ASVAB Mechanical Comprehension Practice Test 391797 Results

Your Results Global Average
Questions 5 5
Correct 0 3.06
Score 0% 61%

Review

1 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 130 lbs. load concentrated at a point 1.0 ft. from the axle?
52% Answer Correctly
260
None of these is correct
0
58
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{130 \times 1.0}{0.5} \)
\( F_e = \frac{130.0}{0.5} \)
\( F_e = 260 \)
2

Normal force is generally equal to the __________ of an object.

61% Answer Correctly

density

coefficient of friction

weight

mass
Solution

Normal force arises on a flat horizontal surface in response to an object's weight pressing it down. Consequently, normal force is generally equal to the object's weight.

3 If you lift a 22 lbs. rock 20 ft. from the ground, how much work have you done?
70% Answer Correctly
220 ft⋅lb
880 ft⋅lb
None of these is correct
440 ft⋅lb
Solution
Work is force times distance. In this case, the force is the weight of the rock so:
\( W = F \times d \)
\( W = 22 \times 20 \)
\( W = 440 \)
4

What type of load is sudden and for a relatively short duration?

69% Answer Correctly

impact load

dynamic load

concentrated load

non-uniformly distributed load
Solution

A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.

5 A mass of air has a pressure of 15.0 psi and a volume of 75 ft.3. If the air is compressed to a new volume of 55 ft.3, what is the new pressure?
56% Answer Correctly
20.5 psi
18.4 psi
22 psi
29.5 psi
Solution

According to Boyle's Law, pressure and volume are inversely proportional:

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

In this problem, V2 = 55 ft.3, V1 = 75 ft.3 and P1 = 15.0 psi. Solving for P2:

P2 = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{15.0 psi}{\frac{55 ft.^3}{75 ft.^3}} \) = 20.5 psi