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ASHA 777 [7]
3 years ago
11

A passenger jet flying at cruising altitude is often traveling in the upper troposphere at approximately 725 km per hour (or 450

mph). Why don't passenger jets burn up as meteors do?
Physics
1 answer:
kobusy [5.1K]3 years ago
7 0

Explanation:

Objects like meteors travel at high rates of the speed and thus, possess large amounts of the kinetic energy.  As the meteors enter Mesosphere, they enter more dense segment of earth's atmosphere with the greater numbers of the gas molecules.  Meteors which encounter the large numbers of the gas molecules at the high rates of the speed and generate large amounts of the friction in process. As a result, enormous heat is built up. This build-up of heat  ultimately causes the fast-traveling meteors to burn up in atmosphere.

<u>While passenger jets travel at very small speed as compared with the speed of the meteors and don not burn. </u>

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6 0
3 years ago
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\&#10;g_0=g'\\&#10;g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\&#10;ag_0=g_0-w^2r\\&#10;w^2r=g_0(a-1)&#10;
Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

5 0
3 years ago
A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle abov
Digiron [165]

Ox:vₓ=v₀

x=v₀t

Oy:y=h-gt²/2

|vy|=gt

tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g

y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°

cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s

Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ

3 0
3 years ago
A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. what is the volume of the ball?
lesantik [10]
Density=mass/volume therefore volume=mass/density; 55g/11.4g/cm^3= 4.82cm^3
6 0
3 years ago
Read 2 more answers
What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface
kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman = \frac{95.08}{100 } = 0.9508

The density of the woman is calculate as;

Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman  = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0
3 years ago
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