| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.64 |
| Score | 0% | 53% |
Torque involves a perpendicular force applied to a lever arm that moves around a center of rotation. Increasing the length of the lever arm will do which of the following?
increase torque |
|
decrease torque |
|
decrease applied force |
|
increase applied force |
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).
Force of friction due to kinetic friction is __________ the force of friction due to static friction.
higher than |
|
opposite |
|
the same as |
|
lower than |
The formula for force of friction (Ff) is the same whether kinetic or static friction applies: Ff = μFN. To distinguish between kinetic and static friction, μk and μs are often used in place of μ.
A wedge converts force applied to its blunt end into force __________ its inclined surface.
parallel to |
|
opposite to |
|
perpendicular to |
|
along |
The wedge is a moving inclined plane that is used to lift, hold, or break apart an object. A wedge converts force applied to its blunt end into force perpendicular to its inclined surface. In contrast to a stationary plane where force is applied to the object being moved, with a wedge the object is stationary and the force is being applied to the plane. Examples of a wedge include knives and chisels.
Specific gravity is a comparison of the density of an object with the density of:
carbon |
|
air |
|
water |
|
oil |
Specific gravity is the ratio of the density of equal volumes of a substance and water and is measured by a hyrdometer.
| 20.3 psi | |
| 13.5 psi | |
| 14.9 psi | |
| 15.5 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 = 30 ft.3, V1 = 45 ft.3 and P1 = 9.0 psi. Solving for P2:
P2 = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{9.0 psi}{\frac{30 ft.^3}{45 ft.^3}} \) = 13.5 psi