ASVAB Mechanical Comprehension Practice Test 93104 Results

Your Results Global Average
Questions 5 5
Correct 0 2.81
Score 0% 56%

Review

1

What type of load doesn't create specific stress points or vary with time?

60% Answer Correctly

impact load

non-uniformly distributed load

static uniformly distributed load

concentrated 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.

2

The mass of an object correlates to the size of the object but ultimately depends on:

67% Answer Correctly

the object's weight

the object's density

gravity

the object's potential energy
Solution

Mass is a measure of the amount of matter in an object.  In general, larger objects have larger mass than smaller objects but mass ultimately depends on how compact (dense) a substance is.

3 What is the power output of a 3 hp engine that's 95% efficient?
40% Answer Correctly
1567.5 \( \frac{ft⋅lb}{s} \)
391.9 \( \frac{ft⋅lb}{s} \)
4702.5 \( \frac{ft⋅lb}{s} \)
285 \( \frac{ft⋅lb}{s} \)
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 3 hp x 550 \( \frac{ft⋅lb}{s} \) = 1650 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{95 \times 1650 \frac{ft⋅lb}{s}}{100} \) \( = \frac{156750 \frac{ft⋅lb}{s}}{100} \) = 1567.5 \( \frac{ft⋅lb}{s} \)
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 220 lbs. load concentrated at a point 1.5 ft. from the axle?
52% Answer Correctly
660
None of these is correct
110
0
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{220 \times 1.5}{0.5} \)
\( F_e = \frac{330.0}{0.5} \)
\( F_e = 660 \)
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
10.2 psi
20.5 psi
23.5 psi
40.9 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