| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.06 |
| Score | 0% | 61% |
Depending on where you apply effort and resistance, the wheel and axle can multiply:
power or distance |
|
force or speed |
|
force or distance |
|
speed or power |
If you apply the resistance to the axle and the effort to the wheel, the wheel and axle will multiply force and if you apply the resistance to the wheel and the effort to the axle, it will multiply speed.
A a seesaw / teeter-totter is an example of which of the following?
second-class lever |
|
inclined plane |
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first-class lever |
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third-class lever |
A first-class lever is used to increase force or distance while changing the direction of the force. The lever pivots on a fulcrum and, when a force is applied to the lever at one side of the fulcrum, the other end moves in the opposite direction. The position of the fulcrum also defines the mechanical advantage of the lever. If the fulcrum is closer to the force being applied, the load can be moved a greater distance at the expense of requiring a greater input force. If the fulcrum is closer to the load, less force is required but the force must be applied over a longer distance. An example of a first-class lever is a seesaw / teeter-totter.
A fixed pulley has a mechanical advantage of:
2 |
|
1 |
|
0 |
|
-1 |
A fixed pulley is used to change the direction of a force and does not multiply the force applied. As such, it has a mechanical advantage of one. The benefit of a fixed pulley is that it can allow the force to be applied at a more convenient angle, for example, pulling downward or horizontally to lift an object instead of upward.
Which of the following statements about drag is false?
the amount of drag depends on the shape of an object |
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the amount of drag depends on the speed of an object |
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drag occurs during movement through a fluid |
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slower objects experience more drag than faster objects |
Drag is friction that opposes movement through a fluid like liquid or air. The amount of drag depends on the shape and speed of the object with slower objects experiencing less drag than faster objects and more aerodynamic objects experiencing less drag than those with a large leading surface area.
| 6.86 ft. | |
| 0 ft. | |
| 3.43 ft. | |
| 8 ft. |
fAdA = fBdB
For this problem, the equation becomes:
30 lbs. x 8 ft. = 70 lbs. x dB
dB = \( \frac{30 \times 8 ft⋅lb}{70 lbs.} \) = \( \frac{240 ft⋅lb}{70 lbs.} \) = 3.43 ft.