First we have to be careful because 'g' is defined as the nominal acceleration due to gravity at sea-level on earth. So g, by definition, is always 9.80665 m/s2.
Now if we ask "At what height from the earth does the acceleration due to the gravity of the earthbecome 0?" then the "classical" answer, from Newton's Universal Law of Gravity, is infinity (but read the last note at the end of this answer). The classical equation for this acceleration is:
GMe/r2
where G is the universal gravitational constant (~ 6.67428x10-11 m3kg-1s-2)
and Me is the mass of the earth (~ 5.9736x1024 kg)
and r is the distance from the earth
So as r increases the acceleration decreases rapidly. This is a prime example of an "inverse square law". At very large distances the acceleration still has a very small but non-zero value, tapering to zero at infinity.
Another question would be "At what height from the earth does the total acceleration due to the gravity of all other objects become 0?". The answer is that it depends on what direction you are considering. Directly between the earth and the moon for instance, then if we ignore the effects of other objects (sun, planets, stars, ...) we can calculate the point where the effect of gravity from the earth and the moon "cancel each other out", giving a net acceleration of zero:
GMe/r2 = GMm/(D-r)2 Where r < D
where Mm is the mass of the moon (~ 0.07349x1024 kg)
and D is the distance between the earth and the moon (~ 3.78x108 m)
First, simplifying the equation (by dividing by G and multiplying r2 and (D-r)2) :
(D-r)2 = r2M (where we'll simply define the mass ratio M = Mm/Me ~ 0.0123)
Expanding:
D2-2Dr+r2 = r2M
Turning into the quadratic equation:
(1-M)r2 -2Dr + D2 = 0
Solving the quadratic equation:
r = 2D +/- (4D2 - 4x(1-M)xD2)-2 / 2(1-M)
Simplifying:
r = D +/- (D2 - (1-M)xD2)-2 / (1-M)
r = D +/- (D2 - D2 + MD2)-2 / (1-M)
r = D x (1 +/- M-2) / (1-M)
Now filling in the values:
r = 3.78x108 x (1 +/- (0.0123)-2) / (1 - 0.0123)
r = 3.78x108 x (1 +/- 0.1109) / 0.9877
r = 3.83x108 x 1.1109 or 3.83x108 x 0.8891
r = 4.25x108 m or 3.41x108 m
Now we can throw away the 1st value because our condition at the beginning was r < D (the magnitudes of the two forces are equal at this distance too, but they are both now in the same direction and so add together rather than subtract to zero.)
So r = 3.41x108 m = 341,000 km ~ 211,889 miles (or about 90% of the distance to the moon)
To put it in perspective that's roughly the total distance you drive in a car over 10 years.
OK so what about directly between the earth and the sun (ignoring moon etc.)?
A quick flip of the values:
D = distance between earth and the sun = 1.496x1011 m
M = ratio of sun's mass to earth's mass = 333,000
r = 1.496x1011 x (1 +/- (333,000)-2) / (1 - 333,000)
r = 1.496x1011 x (1 +/- 577) / -332,999
r = -4.49x105 x 578 or 4.49x105 x 576
r = -2.60x108 m or 2.59x108 m
Here then r = 2.59x108 m = 259,000 km ~ 160,935 miles (or only about 0.17% of the distance to the sun!)
Note that this height result for the sun is less than the height for the moon. So how about when the sun and the moon "gang up", combining their gravitational forces against that of the earth. This would happen at the time of a solar eclipse, with the moon directly between the earth and the sun.
For this case:
GMe/r2 = GMm/(Dm-r)2 + GMs/(Ds-r)2 Where r < Dm
or 1/r2 = Mme/(Dm-r)2 + Mse/(Ds-r)2
This will produce a 4th order polynomial to solve for those interested to try!
Another interesting thing to note is that we should be careful when using this application of the classical Newton's Universal Law of Gravitation further out in the universe. From the discoveries of both the faster than expected speed of stars around galaxies and the unexpected acceleration of the universe itself, it seems there are other factors at work. They may be the entities tentatively called "Dark Matter" and "Dark Energy" respectively or they may be an indication that the Law of Gravitation is only an approximation that works well within the scale of our solar system but requires adjustment at the much larger scales of galaxies and the entire universe.
The Value of the Determinant becomes 0
You mean 'core'. It's 0.
Any formula for the height of a rhombus will depend on the information that you do have. Without that, all that can be said is that, if the sides of the rhombus are x units, then 0 < h < x where the height is h units. If h = 0 then the rhombus degenerates into a flat line, while at h = x it becomes a square.
It is 0 because the height would be zero I think
The ISBN of The Height of the Scream is 0-87054-075-0.
It will depend upon the radius of the base and as such will be any value between 0 cm3 and approx 403.067 cm3. The height and radius are linked by: radius2 + height2 = slant_height2 Volume cone = 1/3π x radius2 x height When radius is 0 cm, height is 10 cm and volume is 0 cm3 When radius is 10 cm, height is 0 cm and volume is 0 cm3 When radius is √(200/3) cm ≈ 8.165 cm and the height is √(100/3) cm ≈ 5.774 cm, the volume is at its maximum of 1/3π200/3√(100/3) cm3 ≈ 403.067cm3
If you are talking about a culture there are Tribes in Sudan where the average height of a man is 6'4'' with some men reaching 7' or more. By nation the Dutch are the tallest with an average height of 6' 0".
Place value 0 in tens place is always 0 because 0 has no value.
the value of 0 is '1'.
The value is 0.
Absolute value changes the sign of a number to positive, so negative one becomes positive one, and positive one stays positive one. Zero has no sign, and is therefore equal to its absolute value. If you use a graphing calculator, or a graphing program (online or installed software), and graph y equal to theabsolute value of x, you will see two lines, y=x from 0 to infinity and y=(-x) from 0 to negative infinity. When x=0, the two lines are the same because the negative of 0 is still 0.
The absoulte value of Zero is 0