Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Gauge pressure vs Absolute help please?
7. The weight of a car of mass 1.20 × 10^3 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 × 10^2 cm^2? (1 atm = 1.01
× 10^5 Pa.)
A) 11.6 × 10^5 Pa
B) 11.6 × 10^4 Pa
C) 2.94 × 10^5 Pa
D) 2.94 × 10^4 Pa
E) 2.94 × 10^3 Pa
The answer is C according to my professor
Here's my question, isn't that the absolute pressure ?
Isn't the total pressure exerted by the tires = to gauge pressure+atm ?
By doing Force / 4A
aren't I getting the total pressure ?
2 Answers
- ?Lv 71 year ago
The atmosphere is pushing on the outside of the tires, so the gauge pressure is the net outward pressure that they can exert. Think of it this way: if the absolute pressure in the tires were 1 atm (gauge pressure = 0), they would do absolutely nothing to support a vehicle. As you well experience when you have a "blowout" !