Physics: Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again? Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20

Divide 56 / 4.75 = 11.7 cm * 3.99999 = 46.8 - 56 = 9.2 you will get that ...

Consider a bird that flies at an average speed

patricklcc07777

19.01.2022

The distance covered by the bird before feeding is $4.55 \times 10^{5}$ m.

Explanation:

As the bird consumes 4 g of fat before flying, the amount of initial food energy ( $E_{F}$ ) stored by it is given by

$E_{F} = 4 g \times 9.4 (food) cal = 37.6 (food) cal$

So the mechanical energy stored by the bird ( $E_{M}$ ) is given by

$E_{M} = E_{F} \times 4186 J = 1.57 \times 10^{5} J$

Given, the power consumed by the bird P = 3.7 W

So, the time () required to consume this power by the bird is

$t = \dfrac{E_{M}}{P} = \dfrac{1.57 \times 10^{5} J}{3.7 W} = 4.24 \times 10 ^{4} s$

As the bird flies at an average speed () of $10.7 ms^{-1}$ , so the distance () covered by the bird before feeding again is given by

$d = v \times t = 10.7 ms^{-1} \times 4.25 \times 10 ^{4} s = 4.55 \times \times 10^{5} m$

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