Auto Information Practice Test 3: Tires and Wheels

Answer: C

Correct answer: Out of the positive terminal, through the lamp, and back to the negative terminal. By convention, diagrams show current leaving the battery's positive terminal, going through the load and returning to negative. Electrons really move the opposite way, but both descriptions give the same voltages, currents and powers as long as you stay consistent.

• A: That is the direction electrons actually drift in the wire, which is electron flow, not conventional current.
• B: Current flows around a single closed loop in one direction; it does not enter the load from both ends.
• C (correct): Conventional current is drawn leaving positive, passing through the load and returning to negative.
• D: That describes alternating current, but a battery circuit is DC and flows one way only.

Answer: A

Correct answer: Fuel injector. Ask whether the part reports information or does work. Sensors measure a condition and send a signal to the ECM; actuators receive a command and act. The fuel injector opens on command, so it is the actuator.

• A (correct): An injector receives a command from the ECM and does physical work by opening to spray fuel, so it is an actuator.
• B: An oxygen sensor reports exhaust oxygen content to the ECM, so it is a sensor.
• C: The MAP sensor reports manifold pressure as a signal, so it is a sensor.
• D: This part reports engine temperature to the ECM, so it is a sensor.

Answer: B

Correct answer: The highest speed the tire is rated for under proper conditions. In a code such as 94H, the number is the load index and the letter is the speed symbol. H corresponds to a maximum rated speed of 130 mph under proper conditions; like the load index, it is a limit, not a goal.

• A: A speed symbol is a limit, not a recommended operating speed.
• B (correct): The speed symbol is a maximum rating, not a target speed to reach.
• C: The letter after the load index is the speed symbol; it says nothing about rubber hardness.
• D: H is not a road-use category; it is the speed symbol for a 130 mph maximum rating.

Answer: A

Correct answer: 6 m/s². Newton's second law says F = ma, so

• A (correct): From F = ma, a = 7,200 N ÷ 1,200 kg = 6 m/s².
• B: This slips a decimal place; 7,200 divided by 1,200 is 6, not 0.6.
• C: This divides mass by force (1,200 ÷ 7,200) instead of force by mass.
• D: This is ten times too large; check the division 7,200 ÷ 1,200 again.

Answer: C

Correct answer: About 251 kPa. Mass per tire:

so N = 460 x 9.81 = 4,512.6 N.

• A: This puts the whole van's weight on one patch instead of splitting it four ways.
• B: This divides the mass in kilograms by the area without converting to newtons.
• C (correct): 1,840 / 4 = 460 kg; 460 x 9.81 = 4,512.6 N; 4,512.6 / 0.018 = about 250,700 Pa, or about 251 kPa.
• D: This misplaces the decimal in the area, using 0.0018 m${}^2$ instead of 0.018 m${}^2$.

Answer: D

Correct answer: Read the codes and engine data before replacing parts. Stumbling or hesitation when the accelerator is pressed points to fuel delivery, ignition or air metering. With the wheels ruled out, reading the stored codes, freeze-frame and live engine data is the least intrusive way to narrow the cause before any part is replaced.

• A: Plugs are only one possibility; replacing them before reviewing data skips the evidence.
• B: The problem follows throttle input, not road speed, so the wheels are an unlikely source.
• C: Clearing the code erases freeze-frame data while doing nothing to locate the fault.
• D (correct): Hesitation under acceleration involves fuel, ignition or air metering, and codes and live data narrow which one.

Answer: C

Correct answer: 300 N·m. Ratio = driven teeth ÷ driving teeth =

a 2.5:1 reduction.

Ideal output torque =

N·m. Check: a larger driven gear turns slower, so its torque must be higher.

• A: This flips the tooth ratio to 16/40 = 0.4, which would make a reduction lower the torque.
• B: This multiplies the torque by the 40 teeth instead of by the gear ratio.
• C (correct): The ratio is 40/16 = 2.5:1, and ideal output torque is 120 × 2.5 = 300 N·m.
• D: This adds the 40 teeth to the torque instead of multiplying by the ratio.

Answer: C

Correct answer: Air forced through the grille by the car's motion cools the radiator. At highway speed the car's own motion pushes air through the grille and radiator. This ram air usually removes plenty of heat, so the fan barely runs; the fan is needed mostly at idle and in slow traffic.

• A: A closed thermostat would cause overheating, not normal temperature.
• B: The engine still makes a great deal of heat on the highway; it just has enough airflow to shed it.
• C (correct): This ram air moves plenty of air through the radiator fins, so the fan is rarely needed.
• D: Engine oil coolers handle only oil heat and are not on every engine; the radiator still does the main job.

Answer: B

Correct answer: 37.5 J. First find the compression from Hooke's law:

Then find the energy.

As a check, E also equals ½Fx = ½ ×

• A: This multiplies force by compression (1,500 × 0.05) and leaves out the one-half.
• B (correct): Hooke's law gives x = 1,500 ÷ 30,000 = 0.05 m, and ½ × 30,000 × 0.05² = 37.5 J.
• C: This uses the compression as 5 (centimeters) instead of 0.05 m.
• D: This squares 0.05 incorrectly as 0.00025 instead of 0.0025.

Answer: B

Correct answer: 10 milliseconds. Convert to revolutions per second: 3000 divided by 60 is 50, so one revolution takes 1/50 of a second, or 20 ms. Each stroke is 180 degrees, half a revolution, so one stroke takes 10 ms.

• A: This is the time for a full crank revolution, which covers two strokes.
• B (correct): At 3000 rpm the crank makes 50 turns a second, so one turn takes 20 ms, and a stroke is half a turn, or 10 ms.
• C: This is the time for one complete four-stroke cycle of two revolutions.
• D: This treats a stroke as a quarter turn of the crank, but each stroke is half a turn, or 180 degrees.