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Investigating Forces & Motion
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Investigating Forces and Motion (1998)(Granada Learning).iso
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topic4
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example.dat
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INI File
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1998-02-10
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2KB
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48 lines
[general]
[page1]
type:0
caption:\
<img src="4p8" align=center><p>\
A snooker ball is struck by a cue and accelerates from rest to 2.0 m/s \
in 0.1 s. Calculate the acceleration of the ball. If the mass of the \
ball is 0.2 kg, calculate the average force on it from the cue.<p>\
To calculate the acceleration, use the equation:<p>\
acceleration = velocity change ÷ time<p>\
<center>= (2.0 - 0.0) ÷ 0.1</center><p>\
<center>= 20 m/s<sup>2</sup></center><p>\
To calculate the force, use the formula:<p>\
<i>F</i> = <i>ma</i> (Newton's second law of motion)<p>\
<center>= 0.2 x 20</center><p>\
<center>= 4.0 N</center><p>
[page2]
type:1
caption:\
Now suppose the same force of 4.0 N is applied to the masses below. \
What accelerations will they experience ? Drag the arrows to match the \
values to the masses.<p>
feedback:\
Correct.<p>\
<i>a</i> = <i>F/m</i> = 4.0 ÷ 1.0 = 4.0 m/s<sup>2</sup><p>\
<i>a</i> = <i>F/m</i> = 4.0 ÷ 0.5 = 8.0 m/s<sup>2</sup><p>\
<i>a</i> = <i>F/m</i> = 4.0 ÷ 0.1 = 40 m/s<sup>2</sup><p>
source:4ex2a, 4ex2b, 4ex2c
target:4ex2a1, 4ex2b1, 4ex2c1
[page3]
type:3
caption:\
Each of the masses on the left has a different force acting on it. \
What acceleration does the force produce? Drag the acceleration values \
from the right into the box next to the correct combination of force \
and mass.<p>
feedback:\
Correct. In each case <I>a</I> = <I>F/m</I>: 5.0/1.0 = 5.0 \
m/s<SUP>2</SUP>; 1.0/2.0 = 0.50 m/s<SUP>2</SUP>; 3.0/5.0 = 1.5 \
m/s<SUP>2</SUP>; 10/0.1 = 100 m/s<SUP>2</SUP>; 0.40/10 = 0.04 \
m/s<SUP>2</SUP>.<p>
source:4ex3a, 4ex3b, 4ex3c, 4ex3d, 4ex3e
target:4ex3d2, 4ex3b2, 4ex3c2, 4ex3e2, 4ex3a2