| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.17 |
| Score | 0% | 63% |
The force amplification achieved by using a tool, mechanical device or machine system is called:
efficiency |
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mechanical advantage |
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power |
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work |
Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. Such a device utilizes input force and trades off forces against movement to amplify and/or change its direction.
Coplanar forces:
act in a common plane |
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have opposite dimensions |
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pass through a common point |
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act along the same line of action |
Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.
A shovel is an example of which class of lever?
third |
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first |
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a shovel is not a lever |
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second |
A third-class lever is used to increase distance traveled by an object in the same direction as the force applied. The fulcrum is at one end of the lever, the object at the other, and the force is applied between them. This lever does not impart a mechanical advantage as the effort force must be greater than the load but does impart extra speed to the load. Examples of third-class levers are shovels and tweezers.
The mechanical advantage of a block and tackle is equal to which of the following?
the number of input forces |
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the number of loads |
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the number of connecting ropes |
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the number of pulleys |
Two or more pulleys used together constitute a block and tackle which, unlike a fixed pulley, does impart mechanical advantage as a function of the number of pulleys that make up the arrangement. So, for example, a block and tackle with three pulleys would have a mechanical advantage of three.
| 20.3 psi | |
| 30.4 psi | |
| 23.3 psi | |
| 18.2 psi |
According to Boyle's Law, pressure and volume are inversely proportional:
\( \frac{P_1}{P_2} \) = \( \frac{V_2}{V_1} \)
In this problem, V2 = 20 ft.3, V1 = 45 ft.3 and P1 = 9.0 psi. Solving for P2:
P2 = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{9.0 psi}{\frac{20 ft.^3}{45 ft.^3}} \) = 20.3 psi