| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.83 |
| Score | 0% | 57% |
Which of the following is the formula for torque?
τ = F/r2 |
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τ = rF |
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τ = r/F |
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τ = F/r |
Torque measures force applied during rotation: τ = rF. Torque (τ, the Greek letter tau) = the radius of the lever arm (r) multiplied by the force (F) applied. Radius is measured from the center of rotation or fulcrum to the point at which the perpendicular force is being applied. The resulting unit for torque is newton-meter (N-m) or foot-pound (ft-lb).
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 |
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density, volume |
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pressure, density |
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pressure, volume |
Boyle's law states that "for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional".
Which of these will have the most impact on the kinetic energy of an object?
its speed |
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its direction |
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its weight |
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its mass |
Kinetic energy is the energy of movement and is a function of the mass of an object and its speed: \(KE = {1 \over 2}mv^2\) where m is mass in kilograms, v is speed in meters per second, and KE is in joules. The most impactful quantity to kinetic energy is velocity as an increase in mass increases KE linearly while an increase in speed increases KE exponentially.
Which class of lever offers no mechanical advantage?
third |
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first |
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second |
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none of these, all levers offer mechanical advantage |
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.
| 16 psi | |
| 18 psi | |
| 12 psi | |
| 10.8 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 = 15 ft.3, V1 = 30 ft.3 and P1 = 6.0 psi. Solving for P2:
P2 = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{6.0 psi}{\frac{15 ft.^3}{30 ft.^3}} \) = 12 psi