ASVAB Mechanical Comprehension Practice Test 494719 Results

Your Results Global Average
Questions 5 5
Correct 0 3.14
Score 0% 63%

Review

1 A mass of air has a pressure of 9.0 psi and a volume of 45 ft.3. If the air is compressed to a new volume of 30 ft.3, what is the new pressure?
56% Answer Correctly
40.5 psi
12.2 psi
13.5 psi
15 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 = 30 ft.3, V1 = 45 ft.3 and P1 = 9.0 psi. Solving for P2:

P2 = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{9.0 psi}{\frac{30 ft.^3}{45 ft.^3}} \) = 13.5 psi

2

Friction between two or more solid objects that are not moving relative to each other is called:

74% Answer Correctly

gravitational friction

dynamic friction

static friction

kinetic friction
Solution

Static friction is friction between two or more solid objects that are not moving relative to each other. An example is the friction that prevents a box on a sloped surface from sliding farther down the surface.

3 How much resistance could a 120 lb. effort force lift using a block and tackle pulley that has 8 ropes supporting the resistance?
82% Answer Correctly
960 lbs.
480 lbs.
1920 lbs.
320 lbs.
Solution

The mechanical advantage (MA) of a block and tackle pulley is equal to the number of times the effort force changes direction. An easy way to count how many times the effort force changes direction is to count the number of ropes that support the resistance which, in this problem, is 8. With a MA of 8, a 120 lbs. effort force could lift 120 lbs. x 8 = 960 lbs. resistance.

4 The radius of the axle is 3, the radius of the wheel is 8, and the blue box weighs 100 lbs. What is the effort force necessary to balance the load?
52% Answer Correctly
24 lbs.
8.01 lbs.
10.67 lbs.
37.45 lbs.
Solution

The mechanical advantage of a wheel and axle is the input radius divided by the output radius:

MA = \( \frac{r_i}{r_o} \)

In this case, the input radius (where the effort force is being applied) is 8 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{8}{3} \) = 2.67

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{100 lbs.}{2.67} \) = 37.45 lbs.

5 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 110 lbs. load concentrated at a point 0.5 ft. from the axle?
52% Answer Correctly
68.5
110
-72
None of these is correct
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{110 \times 0.5}{0.5} \)
\( F_e = \frac{55.0}{0.5} \)
\( F_e = 110 \)