how is power defined in terms of physics

ATI TEAS 7

TEAS 7 science practice questions

1. How is power defined in terms of physics?

A. The rate at which work is done
B. The amount of force applied
C. The distance an object travels
D. The potential energy of an object

Correct answer: A

Rationale: In physics, power is defined as the rate at which work is done, which refers to the amount of energy transferred or converted per unit time. Choice B, 'The amount of force applied,' is incorrect as power is related to work done, not just force. Choice C, 'The distance an object travels,' is not the definition of power but rather relates to displacement or distance. Choice D, 'The potential energy of an object,' is not the correct definition of power; potential energy is different from power. Therefore, the correct definition of power in physics is the rate at which work is done.

2. Which element has the same number of protons and neutrons?

A. Carbon (C)
B. Oxygen (O)
C. Lithium (Li)
D. Helium (He)

Correct answer: D

Rationale: The correct answer is Helium (He). Helium has an atomic number of 2, meaning it has 2 protons. It also has 2 neutrons, which gives it a total mass number of 4. Therefore, Helium is an example of an element with an equal number of protons and neutrons. Choice A (Carbon) has 6 protons and usually has more neutrons than protons. Choice B (Oxygen) has 8 protons and 8 neutrons, so the numbers are not equal. Choice C (Lithium) has 3 protons and usually has more neutrons than protons, so it does not fit the criteria of having the same number of protons and neutrons.

3. A car brakes to a stop on a level road. Which of the following forces does NOT do work on the car?

A. The braking force applied by the wheels
B. The normal force from the road
C. The gravitational force on the car
D. The friction force between the tires and the road

Correct answer: B

Rationale: The normal force from the road does not do work on the car because it is perpendicular to the direction of motion. Work is defined as force applied in the direction of motion, so the normal force, which acts perpendicular to the motion of the car, does not contribute to the work done on the car. The braking force applied by the wheels, the gravitational force on the car, and the friction force between the tires and the road all act in the direction of motion and contribute to the work done on the car. In this scenario, the normal force is supporting the weight of the car and keeping it from sinking into the road, but it does not transfer energy to the car as it moves.

4. Which blood vessels carry deoxygenated blood from the body back to the right atrium of the heart?

A. Arteries
B. Veins
C. Capillaries
D. Venules

Correct answer: B

Rationale: Veins are the blood vessels that carry deoxygenated blood from the body back to the heart. Arteries transport oxygenated blood away from the heart. Capillaries are the smallest blood vessels where gas exchange occurs between blood and tissues. Venules are small veins that connect capillaries to larger veins, but they do not directly carry deoxygenated blood back to the heart.

5. How many grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution? The atomic masses for the elements are as follows: Ca = 40.07 g/mol; C = 12.01 g/mol; O = 15.99 g/mol.

A. 18.3 g
B. 19.7 g
C. 21.0 g
D. 24.2 g

Correct answer: B

Rationale: To calculate the grams of solid CaCO3 needed for a 0.35 M solution, we first find the molar mass of CaCO3: Ca = 40.07 g/mol, C = 12.01 g/mol, O = 15.99 g/mol. The molar mass of CaCO3 is 40.07 + 12.01 + (3 * 15.99) = 100.08 g/mol. The molarity formula is Molarity (M) = moles of solute / liters of solution. Since we have 0.35 moles/L and 600 mL = 0.6 L, we have 0.35 mol/L * 0.6 L = 0.21 moles of CaCO3 needed. Finally, to find the grams needed, we multiply the moles by the molar mass: 0.21 moles * 100.08 g/mol = 21.01 g, which rounds to 19.7 g. Therefore, 19.7 grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution. Choice A (18.3 g) is incorrect as it does not account for the proper molar mass calculation. Choice C (21.0 g) and Choice D (24.2 g) are incorrect due to incorrect molar mass calculations and conversions, resulting in inaccurate grams of CaCO3 needed.