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yuradex [85]
2 years ago
9

I need help asap it’s due today

Physics
1 answer:
jolli1 [7]2 years ago
6 0

For this experiment on a solar collector, the variables include the following:

  • Time
  • Temperature
  • Plastic wrap
  • Black plastic

<h3>What are the variables?</h3>

In Science, a variable can be defined as a measurable attribute or physical quantity that changes or varies across an experiment and as such can take on different numerical values.

For this experiment on a solar collector, there are different variables and these include the following:

  • Time
  • Temperature
  • Plastic wrap
  • Black plastic

Furthermore, you should determine both the greatest and smallest rate of temperature change for each of the trials by subtracting the initial temperature from the final temperature based on the numerical values obtained from this experiment.

In conclusion, the trial with the greatest numerical value has the greatest rate of temperature change while the trial with the smallest numerical value has the smallest rate of temperature change

Read more on variables here: brainly.com/question/632913

#SPJ1

<u>Complete Question:</u>

1. What are the variables in this investigation?

2. Which trial had the greatest rate of temperature change?

3. Which trial had the smallest rate of temperature change?

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Answer:

\kappa = \frac{1}{2 b}

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The equation for kappa ( κ) is

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As b is fixed, we can use kappa as a function of a

\kappa (a) = \frac{a}{a^2 + b^2}

Now, the conditions to find a maximum at a_0 are:

\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0

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Taking the first derivative:

\frac{d}{da} \kappa = \frac{d}{da}  (\frac{a}{a^2 + b^2})

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da}  (\frac{1}{a^2 + b^2} )

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1)  (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da}  (a^2+b^2)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a  (\frac{1}{(a^2 + b^2)^2} ) (2* a)

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\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2}  -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

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This clearly will be zero when

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as both are greater (or equal) than zero, this implies

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\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 -  a^2 ) + (b^2 -  a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2}  )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2  a ) + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

\frac{d^2}{da^2} \kappa = \frac{-2  a}{(a^2 + b^2)^2} + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

We dcan skip solving the equation noting that, if a=b, then

b^2 -  a^2 = 0

at this point, this give us only the first term

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the value of kappa is

\kappa = \frac{b}{b^2 + b^2}

\kappa = \frac{b}{2* b^2}

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