Strange sum (riddle tread)
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10-09-2011, 09:17 AM
 
RE: Strange sum
Here is another quiz.

How would you find out the size and shape of the Earth, the size and distance of the Moon and the size and distance of the Sun, based on the following pieces of information:

1. The sun shines into the very bottom of a deep well on June 21st, on the Northern Hemisphere, 600 miles south of you.

2. During a Lunar eclipse it took the Moon 50 minutes to completely disappear in the Earth’s shadow and 200 minutes to fully emerge again.

The ancients Greek did it by using these pieces of data, 2 sticks and a thumb.

I am sure some of you already know the answers but, until I actually read the story, I am not sure I would have easily figured it out.
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10-09-2011, 12:18 PM
RE: Strange sum (riddle tread)
I like that one Zat, but I've already read it, so I leave it for someone else to figure out.

Here's a fun one:

You are sitting on a square piece of brown carpet, in a round, glass house with a southern view.

A bear walks by outside. What color is the bear?

No guessing here either. You gotta explain how you know the color of the bear.

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10-09-2011, 12:32 PM
RE: Strange sum (riddle tread)
(10-09-2011 12:18 PM)Stark Raving Wrote:  I like that one Zat, but I've already read it, so I leave it for someone else to figure out.

Here's a fun one:

You are sitting on a square piece of brown carpet, in a round, glass house with a southern view.

A bear walks by outside. What color is the bear?

No guessing here either. You gotta explain how you know the color of the bear.

Yay more brain exercises Big Grin

Does it matter what season it is and what time of day/night it is??? Do I have to include the position of the sun/moon in my working out of this???

For no matter how much I use these symbols, to describe symptoms of my existence.
You are your own emphasis.
So I say nothing.

-Bemore.
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10-09-2011, 12:34 PM
RE: Strange sum (riddle tread)
Round glass house with southern view.... north pole....
....white bear.

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10-09-2011, 01:01 PM
RE: Strange sum (riddle tread)
(10-09-2011 12:34 PM)Peterkin Wrote:  Round glass house with southern view.... north pole....
....white bear.

Nicely done.

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10-09-2011, 01:03 PM
RE: Strange sum (riddle tread)
White because you are at the northpole

Behold the power of the force!
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10-09-2011, 01:09 PM
RE: Strange sum (riddle tread)
Ya know, this is quickly becoming my favorite thread!

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10-09-2011, 01:23 PM
 
RE: Strange sum (riddle tread)
Here is a real problem, this time in high school Physics.

Newton's second law says that force equals mass times acceleration: F = m * a

We have an equation with three values in it. I know one of these three: I know what acceleration is and how to measure it.

I have an instinctive ‘feel’ for the concepts of mass and force, but I have no idea how to measure them and without measuring instructions, no concept in Physics has any practical value.

Any time I asked my teachers in high school what mass was, I was told that it is the measure of an object’s inertia (force divided by acceleration). And when I asked what force was, I was again told, predictably, that it was mass times acceleration.

How do we measure force and mass when they are defined by each other?
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10-09-2011, 11:06 PM
RE: Strange sum (riddle tread)
Quote:How would you find out the size and shape of the Earth, the size and distance of the Moon and the size and distance of the Sun, based on the following pieces of information:

1. The sun shines into the very bottom of a deep well on June 21st, on the Northern Hemisphere, 600 miles south of you.

2. During a Lunar eclipse it took the Moon 50 minutes to completely disappear in the Earth’s shadow and 200 minutes to fully emerge again.

Using only the first information, we can deduce the approximate distance to the Sun and the circumference of Earth.

June 21st is a special date. It is the summer solstice, when the sun is at the zenith (highest point in the sky). and since you know the distance the well is from you. All you need to do is just measure the angle of the sun relative to the zenith. Using this angle, you can obtain the circumference of the Earth, and the approximate distance to the Sun.

Not sure about the other points. My guess regarding how they determine the shape of Earth is through the shadows Earth casts on the Moon during a lunar eclipse, and using the relative shape of the Earth to the Moon, you can find the relative sizes?

Quote:Here is a real problem, this time in high school Physics.

Newton's second law says that force equals mass times acceleration: F = m * a

We have an equation with three values in it. I know one of these three: I know what acceleration is and how to measure it.

I have an instinctive ‘feel’ for the concepts of mass and force, but I have no idea how to measure them and without measuring instructions, no concept in Physics has any practical value.

Any time I asked my teachers in high school what mass was, I was told that it is the measure of an object’s inertia (force divided by acceleration). And when I asked what force was, I was again told, predictably, that it was mass times acceleration.

How do we measure force and mass when they are defined by each other?

I thought of two ways to calculate mass without the need to know the value of force, not sure whether they work.

1) We can obtain the mass of an object through heating it and finding the change in temperature. In this case, we use the formulae q=mcT. Assuming the system is closed, we need to find out the amount of energy provided by the heat source (q), the specific heat capacity of the object ©, and the change in temperature, to obtain the mass of the object.

2) Another way utilises the principle of conservation of momentum -
(m1)u1 + (m2)u2 = (m1)v1 +(m2)v2
We need two objects, one with a mass of 1kg and the other of unknown mass. Ensure that the 1kg object is stationary - m1 = 1kg, u1 = 0ms^-2. So, using devices like photo-gates, we can obtain the initial velocity of the unknown mass, and the final velocity of both objects. From there, we can obtain m2, the unknown mass.

3) To calculate force independently from mass, a friend of mine suggested using a spring balance. By hanging an object on the spring balance, and allowing the system to equilibrate, we can obtain the value of force exerted on the object through the change in reading of the spring balance.

I have a feeling I have [/quote]misinterpreted the question wrongly...

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11-09-2011, 05:56 AM
 
RE: Strange sum (riddle tread)
robotworld, you are close on both questions (Astronomy and Physics). I will answer the Astronomy first and wait with the Physics, just in case there are other bidders. Smile

Here it comes, as I read it in Isaac Asimov’s “Chronology of Science and Discovery”, Carl Sagan’s “Cosmos” and Simon Singh’s book: “Big Bang” (not a direct quote):

Apparently, Erathostenes, chief librarian of the Alexandria library, in 240BC heard that in the city of Syen (now called Aswan in Egypt), on June 21st, at exactly noon, sunlight reached the very bottom of a deep and narrow well. He wondered whether it would be the same in Alexandria. On the next June 21st he went outside the library and stuck a perfectly straight pole vertically into the ground.

At high noon, he was surprised to see a definite shadow of the pole, which told him that the sun was not directly overhead, and thus, could not illuminate the bottom of a well where he lived - while a few hundred miles to the south, it did. The only logical explanation was that the Earth must be round, just as the full Moon appeared to be.

Once Erathostenes had digested this portentous thought, he started wondering how big this globe of Earth might be, and how he could find out. Knowing a fair bit of geometry, he realized that all he needed to do was to measure the distance from Alexandria to Syen and the angle of the sun’s rays shining down on his pole at noon on June 21st. Measuring the angle was easy. To measure the distance between the two cities, he hired a man to walk to Syen, counting his steps.

From these two pieces of data, he calculated the circumference of our planet as 24,200 miles or 38,962 km. He must have had a very conscientious walker, because this figure is very close to the value we can measure today with our high-tech instruments from space (24,902 miles or 40,092kms).

Then, he started wondering what else he could learn.

Looking up from the Earth, the Moon and the Sun were the two most visible objects in the sky. How large were they?

Being the chief librarian of the Great Library of Alexandria has its advantages. With a little research he learned that the relative size of the Moon had been calculated 10 years earlier by Aristarchus, who used the lunar eclipse to accomplish this feat. Aristarchus measured the time it took the Moon to completely disappear in the Earth’s shadow (50 minutes) and the time it took to fully emerge again (200 minutes). From these two numbers he determined that the Moon’s diameter had to be one-quarter of the Earth’s.

Since Erathostenes had just calculated the Earth’s diameter, he now knew that the diameter of the Moon had to be about 3200km.

The next piece of the puzzle was the distance of the Moon from Earth. Being in possession of the diameter of the Moon, it was easy to estimate its distance.
He did this by holding up a thumb to cover the full Moon, with his arm straight out in front of him. He knew that the ratio of thumb to arm (about 1:100) had to be the same as the ratio of moon-diameter to moon-distance. Consequently, the distance of the Moon had to be 100 times the diameter of the Moon, or about 320,000 km. The distance of the Moon was also calculated by Hipparchus at circa 150BC by using parallax (observing the Moon’s position against the stars from different positions on Earth) and trigonometry that he had invented.

Two more items could be determined by further observation and logic: the distance of the Sun from Earth and the diameter of the Sun itself.

To find out the Sun’s distance, all Erathostenes had to do was to measure the angle between the lines of sight to the Moon and to the Sun on an exactly half-moon day, when both were visible. Knowing that on half moon-days the Earth, Moon and Sun are on three vertices of a right-angled triangle (as suggested by Aristarchus), he could calculate the distance to the Sun from the angle and the known distance to the Moon. The result was very close to the one we know today: 150,000,000 km.

The last datum, the size of the Sun, fell into place now, by the same reasoning he used to determine the distance to the Moon – except now the Moon took over the role of the thumb (by a lucky coincidence, the Moon exactly covers the Sun during a total solar eclipse). This time he knew the distance of the Sun, instead of its diameter. The result of his calculation was not so accurate, but still, it was the first approximation to the correct value of 1,390,000 km.


Now, weren't those Greeks very very clever?
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