJ
Explanation:
The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:
where,
W = Work done = ?
mg = W = weight of cement = 500 N
h = height covered = 2.5 m
Therefore,
<u>W = 1250 J = 1.25 KJ</u>
Answer:
Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd
Explanation:
The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is
d = 12 + 9 = 21 yd
The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:
Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was
d = 15 - 3 = 12 yd