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
Solving for power out: \( P_{o} = \frac{E \times P_{i}}{100} \)
Knowing that 1 hp = 550 \( \frac{ft⋅lb}{s} \), Pi becomes 9 hp x 550 \( \frac{ft⋅lb}{s} \) = 4950 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{85 \times 4950 \frac{ft⋅lb}{s}}{100} \) \( = \frac{420750 \frac{ft⋅lb}{s}}{100} \) = 4207.5 \( \frac{ft⋅lb}{s} \)
Which of the following would you find in the side view mirror of a car?
convex mirror
A concave (or converging) mirror bulges inward and focuses reflected light on the mirror's focal point where the mirror's angles of incidence converge. In contrast, a convex (or diverging) mirror bulges outward and diffuses the light waves that strike it. A common use of a concave mirror is in a reflecting telescope, a common use of a convex mirror is in the side view mirror of a car.
Which of these is matter than cannot be separated into different types of matter by ordinary chemical methods?
element
An element is matter than cannot be separated into different types of matter by ordinary chemical methods.
Reproductive haploids are also known as:
gametes
Reproductive (haploid) cells known as gametes have half as many (23) pairs of chromosomes as normal (diploid) cells. When the male gamete (sperm) combines with the female gamete (ovum) through meiosis to form a zygote, each gamete supplies half the chromosomes needed to form the normal diploid cells.
A subatomic particle that orbits the nucleus of an atom. It carries a negative electric charge. This desribes which of the following?
electron
An electron is a subatomic particle that orbits the nucleus of an atom. It carries a negative electric charge. Generally, an atom has the same number of negative electrons orbiting the nucleus as it does positive protons inside.
If the green box weighs 55 lbs. and is 3 ft. from the fulcrum, how far from the fulcrum would a 70 lbs. weight need to be placed to balance the lever?
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 db, our missing value, and plugging in our variables yields:
db = \( \frac{R_ad_a}{R_b} \) = \( \frac{55 lbs. \times 3 ft.}{70 lbs.} \) = \( \frac{165 ft⋅lb}{70 lbs.} \) = 2.36 ft.
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 = 10 ft.3, V1 = 30 ft.3 and P1 = 6.0 psi. Solving for P2:
P2 = \( \frac{P_1}{\frac{V_2}{V_1}} \) = \( \frac{6.0 psi}{\frac{10 ft.^3}{30 ft.^3}} \) = 18 psi
If the green box weighs 20 lbs. and is 1 ft. from the fulcrum, how far from the fulcrum would a 70 lbs. force need to be applied to balance the lever?
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 db, our missing value, and plugging in our variables yields:
db = \( \frac{R_ad_a}{R_b} \) = \( \frac{20 lbs. \times 1 ft.}{70 lbs.} \) = \( \frac{20 ft⋅lb}{70 lbs.} \) = 0.29 ft.