. . . ${\itExplain}.$¹

If the heron's wings are fairly massive (which they are), it can lift them up over its head (raising its centre of gravity) and then bring them down quickly. In that short moment, the downward acceleration of gravity doesn't have time to lower the centre of gravity significantly, so the rest of the bird has to rise to compensate the wings coming down. The bird's feet leave the ground. Of course, they come back rather soon, but that was not part of the question.

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. . . field.²

The shark only detects electric fields directly, but when it moves through a magnetic field, an electric field is induced by the HALL EFFECT, and this field is easily detected by the shark.

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. . . air.³

In the water, shrimp snap their tails and make a "popping" sound by cavitation. On the ground are numerous insects, who make noise by a variety of means such as rubbing across rows of bumps on their legs. There are also reptiles like rattlesnakes that warn off intruders with a "buzz". In the air there are also many noisy insects, notably the dread mosquito, whose reasons for warning us she is coming are unknown.

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. . . have?⁴

The eye focuses light using two lenses: the first is the curved cornea, which has a higher index of refraction than air and so acts as a converging lens; the second is the adjustable lens behind the corvea, which has a still higher index of refraction. When the eya is immersed in water (which has nearly the same index of refraction as the eye itself), the curved cornea no longer has much effect, which changes the overall focal length by too much for the adjustable lens to compensate. In order to compensate for this loss, a contact lens would need to have an index of refraction much higher than that of water and be very dramatically convex on the outside; it would probably be rather uncomfortable to wear.

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. . . it?⁵

The Great Flight Diagram says $U = 15 M^1/6$ where $U$ is in m/s and $M$ is in kg. You can plug into this formula to get $U = 32.3$ m/s or just read it off the graph: $U 30$ m/s. However, the metabolic power required to overcome the drag force (which increases with $U$) increases with the mass of the bird so that the maximum velocity the bird can sustain with its metabolism (i.e. the "metabolic velocity" $U_m$) decreases with mass as $U_m = 40 M^-1/4$. For a 100 kg Giant Goose this would be $U_m = 12.65$ m/s, which is well below $U$. The Giant Goose could probably get into the air for a short flight, but then it would have to spend the rest of the day eating to recover the energy squandered in that heroic effort.

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. . . magnitude?⁶

The metabolic rate $$ is the rate [in Watts] at which the animal "burns" fuel with oxygen to "keep its motor running".

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. . . size.⁷

Metabolic rate scales allometrically with body size. Whether you interpret "size" to mean dimensions, volume or mass, the metabolic rate cannot be considered to scale isometrically with size.

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. . . equation.⁸

The resting metabolic rate $_0$ scales with body mass $M$ as

$$\Gamma_0 \approx 4 M^{3/4} \; .$$

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. . . elephant.⁹

A mouse has a lower metabolic rate than an elephant.

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. . . elephant.¹⁰

The mass-specific metabolic rate is the metabolic rate per unit mass, or $/M M^-1/4$. Therefore A mouse has a higher mass-specific metabolic rate than an elephant.

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. . . organism.¹¹

Assuming that the cells are roughly the same size in both organisms, a cell in a smaller organism should have a higher metabolic rate then a cell from a larger organism, because the former has a higher metabolic rate per unit mass, and the mass of the cells are presumed to be the same.

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. . . organism.¹²

This gets interesting. It is believed that the higher efficiency (lower metabolic rate per cell) of the larger organism has to do with the sharing of reseources by many cells in close proximity with each other. If this is true, then the isolated cells in vitro would all have the same metabolic rate.

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. . . principles.¹³

The explanation was offered with the answer in each case.

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. . . birds?¹⁴

Because they need less fuel (and therefore less oxygen) per unit body weight, but their blood volume (which cetaceans use as their primary oxygen store) scales linearly with body mass. Lung capacity (auxiliary oxygen storage) also scales approximately linearly with body mass.

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. . . rate.¹⁵

Many possibilities . . . . See e.g. Sections 2.4.2, 5.3.2 and 6.2.1 in textbook.

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. . . wavelengths.¹⁶

Fish eyes in the deep sea are most sensitive to blue light because light is more strongly attenuated at long wavelengths. (This is opposite to what one might expect from Rayleigh scattering. Presumably this is for the same reason that sea water looks blue at relatively short range: it absorbs red light preferentially.)

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. . . light.¹⁷

Many deep-sea organisms appear red under white light, so that they are invisible with respect to deep-sea ambient light.

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. . . sea?¹⁸

Blue bioluminescence will be seen by all the other critters, so it is good for advertising one's presence (e.g. to make a "lure"). Red bioluminescence will be invisible to most of the other critters, so it makes a good "flashlight" for seeing one's way (especially when looking for reddish prey). Of course, this requires parallel evolution of red-sensitive vision . . . .

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. . . light?¹⁹

Small bumps (smaller in diameter than the wavelength of visible light, but tall enough to make the index of refraction "fade gradually" from $n=1$ of air to that of the wing material) minimize reflection even more effectively than a "quarter-wave plate" nonreflective coating, because they work on a wide range of wavelengths.

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. . . frequencies.²⁰

Sound is more strongly attenuated at high frequencies.

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. . . hearing).²¹

Mammals can generate both infrasonic and ultrasonic sounds; so can insects!

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. . . water.)²²

A bat would not be able to echolocate to detect fish in water because of impedance mismatching.

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. . . water.²³

Impedance mismatching allows a dolphin to prevent sound from penetrating its head uniformly: both bone and air sacks can be used for this purpose.

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. . . Law.²⁴

There is a sound focusing organ called the "melon" within a dolphin's head. This organ has a low sound-velocity core with a high sound-velocity outer shell, which acts to focus sound as predicted by Snell's Law.

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. . . dawn.²⁵

The cold, dense air has a higher index of refraction than the warm air above it, so sound "rays" propagating upward at a small angle will be bent back down by the same phenomenon as "total internal reflection" and thus their intensity will fall off slower than the $1/r^2$ dependence of isotropic radiation. (Recall the whales' "Sofar Channel".) The birds' songs will be heard from further away at dawn.

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. . . source;²⁶

You hear the lower frequencies unattenuated. (The higher frequencies are scattered out to the sides).

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. . . up;²⁷

In this case you hear only the (higher) scattered frequencies.

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. . . left;²⁸

Now it gets interesting. Big fish reflect all but the very longest wavelengths (lowest frequencies) and so you will hear approximately the full "white noise" spectrum with only a few very low frequencies missing, plus the higher frequency sounds scattered by the water itself. This is sort of like the setting sun reflected off some big white clouds, which scatter the sunset's red light down to you, while in between you still see patches of blue sky.

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. . . right.²⁹

The only difference between big fish and little fish is the "cutoff" of scattered frequencies: little fish do not scatter the longer wavelengths (lower frequencies) as well as big fish, so the "missing lows" extend higher than for the big fish.

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. . . produce?³⁰

All muscles exert roughly the same force per unit cross-sectional area, $f_0 = F/A 2 ×10^5$ Pa (about 2 atm). Thus the force is proportional to the cross-sectional area of the muscle.

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. . . ${\slExplain}.$³¹

The cross-sectional area is always perpendicular to the muscle fibers; thus the pennate muscle applies more fibers to make a bigger force. Note however that the same fractional length contraction of the pennate fibers produces a smaller contraction between the tendons of origin and insertion because the fibers contract at an angle to that axis.

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. . . them.³²

Work = force times parallel distance. Thus a normal muscle and a pennate muscle of the same overall size and shape will produce the same work: the extra force of the pennate muscle is exactly offset by its smaller contraction due to the same geometrical factor.

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. . . $}$.³³

Any of the listed possibilities can be achieved, depending upon what sort of cycle the muscle contraction and expansion execute.

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. . . springs.³⁴

True. A Mechanics purist might quibble that a "strut" can support compression forces as well as tension, whereas, "You can't push on a muscle."

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. . . locomotion.³⁵

If you simply "scale up" an animal, its weight scales as the cube of its linear dimensions while the strength of its bones only scales as the square of same. So the bone strength/weight ratio decreases linearly with the size of the animal, for isometric scaling.

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. . . locomotion.³⁶

For starters, they do not scale isometrically. Larger animals tend to evolve disproportionately thicker bones. However, the allometric power is 0.89, not far off linear, so this is (on average) a modest effect. Probably more important (see slides 20-23 of the Running lecture) is the change of gait - they give up running and walk instead. (And they are very, very careful . . . .   :-)

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. . . bird.³⁷

Muscles that swing the limbs back and forth use approximately 25% of the total energy expended by a running bird, according to blood flow measurements on an actual running bird. See Jeremy's lecture.

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. . . way?³⁸

Elastic structures such as tendons and ligaments store energy when stretched and deliver almost all of it back when they contract.

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. . . adjustment?³⁹

The bear will change the stiffness of its legs, either by changing its EMA or via neural control of its leg muscles.

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. . . needs?⁴⁰

The resting metabolism $_0 = H/t$ is given by the fundamental allometric relation $_0 = 4 M^3/4$ where $_0$ is in watts (W) and $M$ is in kg. Thus we expect the blue whale to need $_0 4 ×100,000^0.75 = 22,494$ W. In one day ($24 ×60 ×60 = 86,400$ s, this adds up to $22,494 ×86,400 = 1.94 ×10^9$ J or $1.94 ×10^3$ MJ. At $h = 5$ MJ/kg this requires $1.94 ×10^3/5$ or 389 kg/day .

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. . . speed?⁴¹

For a large body like this, Stokes drag is negligible compared with hydrodynamic drag, for which we have $F_D = 12 C_D A _w v^2$ where $A = (D/2)^2 = 12.57$ m$^2$ and $_w = 1025$ kg/m$^3$ is the density of sea water. Thus $F_D = 12 ×0.2 ×12.57 ×1025 ×2^2$ or $F_D = 5152$ N .

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. . . whale?⁴²

To cruise horizontally at constant speed, the whale must exert a constant forward thrust equal to the above drag force. This requires a mechanical power $P = F_D ×v = 10,304$ W. At 25% efficiency the whale must "burn" 4 times this amount of metabolic "fuel": $_c = 41,218$ W, so that $b = _c/_0 = 1.832$ .

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. . . back?⁴³

This problem is a "ringer" for the Physics enthusiast.
The crudest approximation is to ignore the drag force $F_D$ (which, for an actual cylinder, would cease as soon as the top end of the cylinder emerged from the water's surface - i.e. for $y = -L/2$, if we define $y$ as the height of the center of mass above the surface) and the buoyant force $F_B = _w g A (L/2 - y)$ due to the displaced water (which starts decreasing gradually as soon as the top end of the cylinder emerges from the water's surface) and the thrust $T = Mg + 12 C_D A _w v^2 - _w g A L$ which the whale had to exert in order to swin straight up at its cruising speed.
If all these are neglected, then the problem reduces to a simple trajectory problem: how high does $M$ get after leaving the water with an initial vertical velocity $v$? The answer is familiar: setting the initial kinetic energy equal to the final potential energy gives $12 M v^2 = M g h$ or $h = v^2/2g = 2^2/2×9.81$ or $h = 0.2$ m . Not very high.
What if we include the effects of a dwindling buoyant force? Relative to the above case this decreases the effective weight pulling the whale down, so we would expect to get a little higher value for $h$.
What if we assume the whale "keeps pushing" as it climbs skyward, with the same thrust $T$ as before it reached the surface? Now we have an additional upward force acting to drive the whale higher. How much higher? What are the equations of motion?
This is a fun problem to mull over, but you get full marks for the crude approximation above, and extra credit for noting (on paper) that there is more to the problem than that. I won't spoil your extra fun by giving away the anwswers to the hard parts . . . .

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. . . 500 m?⁴⁴

Neglecting any absorption of energy by the water, the total power is distributed over a sphere of radius $r = 500$ m, which has a total area of $A = 4 r^2 = 3.14 ×10^6$ m$^2$, giving an intensity of $I = P/A = 1.2/3.14 ×10^6$ or $I = 0.382 ×10^-6$ W/m$^2$ anywhere on that sphere.

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. . . frequencies?⁴⁵

Lower frequencies carry further because they are not attenuated by scattering or absorption in the water.

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. . . ${\slExplain}.$⁴⁶

One possibility is a Helmholtz resonator (see Eq. (9.41) on p. 340 in the textbook): a windpipe $= 1.5$ m long and $d = 0.3$ m in diameter attached to a lung of volume $V = 2.0$ m$^3$ should resonate at a frequency $f_0 = (v_s/2)A/V$ where $v_s = 340$ m/s is the speed of sound and $A = (d/2)^2 = 0.0707$ m$^2$ is the cross-sectional area of the windpipe. This gives $f_0 = (340/2)0.0707/(1.5×2)$ or $f_0 = 8.3$ Hz . So this is a plausible mechanism.

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