1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nataly862011 [7]
3 years ago
11

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

assumed constant at 0.0075 lb/ft3 , calculate the elevation of the mountain top.
Physics
1 answer:
stealth61 [152]3 years ago
7 0

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

\rho_m = 846lb/ft^3

g = 32.17405ft/s^2

h_1 = 1in = \frac{1}{12} ft

For the air the defined properties would be

\rho_a = 0.0075lb/ft^3

g = 32.17405ft/s^2

h_2 = ?

We have for equilibrium that

\text{Pressure change in Air}=\text{Pressure change in Mercury}

\rho_m g h_1 = \rho_a g h_2

Replacing,

(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)

Rearranging to find h_2

h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}

h = 9400ft

Therefore the elevation of the mountain top is 9400ft

You might be interested in
Whats the difference between an inner planet and an outer planet
Kryger [21]
The outer planets<span> are further away, larger and made up mostly of gas. The </span>inner planets<span> (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars. After an asteroid belt comes the </span>outer planets<span>, Jupiter, Saturn, Uranus and Neptune.</span>
8 0
3 years ago
Read 2 more answers
If you push a 4-kg mass...
Lena [83]

Answer:

B

Explanation:

F = ma , a = F/m

a1 = F/10 and a2 = F/4

Since Force is constant, a2 will we greater than a1

4 0
3 years ago
Explain Newton's first law of motion for an object in<br> motion.
Anit [1.1K]

Answer:

bshghhxhgdyxhsygfhtgedhrugrugdjifgu

rolling of a ball uses motion his bushy we I his own shaken his known fish of his jus on

6 0
2 years ago
Suppose the electrostatic force between two electrons is F. What is the electrostatic force between an electron and a proton
mars1129 [50]

Answer:

D. −F

Explanation:

the rest of the answers are

2/3F

The force is represented as a positive quantity and is repulsive.

Electrostatic force is inversely proportional to the square of the distance.

The direction of the force changes, and the magnitude of the force quadruples.

hope this helps sorry if i was too late! :)

8 0
3 years ago
A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature
Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

5 0
3 years ago
Other questions:
  • Every 30,000 miles, __________. replace the brakes vacuum the interior service the automatic transmission (if your car has one)
    8·2 answers
  • All animals need oxygen. We get oxygen from the air we breathe. How do fish get theirs?
    14·2 answers
  • How does earths atmispher help it to sustain life
    7·1 answer
  • What is NOT an example of Reflection
    9·2 answers
  • . Two people are pushing a car of mass 2000 kg.
    15·1 answer
  • A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a fricti
    12·1 answer
  • Derive Isothermal process through ideal gas.(anyone plzz!!)​
    5·1 answer
  • Which equation correctly represents the gravitational potential energy of a
    14·2 answers
  • What do both types of bone marrow do that is found in our long bones.<br> (answer asappp)
    9·1 answer
  • How can energy crisis be aberted?​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!