ASVAB Mechanical Comprehension Practice Test 971356 Results

Your Results Global Average
Questions 5 5
Correct 0 3.16
Score 0% 63%

Review

1

The mechanical advantage of connected gears is proportional to which characteristic of the gears?

73% Answer Correctly

circumference

speed

number of teeth

diameter
Solution

The mechanical advantage (amount of change in speed or torque) of connected gears is proportional to the number of teeth each gear has. Called gear ratio, it's the ratio of the number of teeth on the larger gear to the number of teeth on the smaller gear.  For example, a gear with 12 teeth connected to a gear with 9 teeth would have a gear ratio of 4:3.

2

A a seesaw / teeter-totter is an example of which of the following?

69% Answer Correctly

third-class lever

first-class lever

inclined plane

second-class lever
Solution

A first-class lever is used to increase force or distance while changing the direction of the force. The lever pivots on a fulcrum and, when a force is applied to the lever at one side of the fulcrum, the other end moves in the opposite direction. The position of the fulcrum also defines the mechanical advantage of the lever. If the fulcrum is closer to the force being applied, the load can be moved a greater distance at the expense of requiring a greater input force. If the fulcrum is closer to the load, less force is required but the force must be applied over a longer distance. An example of a first-class lever is a seesaw / teeter-totter.

3

Boyle's law defines the relationship between pressure and volume as:

57% Answer Correctly

\(\frac{P_1}{P_2} = \frac{V_1}{V_2}\)

\({P_1}{P_2} = {V_1}{V_2}\)

\(\frac{P_1}{P_2} = {V_1}{V_2}\)

\(\frac{P_1}{P_2} = \frac{V_2}{V_1}\)
Solution

Boyle's law states that "for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional". Expressed as a formula, that's \(\frac{P_1}{P_2} = \frac{V_2}{V_1}\)

4 If the handles of a wheelbarrow are 0.5 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 120 lbs. load concentrated at a point 1.0 ft. from the axle?
52% Answer Correctly
240
-92.8
120
63.2
Solution
This problem describes a second-class lever and, for a second class lever, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance: Fede = Frdr. In this problem we're looking for effort force:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{120 \times 1.0}{0.5} \)
\( F_e = \frac{120.0}{0.5} \)
\( F_e = 240 \)
5 A = 9 ft., the green box weighs 45 lbs., and the blue box weighs 90 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
1.13 ft.
4.5 ft.
0 ft.
1.5 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:

45 lbs. x 9 ft. = 90 lbs. x dB

dB = \( \frac{45 \times 9 ft⋅lb}{90 lbs.} \) = \( \frac{405 ft⋅lb}{90 lbs.} \) = 4.5 ft.