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Investigating Forces & Motion
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Investigating Forces and Motion (1998)(Granada Learning).iso
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topic5
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example.dat
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INI File
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1998-02-16
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2KB
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55 lines
[general]
[page1]
type:4
caption:\
Some of the following five statements about the force of gravity are \
true, others are false. Drag<B> true</B> or <B>false </B>into place \
against each statement.<p>
feedback:\
Correct. The force of gravity is proportional to mass. All objects \
accelerate at the same rate towards the Earth. Objects do have weight \
on the Moon. Gravity holds the Moon in its orbit.<p>\
<I>g</I> = 9.8 m/s<SUP>2</SUP>, so 5.0 x 9.8 = 49 m/s.<p>
true:5ex1d, 5ex1a, 5ex1e
false:5ex1b, 5ex1c
[page2]
type:0
caption:\
<img src="5ex2" align=center><p>\
An air-sea rescue officer jumps into the sea from a stationary \
helicopter hovering 20 m above the water's surface. How long is it \
before he hits the water?<p>\
Sign convention: down is positive.<p>\
Neglecting air resistance, we can use the equation:<p>\
<I>s</I> =<I> ut</I> + ½ <I>at</I><SUP>2</sup><p>\
The known and unknown values are:<p>\
<I>u = 0.0</I><p>\
<I>s</I> = 20 m<p>\
<I>a</I> = <I>g</I> = + 9.8 m/s<SUP>2</SUP><p>\
<I>t = ?<p>\
</I><p>\
Substituting the values, we get:<p>\
20 = 0.0 + ½ x 9.8 x <I>t<SUP>2</SUP></I><p>\
Therefore,<p>\
<img src="r5ex2b"><p>\
= 4.1 </SUP><p>\
Therefore,<p>\
<I>t</I> = 2.0 s.<p>
[page3]
type:0
caption:\
<img src="5ex2a" align=center><p>\
Suppose the helicopter had been travelling at 10 m/s in a straight \
line when the man jumped. Which path would the man have followed into \
the water? How long would it now be before he hits the water?<p>\
The path is a parabola. The man's horizontal velocity is constant as \
he falls. Therefore, when he hits the water he is still directly under \
the moving helicopter.<p>\
Because the downward acceleration is not affected by the horizontal \
movement, the man takes the same time to reach the water's surface \
from the moving helicopter as he does from a stationary helicopter.<p>