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3 years ago

WHAT FEATURE DO ALL VOLCANOES SHARE? PLEASE LIST AT LEAST 3!!!

Physics

2 answers:

tangare [24]3 years ago

8 0

Some features all volcanoes share is pocket of magma, side vents and crater

snow_tiger [21]3 years ago

7 0

Rocky,hot,magma,underground and much more

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The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of

snow_tiger [21]

Answer:

Explanation:

Given

height of grand canyon is $h=1800\ m$

Rock is dropped i.e. initial velocity is zero

$u=0$

Using Equation of motion

$h=ut+\frac{1}{2}at^2$

h=height

u=initial velocity

a=acceleration

t=time

here a=acceleration due to gravity

$1800=0+\frac{1}{2}\times 9.8\times t^2$

$t^2=\frac{2\times 1800}{9.8}$

$t=19.166\ s$

3 0

3 years ago

How would the seasons change if the earth were tilted at 90 degrees instead of 23.5?

PIT_PIT [208]

<span>The 23.5 degree tilt is responsible for the seasons. If the earth had no tilt there would not be seasons. If the earth was tilted by 90 degrees the seasonal changes would be at the most extreme. The Earth's pole would point directly at the sun at a point on the track around the sun. As the Earth revolves around the Sun the pole would alternate twice each year between pointing directly at the sun and being perpendicular to the sun.

I hope this helps you!
xo, Leafling</span>

6 0

3 years ago

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$