ASVAB Mechanical Comprehension Practice Test 282103 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.59
Score	0%	52%

Review

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?

54% Answer Correctly

	decrease applied force
	increase applied force
	decrease torque
	increase torque

Solution

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).

Depending on where you apply effort and resistance, the wheel and axle can multiply:

45% Answer Correctly

	force or speed
	force or distance
	speed or power
	power or distance

Solution

If you apply the resistance to the axle and the effort to the wheel, the wheel and axle will multiply force and if you apply the resistance to the wheel and the effort to the axle, it will multiply speed.

3 What is the power output of a 1 hp engine that's 25% efficient?

39% Answer Correctly

	137.5 \( \frac{ft⋅lb}{s} \)
	0 \( \frac{ft⋅lb}{s} \)
	275 \( \frac{ft⋅lb}{s} \)
	25 \( \frac{ft⋅lb}{s} \)

Solution

\( Efficiency = \frac{Power_{out}}{Power_{in}} \times 100 \)
Solving for power out: \( P_{o} = \frac{E \times P_{i}}{100} \)
Knowing that 1 hp = 550 \( \frac{ft⋅lb}{s} \), P_i becomes 1 hp x 550 \( \frac{ft⋅lb}{s} \) = 550 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{25 \times 550 \frac{ft⋅lb}{s}}{100} \) \( = \frac{13750 \frac{ft⋅lb}{s}}{100} \) = 137.5 \( \frac{ft⋅lb}{s} \)

Drag is a type of:

82% Answer Correctly

	friction
	work
	kinetic energy
	potential energy

Solution

Drag is friction that opposes movement through a fluid like liquid or air. The amount of drag depends on the shape and speed of the object with slower objects experiencing less drag than faster objects and more aerodynamic objects experiencing less drag than those with a large leading surface area.

The advantage of using a third-class lever is that it increases:

37% Answer Correctly

	the speed of the load
	the distance traveled by the load
	the force applied to the load
	the mechanical advantage of the lever

Solution

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.