ASVAB Mechanical Comprehension Practice Test 135486 Results

Your Results Global Average
Questions 5 5
Correct 0 2.77
Score 0% 55%

Review

1 What is the power output of a 7 hp engine that's 40% efficient?
40% Answer Correctly
3080 \( \frac{ft⋅lb}{s} \)
1540 \( \frac{ft⋅lb}{s} \)
385 \( \frac{ft⋅lb}{s} \)
6160 \( \frac{ft⋅lb}{s} \)
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 7 hp x 550 \( \frac{ft⋅lb}{s} \) = 3850 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{40 \times 3850 \frac{ft⋅lb}{s}}{100} \) \( = \frac{154000 \frac{ft⋅lb}{s}}{100} \) = 1540 \( \frac{ft⋅lb}{s} \)
2

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

61% Answer Correctly

high melting point

low corrosive action

chemically stable

low density
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.

3 How much work can a 5 hp engine do in 3 seconds?
52% Answer Correctly
0 ft⋅lb
20 ft⋅lb
1 ft⋅lb
8250 ft⋅lb
Solution
Horsepower (hp) is a common measure of power output for complex machines. By definition, a 1 hp machine does 550 ft⋅lb of work in 1 second: 1 hp = 550 ft⋅lb/s. Substituting the variables for this problem gives us:
\( W = 5 hp \times 550 \frac{ft⋅lb}{s} \times 3s = 8250 ft⋅lb \)
4 If 55 lbs. of force is applied 7 ft. from the fulcrum at the blue arrow and the green box is 3 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?
62% Answer Correctly
256.67 lbs.
128.33 lbs.
513.33 lbs.
42.78 lbs.
Solution

To balance this lever the torques at the green box and the blue arrow must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

Rada = Rbdb

where a represents the green box and b the blue arrow, R is resistance (weight/force) and d is the distance from the fulcrum.

Solving for Ra, our missing value, and plugging in our variables yields:

Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{55 lbs. \times 7 ft.}{3 ft.} \) = \( \frac{385 ft⋅lb}{3 ft.} \) = 128.33 lbs.

5 If the green box weighs 50 lbs. and is 5 ft. from the fulcrum, how much weight would need to be placed at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 2 ft.?
63% Answer Correctly
250 lbs.
31.25 lbs.
375 lbs.
125 lbs.
Solution

To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

Rada = Rbdb

where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.

Solving for Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{50 lbs. \times 5 ft.}{2 ft.} \) = \( \frac{250 ft⋅lb}{2 ft.} \) = 125 lbs.