Free ASVAB Practice Tests

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Sample Practice Test Questions

General Science

When two air masses meet and neither is displaced, what kind of front is present?

stationary

When two air masses meet and neither is displaced, a stationary front is created. Stationary fronts often cause persistent cloudy wet weather.

Shop Information

When using a hacksaw, blades with __________ teeth are more appropriate for cutting thinner materials.

more

A hacksaw has replaceable blades and is used to cut metal. The blade type is chosen based on the material that is to be cut. Blades with larger numbers of teeth per inch are more appropriate for cutting thinner materials.

Automotive Information

Power brakes multiply the force a driver applies to the brake pedal using a __________ connected to the engine intake manifold.

vacuum booster

Power brakes multiply the force a driver applies to the brake pedal using a vacuum booster connected to the engine intake manifold. This provides for much higher hydraulic pressure in the braking system than could be generated by the driver alone. Antilock brakes (ABS) use speed sensors and adjust the brake pressure at each wheel to prevent skidding and allow the driver more steering control in slippery conditions.

Automotive Information

When a car engine is running, what provides electrical current to recharge the battery and power the electrical system?

alternator

Once the engine is running, the alternator provides electrical current to recharge the battery and power the electrical system. The alternator is driven by the engine's crankshaft and produces alternating current (AC) which is then fed through a rectifier bridge to convert it to the direct current (DC) required by the electrical system. A voltage regulator controls the output of the alternator to maintain a consistent voltage (approx. 14.5 volts) in the electrical system regardless of load.

Automotive Information

Engine oil viscosity is rated using the format XW-XX with the number preceding the W indicating viscosity at __________ ℉ and the XX indicating viscosity at __________ ℃.

0, 100

The primary component of the lubrication system is engine oil. Engines require oil blends with different thickness (viscosity) and additives depending on their operating conditions. Viscosity is rated using the format XW-XX with the number preceding the W (winter) rating the oil’s viscosity at 0 ℉ (-17.8 ℃) and the XX indicating viscosity at 100 ℃.

Shop Information

Measuring tools provide a convenient and easy way to compare object dimensions against:

a standard

Measuring tools give easy access to the wide variety of standard dimensions that are used to categorize and classify objects.

General Science

Force is measured in newtons (N) with 1 N being the force required to impart an acceleration of:

1 m/s2 to a mass of 1 kg

Weight is a force that describes the attraction of gravity on an object. Force is measured in newtons (N) with 1 N being the force required to impart an acceleration of 1 m/s2 to a mass of 1 kg.

General Science

A large naturally occurring community of flora and fauna occupying a major habitat is:

a biome

A biome is a large naturally occurring community of flora (plants) and fauna (animals) occupying a major habitat (home or environment).

General Science

Water freezing or boiling is an example of which of the following?

phase transition

A substance undergoes a phase transition when it moves from one state of matter to another, for example, when water freezes or boils.

Mechanical Comprehension
48.89 lbs. If A = 10 ft., B = 1 ft., C = 9 ft., the green box weighs 50 lbs. and the blue box weighs 60 lbs., what does the orange box have to weigh for this lever to balance?
48.89 lbs.
In order for this lever to balance, the torque acting on each side of the fulrum must be equal. So, the torque produced by A must equal the torque produced by B and C. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

fAdA = fBdB + fCdC

For this problem, this equation becomes:

50 lbs. x 10 ft. = 60 lbs. x 1 ft. + fC x 9 ft.

500 ft. lbs. = 60 ft. lbs. + fC x 9 ft.

fC = \( \frac{500 ft. lbs. - 60 ft. lbs.}{9 ft.} \) = \( \frac{440 ft. lbs.}{9 ft.} \) = 48.89 lbs.