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
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HESI A2

HESI A2 Physics Practice Test

1. 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?

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.

2. Marilyn is driving to a wedding. She drives 4 miles south before realizing that she left the gift at home. She makes a U-turn, returns home to pick up the gift, and sets out again driving south. This time, she drives 1 mile out of her way to pick up a friend. From there, they continue 5 miles more to the wedding. Which of these statements is true about Marilyn’s trip?

Correct answer: C

Rationale: Marilyn’s displacement is calculated based on her final position relative to the starting point. She drives 1 mile to pick up her friend, then 5 miles more to the wedding, totaling 6 miles after returning to her home. So, the correct displacement is 8 miles south from her starting point (4 miles to the gift + 4 miles return + 1 mile to the friend + 5 miles to the wedding). The total distance traveled is 14 miles (adding all the distances). Choice A is incorrect because it miscalculates the displacement. Choice B is incorrect as it overestimates both the displacement and distance traveled. Choice D is incorrect as it underestimates the displacement.

3. A key parameter in fluid selection is specific gravity (SG). For a submerged object in a reference fluid (often water), SG = ρ_object / ρ_reference. An object with SG > 1 will:

Correct answer: A

Rationale: When the specific gravity (SG) of an object is greater than 1, it indicates that the object is denser than the reference fluid, which is often water. According to Archimedes' principle, an object with SG > 1 will experience a net buoyant force acting upwards when submerged in the fluid. This is because the buoyant force is greater than the weight of the object, causing it to float. Therefore, the correct answer is A: 'Experience a net buoyant force acting upwards.' Objects with SG < 1 would sink as they are less dense than the fluid, while objects with SG = 1 would be neutrally buoyant, neither sinking nor floating.

4. If the force acting on an object is doubled, how does its acceleration change?

Correct answer: C

Rationale: According to Newton's second law of motion, the acceleration of an object is directly proportional to the force acting on it. Therefore, if the force acting on an object is doubled, its acceleration will also double. This relationship is expressed by the equation F = ma, where F is the force, m is the mass of the object, and a is the acceleration. When the force (F) is doubled, the acceleration (a) will also double, assuming the mass remains constant. Choice A is incorrect because acceleration changes with a change in force. Choice B is incorrect because acceleration and force are directly proportional. Choice D is incorrect because increasing the force acting on an object does not eliminate its acceleration; instead, it results in an increase in acceleration, as per Newton's second law.

5. A 5-cm candle is placed 20 cm away from a concave mirror with a focal length of 15 cm. About what is the image height of the candle in the mirror?

Correct answer: B

Rationale: The magnification formula for a mirror is given by M = -f / (f - d), where f is the focal length of the mirror, and d is the object distance from the mirror. Using the mirror equation and magnification formula, the image height is found to be negative because it is inverted. Plugging in the values (f = 15 cm, d = 20 cm) into the formula gives M = -15 / (15 - 20) = -15 / -5 = 3. The negative sign indicates that the image is inverted. The image height is then calculated by multiplying the magnification by the object height: 3 * 5 cm = 15 cm. Therefore, the correct image height is approximately -15 cm. Choice A (30.5 cm) and Choice D (-30.5 cm) are incorrect as they do not consider the inversion of the image. Choice C (-15 cm) is also incorrect because it neglects the negative sign, which indicates the inversion of the image.

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