HESI A2
HESI Exams Quizlet Physics
1. The drag force (F_d) experienced by an object moving through a fluid depends on:
- A. Object's shape and size only
- B. Fluid properties and object velocity
- C. Depth of submersion only
- D. Buoyant force acting on the object
Correct answer: B
Rationale: The drag force experienced by an object moving through a fluid depends on multiple factors, including the object's shape, size, velocity, and the fluid's properties such as viscosity and density. Choices A, C, and D are incorrect because drag force is not solely determined by the object's shape and size, depth of submersion, or buoyant force acting on the object. The primary factors affecting drag force are the fluid properties and the object's velocity. Therefore, the correct answer is B.
2. 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.
3. Fluids can be categorized based on their shear stress-strain rate relationship. An ideal fluid exhibits:
- A. Zero shear stress at any strain rate
- B. Linear relationship between shear stress and strain rate (Newtonian)
- C. Non-linear relationship between shear stress and strain rate (Non-Newtonian)
- D. High dependence of viscosity on temperature
Correct answer: A
Rationale: An ideal fluid, often referred to as an inviscid fluid, is a theoretical concept used in fluid mechanics to simplify calculations. It is characterized by having zero shear stress at any strain rate. In reality, such fluids do not exist, but they serve as a useful starting point for understanding fluid behavior in idealized situations. Choice B is incorrect because a linear relationship between shear stress and strain rate defines a Newtonian fluid, not an ideal fluid. Choice C is incorrect because a non-linear relationship between shear stress and strain rate characterizes Non-Newtonian fluids, not ideal fluids. Choice D is incorrect because the high dependence of viscosity on temperature is a characteristic seen in real fluids and does not define an ideal fluid.
4. What is the net force acting on the car?
- A. 450 N
- B. 700 N
- C. 1,500 N
- D. 6,300 N
Correct answer: C
Rationale: To determine the net force acting on an object, we need to consider the sum of the forces acting in the same direction and subtract the forces acting in the opposite direction. In this scenario, there is a force of 4,200 N to the right and a force of 2,700 N to the left. By subtracting the leftward force from the rightward force (4,200 N - 2,700 N), we find that the net force acting on the car is 1,500 N to the right. Therefore, choice C, 1,500 N, is the correct answer. Choice A, 450 N, is too small as it does not account for the total forces involved. Choice B, 700 N, is also incorrect as it is not the result of the correct mathematical operation on the given forces. Choice D, 6,300 N, is too large and does not align with the calculation based on the forces provided.
5. A 5-kg block is suspended from a spring, causing the spring to stretch 10 cm from equilibrium. What is the spring constant for this spring?
- A. 4.9 N/cm
- B. 9.8 N/cm
- C. 49 N/cm
- D. 50 N/cm
Correct answer: C
Rationale: The spring constant (k) can be calculated using Hooke's Law formula: F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from equilibrium. In this case, the force applied is equal to the weight of the block, F = mg, where m = mass of the block = 5 kg and g = acceleration due to gravity = 9.8 m/s^2. The displacement x = 10 cm = 0.1 m. Substituting the values, we have: 5 kg * 9.8 m/s^2 = k * 0.1 m. Solving for k gives k = 5 * 9.8 / 0.1 = 49 N/m. Therefore, the spring constant for this spring is 49 N/cm. Choice A (4.9 N/cm) is incorrect because it is one decimal place lower than the correct answer. Choice B (9.8 N/cm) is incorrect as it does not account for the correct calculation based on the given information. Choice D (50 N/cm) is incorrect because it is slightly higher than the accurate value obtained through the calculations.
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