Question

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 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

Answer 1

Answer:

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 ($t$) 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 ($v$) of $10.7 ms^{-1}$, so the distance ($d$) covered by the bird before feeding again is given by

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

Answer 2

The distance of  the bird's flight before him/her feeds again is mathematically given as

d = 4.55* 10^{5} m

What is the distance of the bird's flight before him/her feeds again?

Question Parameter(s):

a bird that flies at an average speed of 10.7 m/s

its body fat reserves at an average rate of 3.70 W

the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories,

Generally, the initial food energy  is mathematically given as

Ex= 4 g*9.4

Ex= 37.6cal

Therefore, the mechanical energy

Em = Ex * 4186

Em = 1.57*10^{5} J

In conclusion, time of flight

$t = \frac{E_{M}}{P} \\\\ t=\frac{1.57 *10^{5} J}{3.7 W}$

t= 4.24*10 ^{4} s

Th distance hence is

d = v* t

d= 10.7 *4.25*10 ^{4}

d = 4.55* 10^{5} m

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