HESI A2
HESI A2 Physics Practice Test
1. The efficiency (η) of a heat engine is defined as the ratio of the net work done (Wnet) by the engine to the heat input (Qh) from the hot reservoir. The relationship is expressed as:
- A. η = Wnet / Qh
- B. η = Qh / Wnet
- C. η = Wnet x Qh
- D. η = (Wnet + Qh) / 2
Correct answer: A
Rationale: The correct formula for efficiency (η) of a heat engine is η = Wnet / Qh. Efficiency is defined as the ratio of the net work done by the engine (Wnet) to the heat input from the hot reservoir (Qh). This formula shows how effectively the engine converts heat into useful work, making choice A the correct answer. Choices B, C, and D present incorrect relationships between efficiency, net work done, and heat input, leading to their incorrectness.
2. When a junked car is compacted, which statement is true?
- A. Its mass increases.
- B. Its mass decreases.
- C. Its density increases.
- D. Its density decreases.
Correct answer: C
Rationale: When a junked car is compacted, its volume decreases while its mass remains the same. As a result, the car's density increases because density is mass divided by volume. Choice A is incorrect because the mass of the car remains the same. Choice B is incorrect because the mass does not decrease. Choice D is incorrect because the density increases as the volume decreases, not decreases.
3. A 25-cm spring stretches to 28 cm when a force of 12 N is applied. What would its length be if that force were doubled?
- A. 31 cm
- B. 40 cm
- C. 50 cm
- D. 56 cm
Correct answer: A
Rationale: When the 12 N force stretches the spring from 25 cm to 28 cm, it causes a length increase of 28 cm - 25 cm = 3 cm. Therefore, each newton of applied force causes an extension of 3 cm / 12 N = 0.25 cm/N. If the force is doubled to 24 N, the spring would extend by 24 N × 0.25 cm/N = 6 cm more than its original length of 25 cm. Thus, the new length of the spring would be 25 cm + 6 cm = 31 cm. Choice A, 31 cm, is the correct answer as calculated. Choices B, C, and D are incorrect as they do not consider the relationship between force and extension in the spring, leading to incorrect calculations of the new length.
4. A rock has a volume of 6 cm3 and a mass of 24 g. What is its density?
- A. 4 g/cm3
- B. 4 cm3/g
- C. 144 g/cm3
- D. 144 cm3/g
Correct answer: A
Rationale: Density is calculated by dividing the mass of an object by its volume. In this case, the mass of the rock is 24 g and its volume is 6 cm3. By dividing 24 g by 6 cm3, we find that the density of the rock is 4 g/cm3. Choice A is the correct answer because density is expressed in units of mass per unit volume (g/cm3). Choice B is incorrect as it represents the reciprocal of density. Choices C and D are significantly higher values and do not match the calculated density of the rock.
5. A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?
- A. 0.55 amperes
- B. 1.83 amperes
- C. 50 amperes
- D. 6,600 amperes
Correct answer: A
Rationale: To calculate the current being drawn, use the formula I = P / V, where I is the current, P is the power in watts, and V is the voltage. Substituting the given values, I = 60 / 110 ≈ 0.55 amperes. Therefore, the current being drawn by the 60-watt lightbulb is approximately 0.55 amperes. Choice B, 1.83 amperes, is incorrect as it does not match the calculated value. Choices C and D, 50 amperes and 6,600 amperes, are significantly higher values and do not align with the expected current draw of a 60-watt lightbulb powered by a 110-volt source.
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