a 0 kg block on a table is given a push so that it slides along the table if the block is accelerated at 6 ms2 what was the force applied to the block
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HESI A2

HESI Exams Quizlet Physics

1. A 0-kg block on a table is given a push so that it slides along the table. If the block is accelerated at 6 m/s2, what was the force applied to the block?

Correct answer: A

Rationale: According to Newton's second law of motion, F=ma. Since the block has a mass of 0 kg, the force applied must be 0 N, as no force is needed to move an object with zero mass.

2. A common example of a shear-thinning (non-Newtonian) fluid is:

Correct answer: B

Rationale: The correct answer is B: Ketchup. Shear-thinning fluids become less viscous under stress. Ketchup is an example of a shear-thinning fluid because its viscosity decreases when it is shaken or squeezed, allowing it to flow more easily. Choice A, Water, is a Newtonian fluid with a constant viscosity regardless of stress. Choice C, Air, is also a Newtonian fluid. Choice D, Alcohol, does not exhibit shear-thinning behavior; it typically has a constant viscosity as well.

3. An incandescent lamp consumes 60 Joules of energy per second. What is the power rating of this lamp?

Correct answer: B

Rationale: Power is defined as energy consumed per unit time. If the lamp consumes 60 Joules of energy per second, the power rating is 60 Watts. Therefore, choice B is correct. Choice A ('1 Watt') is incorrect because the lamp consumes 60 Joules per second, not 1 Joule per second. Choice C ('1/60 Joules') is incorrect as it does not represent the power rating. Choice D ('Impossible to determine without knowing the voltage') is incorrect because power can be calculated using energy consumption per unit time without needing to know the voltage.

4. 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?

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.

5. A system undergoes an isobaric process (constant pressure). In this process, the work done (W) by the system is:

Correct answer: B

Rationale: In an isobaric process (constant pressure), the work done is given by the formula W = PΔV, where P is the pressure and ΔV is the change in volume. If the volume does not change, the work done is zero, not negative. Choice A is incorrect as it states the work done is zero when the volume change is zero, which is the correct condition for zero work. Choice C is incorrect as it incorrectly suggests that the work done is negative in an isobaric process. Choice D is incorrect as the work done in an isobaric process is indeed dependent on the volume change and pressure.

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