D-It was discovered a while ago and it's about 500 light years away from earth
Answer:
gravity increase when distance decrease,mass increase
gravity decrease vwhen distance decrease,mass decrease
<span>Exactly 4(4 - 2*2^(1/3) + 2^(2/3)) feet,
or approximately 12.27023581 feet.
Let's first create an equation to calculate the relative intensity of the light based upon the distance D from the brighter light source. The distance from the dimmer light source will of course be (20-D). So the equation will be:
B = 4/D^2 + 1/(20-D)^2
The minimum and maximum can only occur at those points where the slope of the equation is 0. And you can determine the slope by using the first derivative. So let's calculate the first derivative.
B = 4/D^2 + 1/(20-D)^2
B' = d/dD [ 4/D^2 + 1/(20-D)^2 ]
B' = 4 * d/dD [ 1/D^2 ] + d/dD [ 1/(20-D)^2 ]
B' = 4(-2)D^(-3) + (-2)(20 - D)^(-3) * d/dD [ 20-D ]
B' = -8/D^3 - 2( d/dD [ 20 ] - d/dD [ D ] )/(20 - D)^3
B' = -8/D^3 - 2(0 - 1)/(20 - D)^3
B' = 2/(20 - D)^3 - 8/D^3
Now let's find a zero.
B' = 2/(20 - D)^3 - 8/D^3
0 = 2/(20 - D)^3 - 8/D^3
0 = 2D^3/(D^3(20 - D)^3) - 8(20 - D)^3/(D^3(20 - D)^3)
0 = (2D^3 - 8(20 - D)^3)/(D^3(20 - D)^3)
0 = 2D^3 - 8(20 - D)^3
8(20 - D)^3 = 2D^3
4(20 - D)^3 = D^3
4(8000 - 1200D + 60D^2 - D^3) = D^3
32000 - 4800D + 240D^2 - 4D^3 = D^3
32000 - 4800D + 240D^2 - 5D^3 = 0
6400 - 960D + 48D^2 - D^3 = 0
-6400 + 960D - 48D^2 + D^3 = 0
D^3 - 48D^2 + 960D - 6400 = 0
We now have a simple cubic equation that we can use the cubic formulas to solve.
Q = (3*960 - (-48)^2)/9 = 64
R = (9*(-48)*960 - 27*(-6400) - 2*(-48)^3)/54 = -384
D = Q^3 + R^2 = 64^3 + (-384)^2 = 409600
Since the value D is positive, there are 2 imaginary and 1 real root. We're only interested in the real root.
S = cbrt(-384 + sqrt(409600))
S = cbrt(-384 + 640)
S = cbrt(256)
S = 4cbrt(4)
T = cbrt(-384 - sqrt(409600))
T = cbrt(-384 - 640)
T = cbrt(-1024)
T = -8cbrt(2)
The root will be 4cbrt(4) - 8cbrt(2) + 48/3
So simplify
4cbrt(4) - 8cbrt(2) + 48/3
=4cbrt(4) - 8cbrt(2) + 16
=4(cbrt(4) - 2cbrt(2) + 4)
= 4(4 - 2*2^(1/3) + 2^(2/3))
Which is approximately 12.27023581
This result surprises me. I would expect the minimum to happen where the intensity of both light sources match which would be at a distance of 2/3 * 20 = 13.3333 from the brighter light source. Let's verify the calculated value.
Using the brightness equation at the top we have:
B = 4/D^2 + 1/(20-D)^2
Using the calculated value of 12.27023581, we get
B = 4/D^2 + 1/(20-D)^2
B = 4/12.27023581^2 + 1/(20-12.27023581)^2
B = 4/12.27023581^2 + 1/7.72976419^2
B = 4/150.5586868 + 1/59.74925443
B = 0.026567713 + 0.016736611
B = 0.043304324
And the intuition value of 13.33333333
B = 4/D^2 + 1/(20-D)^2
B = 4/13.33333333^2 + 1/(20-13.33333333)^2
B = 4/13.33333333^2 + 1/6.666666667^2
B = 4/177.7777778 + 1/44.44444444
B = 0.0225 +0.0225
B = 0.045
And the calculated value is dimmer. So intuition wasn't correct.
So the object should be placed 4(4 - 2*2^(1/3) + 2^(2/3)) feet from the stronger light source, or approximately 12.27023581 feet.</span>
Answer:
the basic unit of a chemical element.
Explanation:
pls follow me
Explanation:
The mass of one atom of calcium is the same as the molar mass of the calcium. Also, the atomic mass of calcium is 40.078 g/mol.
