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1 year ago

Can an eagle reach exosphere through flying?

Physics

2 answers:

xz_007 [3.2K]1 year ago

4 0

Answer: no
Because of lack of air

nignag [31]1 year ago

3 0

Simply no.

And the problem wouldn't be only the lack of gravity, it would be also the lack of air.

Flying is an action not to far from swiming: something is moving trough some fluid. Of course that the fluid will incredibly matter, but the basics are the same.

When a eagle flies, it is pushing its wings against the air, and the air is holding that pressure against the wing, and, also for a bunch of anatomic reasons, the eagle is able to fly.

Without air and gravity, it will just keep floating around until one of the gravitational pulls wins the battle and attrack it to a painfull fate.

I hope I helped!

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Monochromatic light falls on two very narrow slits 0.048 mm apart. successive fringes on a screen 5.00 m away are 6.5 cm apart n

atroni [7]

In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
$y= \frac{m \lambda D}{d}$
where D=5.00 m is the distance of the screen from the slits, and
$d=0.048 mm=0.048 \cdot 10^{-3}m$ is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
$\lambda = \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}= 6.24 \cdot 10^{-7}m$

And from the relationship between frequency and wavelength, $c=\lambda f$ , we can find the frequency of the light:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz$

4 0

2 years ago

. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass

alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

3 0

2 years ago

Which statement about oceans is incorrect?

Hitman42 [59]

It is C because less than one percent of water is fresh water

7 0

2 years ago

A swimming pool is 4.0 m in depth; a swimmer at this depth feels discomfort in the ear. Calculate the net force on a 0.50-cm-dia

Mashcka [7]

The net force on a 0.50-cm-diameter eardrum is mathematically given as

F= 0.76969 N

<h3>What is the net force on a 0.50-cm-diameter eardrum?</h3>

Generally, the equation for Pressure is mathematically given as

P = ρgh

Therefore

P= 1000*9.8*4

P= 39200 Pa

Where

A= pi*(0.005/2)^2

Generally, the equation for Net force is mathematically given as

F = PA

F= 39200 *( pi*(0.005/2)^2)

F= 0.76969 N

In conclusion, The net force is

F= 0.76969 N