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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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natali 33 [55]

Opportunity cost refers to what you have to give up to buy what you want in terms of other goods or services. When economists use the word “cost,” we usually mean opportunity cost.

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3 years ago

Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at

beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0

3 years ago

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NISA [10]

Answer:

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Explanation:

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8 0

3 years ago

How do you change current in a circuit without changing the voltage?

Sunny_sXe [5.5K]

Decrease the reactance is the correct answer i believe

Explanation:

4 0

3 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

5 0

2 years ago