ps12a31 - Ph 111.01 November 16, 2008 PS 12a (31) Problem...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Ph 111.01 November 16, 2008 PS 12a (31) Problem 1. An airtight box has a removable (massless) lid of area 2 ·10 -2 m 2 . A lid is placed on it while it is on top of a mountain (where P atm = 0.87 ·10 5 Pa). It is then taken to sea level, where P atm = 1.013·10 5 Pa. How much force will be required to remove the lid? N (How would you do this if the lid had mass, and this value was given?) Solution. The force down on the top is due to pressure at sea level, and the force up on the top (from the inside) is equal to pressure at the top of the mountain. The force required to remove the lid is thus equal to () ( ) ( ) on lid due to pressure 12 sea level mountain 25 5 m 1.013 10 Pa 0.87 10 Pa N FF p xx =− A F 1 =p sealevel A F 2 =p mountain A p sealevel p mountain to lift lid ( 0.02 286 Ap = If the mass of the lid is not negligible, its weight must be added to the force just calculated. Problem 2. In the figure below, S is a small loudspeaker driven by an audio oscillator with a frequency that is varied from 1000 Hz to 2000 Hz, and D is a cylindrical pipe with two open ends and
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

ps12a31 - Ph 111.01 November 16, 2008 PS 12a (31) Problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online