ASVAB Mechanical Comprehension Practice Test 664119 Results

Your Results Global Average
Questions 5 5
Correct 0 2.99
Score 0% 60%

Review

1

The mechanical advantage of a wheel and axle is equal to the:

61% Answer Correctly

length of the axle

difference in the lengths of the axles

ratio of the diameters of the wheels

difference in the diameters of the wheels
Solution

A wheel and axle uses two different diameter wheels mounted to a connecting axle. Force is applied to the larger wheel and large movements of this wheel result in small movements in the smaller wheel. Because a larger movement distance is being translated to a smaller distance, force is increased with a mechanical advantage equal to the ratio of the diameters of the wheels. An example of a wheel and axle is the steering wheel of a car.

2 If the handles of a wheelbarrow are 0.5 ft. from the wheel axle and the load is concentrated at a point 1.5 ft. from the axle, how many pounds of load will a 130 lbs. force lift?
47% Answer Correctly
-80
43.3
249.6
195
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 resistance force:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{130 \times 0.5}{1.5} \)
\( F_r = \frac{65.0}{1.5} \)
\( F_r = 43.3 \)
3 What's the mechanical advantage of a wedge that's 3 inches wide and 15 inches long?
83% Answer Correctly
4.5
5
8
3
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

4 If the handles of a wheelbarrow are 2.0 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 270 lbs. load concentrated at a point 1.5 ft. from the axle?
52% Answer Correctly
219.5
23.1
202.5
22
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{270 \times 1.5}{2.0} \)
\( F_e = \frac{405.0}{2.0} \)
\( F_e = 202.5 \)
5

Specific gravity is a comparison of the density of an object with the density of:

57% Answer Correctly

carbon

water

oil

air
Solution

Specific gravity is the ratio of the density of equal volumes of a substance and water and is measured by a hyrdometer.