how do you determine the velocity of a wave

HESI A2

HESI A2 Physics

1. How do you determine the velocity of a wave?

A. Multiply the frequency by the wavelength.
B. Add the frequency and the wavelength.
C. Subtract the wavelength from the frequency.
D. Divide the wavelength by the frequency.

Correct answer: A

Rationale: The velocity of a wave can be determined by multiplying the frequency of the wave by the wavelength. This relationship is given by the formula: velocity = frequency × wavelength. By multiplying the frequency by the wavelength, you can calculate the speed at which the wave is traveling. This formula is derived from the basic wave equation v = f × λ, where v represents velocity, f is frequency, and λ is wavelength. Therefore, to find the velocity of a wave, one must multiply its frequency by its wavelength. Choices B, C, and D are incorrect. Adding, subtracting, or dividing the frequency and wavelength does not yield the correct calculation for wave velocity. The correct formula for determining wave velocity is to multiply the frequency by the wavelength.

2. How do a scalar quantity and a vector quantity differ?

A. A scalar quantity has both magnitude and direction, and a vector does not.
B. A scalar quantity has direction only, and a vector has only magnitude.
C. A vector has both magnitude and direction, and a scalar quantity has only magnitude.
D. A vector has only direction, and a scalar quantity has only magnitude.

Correct answer: C

Rationale: The correct answer is C. The main difference between a scalar quantity and a vector quantity lies in the presence of direction. A vector quantity has both magnitude and direction, while a scalar quantity has magnitude only, without any specified direction. Examples of scalar quantities include distance, speed, temperature, and energy, whereas examples of vector quantities include displacement, velocity, force, and acceleration. Choices A, B, and D are incorrect because they incorrectly describe the characteristics of scalar and vector quantities.

3. Which object below has the same density?

A. A block with a mass of 6.5 grams and a volume of 16.25 cm3
B. A block with a mass of 80 grams and a volume of 32 cm3
C. A block with a mass of 48 grams and a volume of 22 cm3
D. A block with a mass of 100 grams and a volume of 250 cm3

Correct answer: A

Rationale: Density is calculated by dividing the mass of an object by its volume. The density of object A is 6.5 g / 16.25 cm3 = 0.4 g/cm3. The density of object B is 80 g / 32 cm3 = 2.5 g/cm3. The density of object C is 48 g / 22 cm3 = 2.18 g/cm3. The density of object D is 100 g / 250 cm3 = 0.4 g/cm3. Objects A and D have the same density of 0.4 g/cm3. Therefore, the correct answer is A as it has the same density as object D, making them the only objects with a density of 0.4 g/cm3.

4. The first law of thermodynamics is a principle of energy conservation. It states that:

A. Energy can be created or destroyed.
B. The total entropy of an isolated system always decreases.
C. Energy can neither be created nor destroyed, only transferred or transformed.
D. The temperature of a system is directly proportional to its entropy.

Correct answer: C

Rationale: The first law of thermodynamics states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another, ensuring energy conservation in any system. Choice A is incorrect because it goes against the principle of energy conservation. Choice B is incorrect as it refers to the second law of thermodynamics, which states that the total entropy of an isolated system always increases. Choice D is incorrect because the temperature of a system is not directly proportional to its entropy.

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.