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Shalnov [3]
3 years ago
10

A model used for the yield y of an agricultural crop as a function of the nitrogen level n in the soil (measured in appropriate

units) is y = kn 1 + n2 where k is a positive constant. what nitrogen level gives the best yield?
Physics
1 answer:
pychu [463]3 years ago
5 0

<span>The maxima of an equation can be obtained by taking the 1st derivative of the equation then equate it to 0.</span>The value of N that result in best yield is when dy/dn = 0.

Taking the 1st derivative of the equation y=(kn)/(9+n^2) :<span>
</span>

By using the quotient rule the form of the equation is:<span>
y = g(n) / h(n) 
where:</span>

g(n) = kn    --->    g'(n) = k 

<span> <span>h(n) = 9 + n^2     --->    h'(n) = 2n </span>
dy/dn is defined as:
<span>dy/dn = [h(n) * g'(n) - h'(n) * g(n)] / h(n)^2 
dy/dn = [(9 + n^2)(k) - (kn)(2n)] / (9 + n^2)^2 
dy/dn = (9k + kn^2 - 2kn^2) / (9 + n^2)^2 
dy/dn = (9k - kn^2) / (9 + n^2)^2 
dy/dn = k(9 - n^2) / (9 + n^2)^2 

<span>Equate dy/dn = 0, then solve for n 
k(9 - n^2) / (9 + n^2)^2 = 0 
k(9 - n^2) = 0 
9 - n^2 = 0 
n^2 = 9 
n = sqrt(9) 
n = 3 

<span>Answer: The nitrogen level that gives the best yield of agricultural crops is 3 units.</span></span></span></span>

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Answer:

The magnitude of the emf induced in the coil is 60 mV.

Explanation:

We have,

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Now, induced emf becomes :

\epsilon=N\dfrac{d(BA\cos \theta)}{dt}\\\\\epsilon=NA\dfrac{dB}{dt}\cos \theta\\\\\epsilon=2\times (0.24)^2\times \dfrac{6\times 10^{-3}}{10\times 10^{-3}}\times \cos (30)\\\\\epsilon=0.0598\ V\\\\\epsilon=59.8\ V

or

\epsilon=60\ mV

So, the magnitude of the emf induced in the coil is 60 mV. Hence, this is the required solution.

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