ASVAB Mechanical Comprehension Practice Test 337086 Results

Your Results Global Average
Questions 5 5
Correct 0 2.92
Score 0% 58%

Review

1

Which of the following surfaces would have the lowest coefficient of friction?

85% Answer Correctly

tile

ice

leather

concrete


Solution

Coefficient of friction (μ) represents how much two materials resist sliding across each other.  Smooth surfaces like ice have low coefficients of friction while rough surfaces like concrete have high μ.


2

Force of friction due to kinetic friction is __________ the force of friction due to static friction.

40% Answer Correctly

higher than

the same as

opposite

lower than


Solution

The formula for force of friction (Ff) is the same whether kinetic or static friction applies: Ff = μFN. To distinguish between kinetic and static friction, μk and μs are often used in place of μ.


3

Which of the following is not a characteristic of a ceramic?

61% Answer Correctly

high melting point

low density

low corrosive action

chemically stable


Solution

Ceramics are mixtures of metallic and nonmetallic elements that withstand exteme thermal, chemical, and pressure environments. They have a high melting point, low corrosive action, and are chemically stable. Examples include rock, sand, clay, glass, brick, and porcelain.


4

Which of the following represents how much two materials resist sliding across each other?

53% Answer Correctly

static friction

kinetic friction

coefficient of friction

normal friction


Solution

Coefficient of friction (μ) represents how much two materials resist sliding across each other.  Smooth surfaces like ice have low coefficients of friction while rough surfaces like concrete have high μ.


5 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 25 ft.3, what is the new pressure?
56% Answer Correctly
24.3 psi
18.2 psi
14.6 psi
16.2 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 = 25 ft.3, V1 = 45 ft.3 and P1 = 9.0 psi. Solving for P2:

P2 = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{9.0 psi}{\frac{25 ft.^3}{45 ft.^3}} \) = 16.2 psi