Free ASVAB Practice Tests
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1,557 Questions, Problems & Flash CardsHuge database of 668 multiple-choice questions, 135 math and algebra problems, and 754 flash cards to help you prepare for the ASVAB.
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Sample Practice Test Questions
Which of the following is the preferred method for splicing wires?
soldering
Splicing is permanently joining two wires together. Splicing can be done with screw-on connectors or wire crimps but the preferred method for splicing is soldering. Soldering takes the most effort but results in a connection that is electrically and mechanically identical to the original wire.
If the green box weighs 20 lbs. and is 1 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 8 ft.?
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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{20 lbs. \times 1 ft.}{8 ft.} \) = \( \frac{20 ft⋅lb}{8 ft.} \) = 2.5 lbs.
Opposed cylinder designs feature two rows of cylinders that are placed at what angle to each other?
180°
Opposed cylinder designs place cylinders flat in two rows 180° opposite each other with a single crankshaft running between them.
If angle a = 68° and angle b = 39° what is the length of angle c?
The sum of the interior angles of a triangle is 180°:
180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 68° - 39° = 73°
A joint created by soldering is:
weaker than a weld
A joint created by soldering is not as strong as a weld and not as strong as the original metal.
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?
45 lbs
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