Nursing Elites

1. Two objects attract each other with a gravitational force of 12 units. If the distance between them is halved, what is the new force of attraction between the two objects?

• A. 3 units
• B. 6 units
• C. 24 units
• D. 48 units

Correct answer: C

Rationale: The gravitational force between two objects is inversely proportional to the square of the distance between them. When the distance is halved, the new force of attraction will be 12 units x (1/(0.5)^2) = 12 units x 4 = 24 units. Therefore, the correct answer is C. Choice A and B are incorrect as they do not consider the inverse square law of gravitational force. Choice D is incorrect as reducing the distance between the objects does not lead to a squared increase in force.

2. An object with a mass of 45 kg has momentum equal to 180 kg⋅m/s. What is the object’s velocity?

• A. 4 m/s
• B. 8.1 km/s
• C. 17.4 km/h
• D. 135 m/s

Correct answer: A

Rationale: The momentum of an object is calculated by multiplying its mass and velocity. Mathematically, momentum = mass x velocity. Given that the mass is 45 kg and the momentum is 180 kg⋅m/s, we can rearrange the formula to solve for velocity: velocity = momentum / mass. Plugging in the values, velocity = 180 kg⋅m/s / 45 kg = 4 m/s. Therefore, the object's velocity is 4 m/s. Choices B, C, and D are incorrect because they do not align with the correct calculation based on the given mass and momentum values.

3. The operating principle of a metal detector relies on:

• A. The static presence of a permanent magnet
• B. The electromotive force induced by a changing magnetic field
• C. The high electrical conductivity of most metals
• D. The unique thermal signature of metallic objects

Correct answer: B

Rationale: The correct answer is B. Metal detectors work based on the principle of electromotive force induced by a changing magnetic field. When a metal object comes into contact with the detector's magnetic field, it disrupts the field, inducing a current in the metal that can be detected. This principle allows metal detectors to identify the presence of metallic objects without relying on the static presence of a permanent magnet, the high electrical conductivity of metals, or the thermal signature of the objects. Choice A is incorrect because metal detectors do not rely on a static magnet but on the interaction of metals with a changing magnetic field. Choice C is incorrect because while metals do have high electrical conductivity, this is not the principle underlying metal detectors. Choice D is incorrect because metal detectors do not operate based on the thermal signature of objects, but rather on their interaction with magnetic fields.

4. Sublimation is the change in matter from solid to gas or gas to solid without passing through a liquid phase. Outside of the laboratory, which solid provides the best example of this?

• A. Iron
• B. Silver
• C. Salt crystal
• D. Dry ice

Correct answer: D

Rationale: Dry ice (solid carbon dioxide) provides the best example of sublimation outside of the laboratory. When dry ice is exposed to normal atmospheric conditions, it changes directly from a solid to a gas without passing through a liquid phase. This process is commonly observed in everyday situations such as creating 'smoke' or 'fog' effects. Choices A, B, and C (Iron, Silver, and Salt crystal) do not undergo sublimation. Iron and Silver melt and then vaporize, while Salt crystal dissolves in water, and the resulting solution evaporates, which involves a liquid phase.

5. A rock has a volume of 6 cm3 and a mass of 24 g. What is its density?

• A. 4 g/cm3
• B. 4 cm3/g
• C. 144 g/cm3
• D. 144 cm3/g

Correct answer: A

Rationale: Density is calculated by dividing the mass of an object by its volume. In this case, the mass of the rock is 24 g and its volume is 6 cm3. By dividing 24 g by 6 cm3, we find that the density of the rock is 4 g/cm3. Choice A is the correct answer because density is expressed in units of mass per unit volume (g/cm3). Choice B is incorrect as it represents the reciprocal of density. Choices C and D are significantly higher values and do not match the calculated density of the rock.

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