ASVAB Mechanical Comprehension Practice Test 423088 Results

Your Results Global Average
Questions 5 5
Correct 0 2.98
Score 0% 60%

Review

1

Gear ratio indicates which of the following about two connected gears?

59% Answer Correctly

mechanical advantage

power conversion

efficiency

work done
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 What is the power output of a 1 hp engine that's 65% efficient?
40% Answer Correctly
178.8 \( \frac{ft⋅lb}{s} \)
119.2 \( \frac{ft⋅lb}{s} \)
89.4 \( \frac{ft⋅lb}{s} \)
357.5 \( \frac{ft⋅lb}{s} \)
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 \( \frac{ft⋅lb}{s} \), Pi becomes 1 hp x 550 \( \frac{ft⋅lb}{s} \) = 550 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{65 \times 550 \frac{ft⋅lb}{s}}{100} \) \( = \frac{35750 \frac{ft⋅lb}{s}}{100} \) = 357.5 \( \frac{ft⋅lb}{s} \)
3

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

73% Answer Correctly

circumference

number of teeth

speed

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.

4 A = 4 ft., the green box weighs 5 lbs., and the blue box weighs 20 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
1 ft.
20 ft.
0 ft.
3 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:

5 lbs. x 4 ft. = 20 lbs. x dB

dB = \( \frac{5 \times 4 ft⋅lb}{20 lbs.} \) = \( \frac{20 ft⋅lb}{20 lbs.} \) = 1 ft.

5

Which of the following will increase the mechanical advantage of a second-class lever?

55% Answer Correctly

move the object being lifted closer to the fulcrum

move the object being lifted farther away from the fulcrum

move the fulcrum between the force and the object being lifted

decrease the length of the lever
Solution

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.