| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.87 |
| Score | 0% | 57% |
Which class of lever is used to increase force on an object in the same direction as the force is applied?
third |
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first |
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all of these |
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second |
A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.
| 0 \( \frac{ft⋅lb}{s} \) | |
| 130.6 \( \frac{ft⋅lb}{s} \) | |
| 522.5 \( \frac{ft⋅lb}{s} \) | |
| 1045 \( \frac{ft⋅lb}{s} \) |
A shovel is an example of which class of lever?
third |
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a shovel is not a lever |
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second |
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first |
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.
What is the first step to solving a problem where multiple forces are acting on an object?
calculate kinetic energy |
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calculate the net force |
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calculate the total force |
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calculate potential energy |
In mechanics, multiple forces are often acting on a particular object and, taken together, produce the net force acting on that object. Like force, net force is a vector quantity in that it has magnitude and direction.
Hydraulics is the transmission of force through the use of which of the following?
liquids |
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torque |
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air pressure |
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gear systems |
Hydraulics is the transmission of force through the use of liquids. Liquids are especially suited for transferring force in complex machines because they compress very little and can occupy very small spaces. Hydraulic pressure is calculated by dividing force by the area over which it is applied: P = F/A where F is force in pounds, A is area in square inches, and the resulting pressure is in pounds per square inch (psi).