<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:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Answer:
The dissolving power of water is very important for life on Earth. Wherever water goes, it carries dissolved chemicals, minerals, and nutrients that are used to support living things. Because of their polarity, water molecules are strongly attracted to one another, which gives water a high surface tension.
Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter
Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.
The algebraic expression would be:
<em>ppb [=] micrograms of compound/liter of solution</em>
We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.
For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.
The respective procedure is in a attached file.
