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Math Help - Problem solving

  1. #1
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    Problem solving

    I have no idea about this problem solving, don't know where to start or how to go about it.. Could someone please help me with these questions? Any help would be HUGELY appreciated Thank you!

    1. )You are watching the sunset in July on a west-facing beach in a country on the equator (say
    Gabon). Behind the beach there is a cliff, 30 meters high. As soon as the sun has disappeared
    behind the horizon you start a timer and run to the top of the cliff, where the sun is still visible.
    You watch the sunset again from the cliff and stop the timer when the sun disappears behind
    the horizon. The reading on the timer is 42.2 seconds. Use this information to estimate the radius of the earth.
    2.) Give estimates of the following quantities Justify your answer!
    (a) The number of (working) hairdressers in Edinburgh,
    (b) The number of grains of sand on (mainland) Spanish beaches (not counting sand which is
    permanently covered with water),
    (c) The energy that could be generated per year if a every Munro was equipped with a
    hydroelectric power plant.
    (d) The total annual domestic energy consumption in Scotland
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  2. #2
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    Welcome to the forum.

    The number of grains of sand on (mainland) Spanish beaches (not counting sand which is
    permanently covered with water)
    This is a so-called Fermi question. It's difficult to give a precise answer, but it's possible to estimate it up to the order of magnitude (or several). A solution proceeds by making a series of reasonable assumptions, and it often happens that an overestimation in one of them is compensated by an underestimation in another.

    According to this Wikipedia page, the length of the Spanish coast is about 5000 km = 5 * 10^6 m (I would have guessed 1000 km at first; oh well). Let's assume that the beach is on average 10 m wide and 20 cm deep. Then the total beach volume is 5 * 10^6 * 10 * 0.2 = 10^7 m^3. Also, let's assume a grain of send is 1 mm^3. Then the number of grains is 10^7 * 10^9 = 10^16. Of course, I could be way off.

    (a) The number of (working) hairdressers in Edinburgh
    Let's follow the solution of this classic Fermi question. There are about 5 * 10^5 people in Edinburgh (hmm, I would have guessed at least 10^6). Let's say an individual gets a haircut every 3 weeks on average (much more rarely for me). Then there are 5 * 10^5 haircuts in about 20 days, i.e., about 2 * 10^4 haircuts a day. Suppose a hairdresser can serve 4 clients an hour, which makes about 30 clients per day. Then it would take about 600 hairdressers.

    Finally, an anecdote about estimating quantities up to the order of magnitude. They say that the famous Russian physicist Lev Landau was getting his salary one day. Having received the cash, he moved aside and started carefully counting the money. Someone in the line to the cashier said, "Mr. Landau, you taught yourself that magnitudes in physics have sense only up to the order of magnitude." Landau replied: "Money is located in the exponent."
    Last edited by emakarov; January 26th 2011 at 12:54 PM. Reason: Spelling.
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  3. #3
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    1. )You are watching the sunset in July on a west-facing beach in a country on the equator (say
    Gabon). Behind the beach there is a cliff, 30 meters high. As soon as the sun has disappeared
    behind the horizon you start a timer and run to the top of the cliff, where the sun is still visible.
    You watch the sunset again from the cliff and stop the timer when the sun disappears behind
    the horizon. The reading on the timer is 42.2 seconds. Use this information to estimate the radius of the earth.
    In the future, such problems should be posted to Geometry, Trigonometry or Calculus forums.



    According to the problem statement, the Earth turned \alpha radians in 42.2 seconds. Therefore, \alpha=2\pi\cdot42.2/86400. Also, \cos\alpha=R/(R+h)=1/(1+h/R) where h = 30 m. The angle \alpha, as well as h/R, is very small. From calculus, \cos\alpha\approx1-\alpha^2/2 and 1/(1+h/R)\approx1-h/R. Therefore, \alpha^2/2\approx h/R, from where R can be found.
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