HESI A2
HESI A2 Physics Practice Test
1. What force was applied to the object that was moved if 100 Nâ‹…m of work is done over 20 m?
- A. 5 N
- B. 80 N
- C. 120 N
- D. 2,000 N
Correct answer: A
Rationale: Work is calculated using the formula Work = Force x Distance. Given that 100 Nâ‹…m of work is done over 20 m, we can rearrange the formula to solve for Force. Force = Work / Distance. Plugging in the values, we get Force = 100 Nâ‹…m / 20 m = 5 N. Therefore, the force applied to the object that was moved is 5 N. Choice B (80 N) is incorrect because it doesn't match the calculated force of 5 N. Choice C (120 N) is incorrect as it is higher than the calculated force. Choice D (2,000 N) is incorrect as it is significantly higher than the correct force of 5 N.
2. An object with a charge of 3 μC is placed 30 cm from another object with a charge of 2 μC. What is the magnitude of the resulting force between the objects?
- A. 0.6 N
- B. 0.18 N
- C. 180 N
- D. 9 × 10−12 N
Correct answer: B
Rationale: To find the magnitude of the resulting force between two charges, we use Coulomb's Law: F = k × (|q1 × q2|) / r² Where: F is the force k is Coulomb’s constant (8.99 × 10â¹ N·m²/C²) q1 and q2 are the charges r is the distance between the charges Plugging in the values: F = (8.99 × 10â¹) × (3 × 10â»â¶) × (2 × 10â»â¶) / (0.3)² = 0.18 N. Therefore, the magnitude of the resulting force is 0.18 N.
3. A 110-volt appliance draws 0 amperes. How many watts of power does it require?
- A. 0 watts
- B. 108 watts
- C. 112 watts
- D. 220 watts
Correct answer: A
Rationale: When a 110-volt appliance draws 0 amperes, it means that the power consumption is zero as well. The formula to calculate power is P = V x I, where P is power in watts, V is voltage in volts, and I is current in amperes. Since the current is 0 amperes, the power required by the appliance is also 0 watts. Therefore, the correct answer is 0 watts. Choice B, 108 watts, is incorrect because there is no current drawn. Choice C, 112 watts, and choice D, 220 watts, are incorrect as well since the appliance is not consuming any power when drawing 0 amperes.
4. When a fluid encounters a bluff body (e.g., a car), the flow can separate behind the object, creating a region of low pressure. This phenomenon is known as:
- A. Cavitation
- B. Boundary layer separation
- C. Bernoulli effect per se
- D. Drag crisis
Correct answer: B
Rationale: The correct answer is B: Boundary layer separation. Boundary layer separation occurs when the flow of fluid detaches from the surface of a bluff body, leading to a low-pressure region behind the object. This separation creates a wake region with reduced pressure. Choice A, Cavitation, refers to the formation of vapor bubbles in a fluid and is not relevant in this context. Choice C, Bernoulli effect per se, does not specifically describe the phenomenon of flow separation behind a bluff body. Choice D, Drag crisis, is not the term used to describe the creation of a low-pressure region due to flow separation.
5. Certain non-Newtonian fluids exhibit shear thickening behavior. In this case, the fluid's viscosity:
- A. Remains constant with increasing shear rate
- B. Decreases with increasing shear rate (shear thinning)
- C. Increases with increasing shear rate
- D. Depends solely on the applied pressure
Correct answer: C
Rationale: When a non-Newtonian fluid exhibits shear thickening behavior, its viscosity increases with increasing shear rate. This means that as more force is applied to the fluid, its resistance to flow also increases, resulting in a higher viscosity. This phenomenon is opposite to shear thinning, where viscosity decreases with increasing shear rate. Therefore, in the case of shear thickening behavior, the correct answer is that the fluid's viscosity increases with increasing shear rate. Choices A, B, and D are incorrect because shear thickening behavior specifically involves an increase in viscosity with increasing shear rate, not remaining constant, decreasing, or depending on applied pressure.
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