Question

Notify Me Bookmark

As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury barometer drops by 25.6 mm from the normal height, what is the atmospheric pressure (in Pa)? Normal atmospheric pressure is 1.013×105 Pa and the density of mercury is 13.6 g/cm3. Pa

As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury barometer drops by 25.6 mm from the normal height, what is the atmospheric pressure (in Pa)? Normal atmospheric pressure is 1.013×105 Pa and the density of mercury is 13.6 g/cm3. Pa

Image text
As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury barometer drops by 25.6 m m from the normal height, what is the atmospheric pressure (in Pa)? Normal atmospheric pressure is 1.013 × 10 5 Pa and the density of mercury is 13.6 g / c m 3 . ◻ P a

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

You are visiting your relatives in the Los Angeles basin on a warm day in late September. Everyone is talking about how windy and warm it is because of the Santa Ana conditions. In these conditions, winds that normally blow from west to east reverse direction, so that air comes westward over the mountains that surround Los Angeles and down into the basin. As a packet of air descends from a location in the mountains into the basin, it is compressed adiabatically, causing it to become hotter. Suddenly, the conversation stops when someone remembers you are taking physics, and everyone turns to look at you. Your uncle asks you on behalf of the rest of your family to show them a calculation that explains why the Santa Ana winds are so warm. He says that he has been to the Cajon Pass, with an elevation of 1,151 m. Every time he has been there, he has noticed strong winds, especially during Santa Ana conditions, when the air to the east and north of southern California pours westward through the passes. Your aunt is looking on her smartphone and tells you that the atmospheric pressure varies with elevation nearly linearly between sea level and low mountain altitudes, with a reduction of 12 Pa for each meter of increased elevation. Your uncle says that the air temperature was 67.3∘F when he was in the Cajon Pass. Everyone starts clamoring and begging you for a calculation that would show the temperature of the air in a Santa Ana wind on that day in the Los Angeles basin. What temperature do you calculate (in ∘F) ? (Assume air is a diatomic gas and that the Los Angeles basin has a pressure of 1 atm. ) ∘F

Solution
0 0

Find the pressure amplitude (in Pa ) of a 1.80 kHz sound wave in air, given that the displacement amplitude is 1.70×10−8 m. [Note: The equilibrium density of air is ρ = 1.20 kg/m3 and the speed of sound in air is v = 343 m/s. Pressure variations ΔP are measured relative to atmospheric pressure, 1.013×105 Pa. ] Pa

Solution
0 0

This problem deals with several aspects of an oscilloscope. You have an 14700 volt supply (Vsupply) for accelerating electrons to a speed adequate to make the front phosphor-coated screen glow when the electrons hit it. Once the electron has emerged from the accelerating region, it coasts through a vacuum at nearly constant speed. You can apply a potential difference of plus or minus 35 volts (Vplates) across the deflection plates to steer the electron beam up or down on the screen to paint a display (other deflection plates not shown in the diagram are used to steer the beam horizontally). Each of the two deflection plates is a thin metal plate of length L = 8 cm and width (into the diagram) 4 cm. The distance between the deflection plates is s = 5 mm. The distance from the deflection plates to the screen is d = 34 cm. When there is a 35 volt potential difference between the deflection plates, what is the deflection y of the electron beam where it hits the screen? An approximate treatment is fine, but state your assumptions. As is usually the case, it pays to carry out all of your calculations algebraically and only evaluate the final algebraic result numerically. Note the exaggerated vertical scale: the deflection is actually small compared to the distance to the screen. When leaving the accelerating plates, the electron's speed is close to the speed of light. Therefore, you must use K = (γ − 1)mc2 where γ = 1 /1 − (v/c)2 and c is the speed of light, 3×108 m/s. m
Solution
0 0