A hummingbird flies 2.0 m along a straight path at a height of 5.3 m above the ground. Upon spotting a flower below, the humming
bird drops directly downward 2.4 m to hover in front of the flower. a) What is the magnitude of the hummingbird’s total displacement? Answer in units of m.
1 answer:
To solve this problem we will apply the concepts of Pythagoras to find the net path. An easy way is to graph the path to trace the path, which resembles that of a right triangle. The information provided is equivalent to that of the two short sides of the triangle, so we will find the longest side by the Pythagorean theorem. In this way,
Therefore the magnitude of the hummingbird’s total displacement is 3.12m
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Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
