Nursing Elites

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?

• A. 31 cm
• B. 40 cm
• C. 50 cm
• D. 56 cm

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. 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:

• A. Experience a net buoyant force acting upwards
• B. Experience a net buoyant force acting downwards
• C. Remain neutrally buoyant
• D. Require knowledge of the object's volume for buoyancy determination

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.

3. What is the SI unit for quantifying the transfer of energy due to an applied force?

• A. Newton (N)
• B. Meter per second (m/s)
• C. Joule (J)
• D. Kilogram (kg)

Correct answer: C

Rationale: The correct answer is C: Joule (J). The joule is the SI unit used to quantify the transfer of energy due to an applied force. It is defined as the work done when a force of one newton is applied over a distance of one meter. Newton (N) is the unit of force, not energy transfer. Meter per second (m/s) is the unit of speed, not energy transfer. Kilogram (kg) is the unit of mass, not energy transfer. Therefore, the correct unit for quantifying the transfer of energy due to an applied force is the joule (J).

4. In physics, the relationship between acceleration and force is expressed in ___________.

• A. Newton’s first law of motion
• B. Newton’s second law of motion
• C. Newton’s third law of motion
• D. none of Newton’s laws of motion

Correct answer: B

Rationale: The relationship between acceleration and force is expressed in Newton’s second law of motion. This law states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object's mass. Mathematically, this relationship is represented as F = ma, where F is the force, m is the mass of the object, and a is the acceleration. Choice A, Newton’s first law of motion, also known as the law of inertia, states that an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an external force. Choice C, Newton’s third law of motion, states that for every action, there is an equal and opposite reaction, focusing on the interaction between two objects. Choice D is incorrect because the relationship between acceleration and force is indeed described by one of Newton’s laws of motion, specifically the second law.

5. A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?

• A. 0.55 amperes
• B. 1.83 amperes
• C. 50 amperes
• D. 6,600 amperes

Correct answer: A

Rationale: To calculate the current being drawn, use the formula I = P / V, where I is the current, P is the power in watts, and V is the voltage. Substituting the given values, I = 60 / 110 ≈ 0.55 amperes. Therefore, the current being drawn by the 60-watt lightbulb is approximately 0.55 amperes. Choice B, 1.83 amperes, is incorrect as it does not match the calculated value. Choices C and D, 50 amperes and 6,600 amperes, are significantly higher values and do not align with the expected current draw of a 60-watt lightbulb powered by a 110-volt source.

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