ASVAB Mechanical Comprehension Practice Test 32711 Results

Your Results Global Average
Questions 5 5
Correct 0 3.32
Score 0% 66%

Review

1

Drag is a type of:

82% Answer Correctly

kinetic energy

work

friction

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.

2

The force exerted on an object due to gravity is called:

70% Answer Correctly

density

mass

weight

potential energy
Solution

Mass is an intrinsic property of matter and does not vary. Weight is the force exerted on the mass of an object due to gravity and a specific case of Newton's Second Law of Motion. Replace force with weight and acceleration with acceleration due to gravity on Earth (g) and the result is the formula for weight: W = mg or, substituting for g, weight equals mass multiplied by 9.8 m/s2.

3

Which of the following is the formula for torque?

61% Answer Correctly

τ = F/r

τ = r/F

τ = rF

τ = F/r2
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).

4 The radius of the axle is 3, the radius of the wheel is 6, and the blue box weighs 45 lbs. What is the effort force necessary to balance the load?
53% Answer Correctly
22.5 lbs.
5 lbs.
3 lbs.
9 lbs.
Solution

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 6 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{6}{3} \) = 2.0

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{45 lbs.}{2.0} \) = 22.5 lbs.

5 A = 9 ft., the green box weighs 10 lbs., and the blue box weighs 15 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
6 ft.
24 ft.
135 ft.
2 ft.
Solution
In order for this lever to balance, the torque acting on side A must equal the torque acting on side B. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

fAdA = fBdB

For this problem, the equation becomes:

10 lbs. x 9 ft. = 15 lbs. x dB

dB = \( \frac{10 \times 9 ft⋅lb}{15 lbs.} \) = \( \frac{90 ft⋅lb}{15 lbs.} \) = 6 ft.