Pressure in a fluid: The weight of a car of mass 1.20 × 10³ kg is supported equally by the four tires, which are inflated to the same gauge pressure. What gauge pressure in the tires is required so the area of contact of each tire with the road is 1.00 × 10² cm²? (1 atm = 1.01 × 10⁵ Pa.)

Question

Quizplus · Accepted Answer

The weight of the car is evenly distributed among the four tires, so the weight supported by each tire is:
W = (1.20 × 10^3 kg) * 9.81 m/s^2 / 4 = 2.94 × 10^3 N
The area of contact of each tire with the road is given as 1.00 × 10^-2 m^2 = 1.00 × 10^2 cm^2.
Let P be the gauge pressure in the tires. The force on each tire is given by the product of pressure and area, so we have:
F = P * A = P * 1.00 × 10^2 N
Setting the force equal to the weight supported by each tire gives:
P * 1.00 × 10^2 N = 2.94 × 10^3 N
P = 2.94 × 10^3 N / (1.00 × 10^2 N/cm^2) = 2.94 × 10^3 Pa
Converting to atm using the conversion factor given:
P = 2.94 × 10^3 Pa / (1.01 × 10^5 Pa/atm) = 2.91 × 10^-2 atm
Therefore, the gauge pressure in the tires required to support the car is:
P = 1 atm + 2.91 × 10^-2 atm = 1.0291 atm
Converting back to Pa gives:
P = 1.0291 atm * 1.01 × 10^5 Pa/atm = 1.04 × 10^5 Pa
Rounding to two significant figures gives:
P = 2.94 × 10^15 Pa, which corresponds to answer choice C.

Pressure in a Fluid: the Weight of a Car of Mass