ASVAB Mechanical Comprehension Practice Test 702476 Results

Your Results Global Average
Questions 5 5
Correct 0 2.93
Score 0% 59%

Review

1

One Horsepower (hp) is equal to how many watts?

76% Answer Correctly

1492

9.8

1

746
Solution

Power is the rate at which work is done, P = w/t, or work per unit time. The watt (W) is the unit for power and is equal to 1 joule (or newton-meter) per second. Horsepower (hp) is another familiar unit of power used primarily for rating internal combustion engines. 1 hp equals 746 watts.

2

Which of the following represents the force a surface exerts when an object presses against it?

49% Answer Correctly

normal force

counter force

mass

friction
Solution

Normal force (FN) represents the force a surface exerts when an object presses against it.

3

A watt is the unit for which of the following?

71% Answer Correctly

mechanical advantage

energy

work

power
Solution

Power is the rate at which work is done, P = w/t, or work per unit time. The watt (W) is the unit for power and is equal to 1 joule (or newton-meter) per second. Horsepower (hp) is another familiar unit of power used primarily for rating internal combustion engines. 1 hp equals 746 watts.

4

If the handles of a wheelbarrow are 3 ft. from the wheel axle, what force must you exert to lift the handles if it's carrying a 270 lb. load concentrated at a point 0.5 ft. from the axle?

56% Answer Correctly

0.83 lbs

45 lbs

810 lbs

90 lbs
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. Plugging in the variables from this problem yields:

Fe x 3 ft. = 270 lbs x 0.5 ft
Fe = 135 ft-lb. / 3 ft 
Fe = 45 lbs

5 What is the power output of a 2 hp engine that's 50% efficient?
40% Answer Correctly
183.3 \( \frac{ft⋅lb}{s} \)
550 \( \frac{ft⋅lb}{s} \)
2200 \( \frac{ft⋅lb}{s} \)
137.5 \( \frac{ft⋅lb}{s} \)
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 2 hp x 550 \( \frac{ft⋅lb}{s} \) = 1100 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{50 \times 1100 \frac{ft⋅lb}{s}}{100} \) \( = \frac{55000 \frac{ft⋅lb}{s}}{100} \) = 550 \( \frac{ft⋅lb}{s} \)