HESI A2
HESI A2 Physics
1. Which vehicle has the greatest momentum?
- A. A 9,000-kg railroad car traveling at 3 m/s
- B. A 2,000-kg automobile traveling at 24 m/s
- C. A 1,500-kg MINI Coupe traveling at 29 m/s
- D. A 500-kg glider traveling at 89 m/s
Correct answer: D
Rationale: The momentum of an object is calculated by multiplying its mass by its velocity. The momentum formula is p = m × v, where p is momentum, m is mass, and v is velocity. Comparing the momentum of each vehicle: A: 9,000 kg × 3 m/s = 27,000 kg·m/s B: 2,000 kg × 24 m/s = 48,000 kg·m/s C: 1,500 kg × 29 m/s = 43,500 kg·m/s D: 500 kg × 89 m/s = 44,500 kg·m/s. Therefore, the glider (500-kg) traveling at 89 m/s has the greatest momentum of 44,500 kg·m/s, making it the correct choice. Options A, B, and C have lower momentum values compared to option D, proving that the 500-kg glider traveling at 89 m/s has the highest momentum among the given vehicles.
2. Enthalpy (H) is a thermodynamic property defined as the sum of a system's internal energy (U) and the product of its pressure (P) and volume (V). The relationship between these is:
- A. H = U + PV
- B. H = U - PV
- C. H = U / PV
- D. H = PV / U
Correct answer: A
Rationale: Enthalpy (H) is defined as H = U + PV, where U represents internal energy, P is pressure, and V is volume. Enthalpy includes both the internal energy of a system and the energy required to create space for the system against an external pressure. Therefore, the correct relationship between enthalpy, internal energy, pressure, and volume is H = U + PV. Choice B is incorrect as subtracting PV would not account for the work done against pressure. Choice C is incorrect as dividing U by PV doesn't represent the definition of enthalpy. Choice D is incorrect as dividing PV by U is not the correct relationship based on the definition of enthalpy.
3. Fluid dynamics is a subfield of fluid mechanics concerned with:
- A. Equilibrium properties of fluids at rest (Fluid Statics)
- B. The motion and behavior of fluids under various conditions
- C. Phase transitions of fluids between liquid, gas, and solid states
- D. Engineering applications of fluids (related but broader than fluid dynamics)
Correct answer: B
Rationale: Fluid dynamics is the study of fluids in motion and their behavior under different conditions, including how they flow, mix, and interact with their surroundings. It focuses on the dynamic aspects of fluids rather than their static properties when at rest, which is the realm of fluid statics. Phase transitions of fluids between liquid, gas, and solid states are more related to thermodynamics than fluid dynamics. While engineering applications involve fluid dynamics, the field itself is more specialized in studying the movement and behavior of fluids.
4. Amanda uses 100 N of force to push a lawnmower around her lawn. If she mows 20 rows measuring 30 meters each, how much work does she do?
- A. 3,000 N⋅m
- B. 6,000 N⋅m
- C. 60,000 N⋅m
- D. The answer cannot be determined from the information given.
Correct answer: C
Rationale: The work done by Amanda pushing the lawnmower is calculated by multiplying the force applied (100 N) by the distance over which the force is applied (the total distance mowed). Since Amanda mows 20 rows, each measuring 30 meters, the total distance mowed is 20 rows x 30 meters/row = 600 meters. Therefore, the work done is 100 N x 600 m = 60,000 N⋅m. Option A and B are incorrect as they do not account for the total distance mowed. Option D is incorrect as the work done can be accurately calculated based on the information provided.
5. The specific heat capacity of tin is 217 J/(g°C). Which of these materials would require about twice as much heat as tin to increase the temperature of a sample by 1°C?
- A. Copper [0.3844 J/(g°C)]
- B. Iron [0.449 J/(g°C)]
- C. Gold [0.1291 J/(g°C)]
- D. Aluminum [0.904 J/(g°C)]
Correct answer: D
Rationale: The correct answer is D: Aluminum. The specific heat capacity of aluminum is 0.904 J/(g°C), which is approximately 4 times that of tin. For a material to require about twice as much heat as tin to increase the temperature by 1°C, it should have a specific heat capacity roughly double that of tin. Therefore, aluminum fits this criterion better than the other options. Gold has a much lower specific heat capacity than tin, so it would require less, not more, heat to increase the temperature by 1°C. Copper and Iron also have specific heat capacities lower than tin, making them incorrect choices for requiring twice as much heat as tin.
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