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Allushta [10]
3 years ago
10

The lightest and heaviest flying birds are the bee hummingbird of Cuba, which weighs about 1.6 grams, and the great bustard of E

urope and Asia, which can weigh as much as 21 kilograms. Show that the bee hummingbird produces about 0.016 newton of lift when it flies, whereas the great bustard produces about 205.8 newtons of lift. Which species would you expect to have proportionally larger wings? Why?
Physics
1 answer:
VMariaS [17]3 years ago
7 0

Answer:

for the birds to be able to stay vertical in flight without falling down to earth, they must produce a lift that will counteract their weight

for the small bee humming bird,

mass = 1.6 g = 1.6 x 10^{-3}  kg

weight of the bird under acceleration due to gravity = mg

where g = acceleration due to gravity = 9.81 m/s^2

weight of the bird = 1.6 x 10^{-3}  x 9.806 = 0.0156 ≅ 0.016 N

for this bird to maintain flight, the least lift upward, it must generate must be equal to its weight downwards, i.e

lift = weight

therefore,

lift = <em>0.016 N</em>

<em></em>

For the bustard of Europe and Asia,

mass = 21 kg

weight of the bird under acceleration due to gravity = mg

weight of the bird = 21 x 9.806 = 205.9 N

lift = weight =  <em>205.9 N</em>

<em></em>

<em>lift generated is proportional to the wing surface area according to the lift equation</em>

L = Cs x p x \frac{v}{2} x S

where L = lift

C = lift coefficient

p = density of air

v = relative velocity of bird and air

S = surface are of the wing.

<em>The great bustard will have a proportionally larger wing area to hold its weight in flight</em>

<em></em>

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Answer:

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Explained

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4 years ago
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plat
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U_f = (U_o)/2)

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From earlier, we saw that C' = 2C.

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6 0
4 years ago
The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr
WARRIOR [948]
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
y_n= \frac{n \lambda D}{a} (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
\lambda is the light's wavelength
D is the distance of the screen from the slit
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In our problem, 
D=37.0 cm=0.37 m
\lambda=530 nm=5.3 \cdot 10^{-7} m
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y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m (2)

If we use the formula to rewrite y_5, y_1, eq.(2) becomes
\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m
Which we can solve to find a, the width of the slit:
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