| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.56 |
| Score | 0% | 71% |
What defines the mechanical advantage of a first class lever?
output force |
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output distance |
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position of the fulcrum |
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input force |
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.
The force amplification achieved by using a tool, mechanical device or machine system is called:
power |
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mechanical advantage |
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work |
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efficiency |
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.
| 2100 lbs. | |
| 1050 lbs. | |
| 3150 lbs. | |
| 1053 lbs. |
The mechanical advantage (MA) of a block and tackle pulley is equal to the number of times the effort force changes direction. An easy way to count how many times the effort force changes direction is to count the number of ropes that support the resistance which, in this problem, is 10. With a MA of 10, a 105 lbs. effort force could lift 105 lbs. x 10 = 1050 lbs. resistance.
On Earth, acceleration due to gravity (g) is approximately __________.
6.67 x 10-11 m/s2 |
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1 m/s |
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1 m/s2 |
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9.8 m/s2 |
Newton's Law of Univeral Gravitation defines the general formula for the attraction of gravity between two objects: \(\vec{F_{g}} = { Gm_{1}m_{2} \over r^2}\) . In the specific case of an object falling toward Earth, the acceleration due to gravity (g) is approximately 9.8 m/s2.
| 19.3 psi | |
| 20.3 psi | |
| 10.1 psi | |
| 21.8 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