ASVAB Mechanical Comprehension Practice Test 10132 Results

Your Results Global Average
Questions 5 5
Correct 0 3.13
Score 0% 63%

Review

1 What's the mechanical advantage of a wedge that's 3 inches wide and 15 inches long?
83% Answer Correctly
4.5
5
7.5
6.5
Solution

The mechanical advantage (MA) of a wedge is its length divided by its thickness:

MA = \( \frac{l}{t} \) = \( \frac{15 in.}{3 in.} \) = 5

2 What is the power output of a 6 hp engine that's 70% efficient?
39% Answer Correctly
577.5 \( \frac{ft⋅lb}{s} \)
4620 \( \frac{ft⋅lb}{s} \)
2310 \( \frac{ft⋅lb}{s} \)
0 \( \frac{ft⋅lb}{s} \)
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 6 hp x 550 \( \frac{ft⋅lb}{s} \) = 3300 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{70 \times 3300 \frac{ft⋅lb}{s}}{100} \) \( = \frac{231000 \frac{ft⋅lb}{s}}{100} \) = 2310 \( \frac{ft⋅lb}{s} \)
3

Potential energy is energy that has the potential to be converted into what?

80% Answer Correctly

power

heat

work

 kinetic energy
Solution

Potential energy is the energy of an object by virtue of its position relative to other objects. It is energy that has the potential to be converted into kinetic energy.

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 0.5 ft. from the axle?
52% Answer Correctly
120
0
103.1
60
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{120 \times 0.5}{0.5} \)
\( F_e = \frac{60.0}{0.5} \)
\( F_e = 120 \)
5

Which of the following is the formula for hydraulic pressure?

58% Answer Correctly

P = F/A

P = FA

P = F/A2

P = FA2
Solution

Hydraulics is the transmission of force through the use of liquids. Liquids are especially suited for transferring force in complex machines because they compress very little and can occupy very small spaces. Hydraulic pressure is calculated by dividing force by the area over which it is applied: P = F/A where F is force in pounds, A is area in square inches, and the resulting pressure is in pounds per square inch (psi).