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3 years ago

As a pelican flies through the air, it flaps its wings, thereby pushing down on the air below. What is the reaction force?

Physics

2 answers:

madam [21]3 years ago

7 0

Answer:

the reaction force in this situation would be B

Explanation:

The action is the wings pushing down whilst the reaction is the air pushing up which allow the bird to fly .

plz mark brainliest to help me lvl up :P

IrinaVladis [17]3 years ago

6 0

Explanation to help get you the answer, not straight-up though:

Both fixed and flapping wings work by deflecting air downwards, says Professor Geoffrey Spedding, Chairman of Aerospace and Mechanical Engineering at the University of Southern California. " [When air is forced downwards] the reaction force is upwards, and this lift force supports the bird's weight. The bird’s wing pushes against the air and the air pushes back with an equal and opposite force. Unfortunately, with every down stroke there must be an up stroke, what goes up must come down. If the bird had an equal upward stroke to its downward stroke, the net force of the bird’s wing flaps would equal zero.

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Hope my knowledge helped you get an answer :3

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A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the

seropon [69]

Answer:

(a) emf = 0.507 V

(b) emf = 0.0507 V

Explanation:

Given;

number of turns of the coil, N = 40 turns

diameter of the coil, d = 11 cm

radius of the coil, r = 5.5 cm = 0.055 m

magnitude of the magnetic field, B = 0.4 T

The magnitude of the induced emf is calculated as;

$emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}$

(a) when the time, t = 0.3 s

$emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V$

(b) when the time, t = 3.0 s

$emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V$

$emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V$

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3 years ago

An electric fence displays a warning sign about the voltage and amperage of the fence. How would amperage and voltage affect the

Lorico [155]

V = volts
I = amps 
P = rate or energy transfer (power) 
and 
P = V * I

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3 years ago

There are ___________ of galaxies in the universe containing __________ of stars in each galaxy.

aliya0001 [1]

There are Billions and billions of galaxies in the universe containing Trillions and trillions of stars in each galaxy.

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4 years ago

Read 2 more answers

A roller coaster is hard to stop because it has a lot of inertia. Infer using newtons first law

olchik [2.2K]

Lol i think i had the same worksheet!! lol anyway look up newtons first law and it will be pretty self explanatory from there

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4 years ago

When blueshift occurs,the preceived frequency of the wave would be?

LiRa [457]

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency $f\; {\rm Hz}$ at the source. In other words, the source of the wave sends out a peak after every $(1/f)\; {\text{seconds}}$ .

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of $(1/f)\; {\text{seconds}}$ . It would appear to the observer that consecutive peaks arrive every $(1/f)\; {\text{seconds}}\!$ . That would correspond to a frequency of $f\; {\rm Hz}$ .

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of $v \; {\rm m \cdot s^{-1}}$ .

Again, the source of this wave would send out a peak after each period of $(1/f)\; {\text{seconds}}$ . However, by the time the source sends out the second peak, the source would have been $v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}$ closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than $(1/f)\; {\text{seconds}}$ . The observed frequency of this wave would be larger than the original $f\; {\rm Hz}$ .

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3 years ago