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2 years ago

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!

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Wegener proposed the continental drift hypotheses suggesting that

kipiarov [429]

There was a supercontinent called Pangea

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3 years ago

If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s

Kazeer [188]

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

= E x Q/ (Q/2)

= 2E

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2 years ago

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3 years ago

A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant

Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0

3 years ago

An astronaut goes out for a space walk. Her mass (including space suit, oxygen tank, etc.) is 100 kg. Suddenly, disaster strikes

Marina CMI [18]

Answer:

Part A:

Unknown variables:

velocity of the astronaut after throwing the tank.

maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen.

Known variables:

velocity and mass of the tank.

mass of the astronaut after and before throwing the tank.

maximum time it can take the astronaut to return to the spacecraft.

Part B:

To obtain the velocity of the astronaut we use this equation:

-(momentum of the oxygen tank) = momentum of the astronaut

-mt · vt = ma · vt

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

To obtain the maximum distance the astronaut can be away from the spacecraft we use this equation:

x = x0 + v · t

Where:

x = position of the astronaut at time t.

x0 = initial position.

v = velocity.

t = time.

Part C:

The maximum distance the astronaut can be away from the spacecraft is 162 m.

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the oxygen tank when it is thrown away must be equal to the momentum of the astronaut but in opposite direction. In other words, the momentum of the system astronaut-oxygen tank is the same before and after throwing the tank.

The momentum of the system before throwing the tank is zero because the astronaut is at rest:

Initial momentum = m · v

Where m is the mass of the astronaut plus the equipment (100 kg) and v is its velocity (0 m/s).

Then:

initial momentum = 0

After throwing the tank, the momentum of the system is the sum of the momentums of the astronaut plus the momentum of the tank.

final momentum = mt · vt + ma · va

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

Since the initial momentum is equal to final momentum:

initial momentum = final momentum

0 = mt · vt + ma · va

- mt · vt = ma · va

Now, we have proved that the momentum of the tank must be equal to the momentum of the astronaut but in opposite direction.

Solving that equation for the velocity of the astronaut (va):

- (mt · vt)/ma = va

mt = 15 kg

vt = 10 m/s

ma = 100 kg - 15 kg = 85 kg

-(15 kg · 10 m/s)/ 85 kg = -1.8 m/s

The velocity of the astronaut is 1.8 m/s in direction to the spacecraft.

Let´s place the origin of the frame of reference at the spacecraft. The equation of position for an object moving in a straight line at constant velocity is the following:

x = x0 + v · t

where:

x = position of the object at time t.

x0 = initial position.

v = velocity.

t = time.

Initially, the astronaut is at a distance x away from the spacecraft so that

the initial position of the astronaut, x0, is equal to x.

Since the origin of the frame of reference is located at the spacecraft, the position of the spacecraft will be 0 m.

The velocity of the astronaut is directed towards the spacecraft (the origin of the frame of reference), then, v = -1.8 m/s

The maximum time it can take the astronaut to reach the position of the spacecraft is 1.5 min = 90 s.

Then:

x = x0 + v · t

0 m = x - 1.8 m/s · 90 s

Solving for x:

1.8 m/s · 90 s = x

x = 162 m

The maximum distance the astronaut can be away from the spacecraft is 162 m.

6 0

3 years ago