| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.79 |
| Score | 0% | 56% |
| 3 | |
| 4 | |
| -1 | |
| 1.33 |
The mechanical advantage of a wheel and axle is the input radius divided by the output radius:
MA = \( \frac{r_i}{r_o} \)
In this case, the input radius (where the effort force is being applied) is 4 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{4}{3} \) = 1.33
| 90 lbs. | |
| 7.5 lbs. | |
| 1.88 lbs. | |
| 2 lbs. |
To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:
Rada = Rbdb
where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.Solving for Rb, our missing value, and plugging in our variables yields:
Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{15 lbs. \times 3 ft.}{6 ft.} \) = \( \frac{45 ft⋅lb}{6 ft.} \) = 7.5 lbs.
| 1.33 | |
| 1 | |
| 0.75 | |
| 4 |
The mechanical advantage of a wheel and axle is the input radius divided by the output radius:
MA = \( \frac{r_i}{r_o} \)
In this case, the input radius (where the effort force is being applied) is 3 and the output radius (where the resistance is being applied) is 4 for a mechanical advantage of \( \frac{3}{4} \) = 0.75
According to Boyle's law, for a fixed amount of gas kept at a fixed temperature, which of the following are inversely proportional?
volume, mass |
|
density, volume |
|
pressure, volume |
|
pressure, density |
Boyle's law states that "for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional".
| 175 ft. | |
| 2.62 ft. | |
| 0 ft. | |
| 7.86 ft. |
To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:
Rada = Rbdb
where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.Solving for da, our missing value, and plugging in our variables yields:
da = \( \frac{R_bd_b}{R_a} \) = \( \frac{55 lbs. \times 5 ft.}{35 lbs.} \) = \( \frac{275 ft⋅lb}{35 lbs.} \) = 7.86 ft.