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hram777 [196]
2 years ago
15

Can an eagle reach exosphere through flying?

Physics
2 answers:
xz_007 [3.2K]2 years ago
4 0
Answer: no
Because of lack of air
nignag [31]2 years 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!

You might be interested in
A Newtonian fluid with a viscosity  drains through the space between two large parallel plates as shown in the figure. The gap
Rina8888 [55]

Answer:

well

Explanation:

basically it's like

8 0
2 years ago
A projectile of mass m is launched with an initial velocity vector v i making an angle θ with the horizontal as shown below. The
sergeinik [125]
Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be: 
<span>L = R * m * vi * cos(90 - theta) </span>

<span>cos(90 - theta) is just sin(theta) </span>
<span>and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g </span>

<span>so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g </span>

<span>We can combine the two vi terms and get: </span>

<span>L = vi^3 * m * sin(theta) * sin(2*theta) / g </span>

<span>What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. </span>

<span>Now, for the first part... </span>

<span>There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. </span>

<span>Okay, so let's get back to what we know: </span>

<span>L = d * m * v * cos(phi) </span>

<span>where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... </span>

<span>m is still m. </span>
<span>v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) </span>
<span>d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g </span>

<span>That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. </span>

<span>The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: </span>

<span>h = 1/2 * (vi * sin(theta))^2 / g </span>

<span>So, there's the rise. So, our *slope* is rise/run, so </span>

<span>slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] </span>

<span>The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: </span>

<span>slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ] </span>

<span>(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). </span>

<span>Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... </span>

<span>slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4 </span>

<span>Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... </span>

<span>alpha = arctan( tan(theta) / 4 ) </span>

<span>Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha </span>

<span>so, phi = 180 - arctan( tan(theta) / 4 ) </span>

<span>Now, we go back to our original formula and plug it ALL in... </span>

<span>L = d * m * v * cos(phi) </span>

<span>becomes... </span>

<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 - arctan( tan(theta) / 4 ) ) ] </span>

<span>Now, cos(180 - something) = cos(something), so we can simplify a little bit... </span>

<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ] </span>
3 0
2 years ago
Read 2 more answers
The aqueous humor in a person's eye is exerting a force of 0.242 N on the 1.21 cm2 area of the cornea. What pressure is this in
Virty [35]

Answer:

<h2><em>15.00124mmHg</em></h2>

Explanation:

Pressure is defined as the ratio of force applied to an object to its area.

Pressure = Force/Area

Given parameters

Force = 0.242N

Area = 1.21cm²

Required parameters

Pressure = 0.242/1.21

Pressure = 0.2N/cm²

Using the conversion to convert the pressure to mmHg

1N/cm² = 75.0062mmHg

0.2N/cm² = y

y = 0.2 * 75.0062

y = 15.00124mmHg

<em>Hence the pressure in mmHg is 15.00124mmHg</em>

4 0
2 years ago
Two blocks, joined by a string, have masses of 6.0 and 9.0 kg. They rest on a frictionless, horizontal surface. A second string,
Lynna [10]

Answer:

T= 27 N

Explanation:

Assuming that the string joining both masses is massless  and inextensible, both masses accelerate at the same rate.

So, we can treat to both masses as a single system, and apply Newton's 2nd Law to both masses.

In this way, we can get the value of the acceleration without taking into account the tension in the string, as it is an internal force (actually a action-reaction pair).

Newton's 2nd law is a vector equation, so we can decompose the forces along perpendicular axis in order to convert it in two algebraic equations.

We can choose one axis as parallel to the horizontal surface (we call it x-axis, being the positive direction the one of  the movement of the blocks due to the horizontal force applied to the 6.0 kg block), and the other, perpendicular to it, so it is vertical (we call y-axis, being the upward direction the positive one).

Taking into account the forces acting  on both masses, we can write both equations as follows:

Fy = N- (m₁+m₂)*g = 0 (as there is no movement in the vertical direction)

Fx = Fh = (m₁ + m₂) * a ⇒ 45 N = 15.0 kg * a

⇒ a = 45 N / 15.0 kg = 3 m/s²

Now, in order to get the value of the tension T, we can choose as our system, to any mass, and apply Newton's 2nd Law again.

If we choose to the mass of 6.0 kg, in the horizontal direction, there are two forces acting on it, in opposite directions: the  horizontal applied force of 45 N, and the tension in the string that join both masses.

The difference of both forces, must be equal to the mass (of this block only) times the acceleration, as follows:

F- T = m₂* a ⇒ 45 N - T = 6.0 kg * 3 m/s²

⇒ T = 45 N -18 N = 27 N

We could have arrived to the same result taking the 9.0 Kg as our system, as the only force acting in the horizontal direction is just the tension in the string that we are trying to find out, as follows:

F = m₁*a = 9.0 kg* 3 m/s² = 27 N

4 0
3 years ago
If earth had no atmosphere , would a falling object ever reach terminal velocity?
Bas_tet [7]
The right answer is no.
6 0
3 years ago
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