They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.
a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.
Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

relative to the direction of the river.
b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

so, Dave's landing point is 540 m downstream.
d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).
Answer:
Average velocity of the girl is 1.25 m/s.
Explanation:
<span>Puffins have a predator-prey relationship with salmon.
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Answer:
103.4° or S76.6°E
Explanation:
The direction N43°E is perpendicular to the direction south-east when the plane turn 90° and heads in the south-east direction.
Since the distance 1/2 mile N43°E is perpendicular to the distance 1 mile south-east, we have a right angled triangle.
So, the angle θ between the aircraft's new position and old position is gotten from tanθ = 1 ÷ 1/2 = 2
θ = tan⁻¹(2) = 63.43°
So, the total angle from North to its new position is 40° + 63.43° = 103.43°
Since we need the south-east bearing, the angle from south is 180° - 103.43° = 76.57° ≅ 76.6°
So, our bearing is 103.4° or S76.6°E