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2 years ago

Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?

Physics

1 answer:

rewona [7]2 years ago

5 0

Hey there!

The pressure under a liquid column can be , calculated using the following formula :

P = p x g x h

P atm = 1.013 x 10⁵ Pa

g = 9.8 m/s²

h = ?

h = P / ( p x g ) =

h= ( 1.013 x 10⁵ Pa ) / ( 900 x 9.8 ) =

h = ( 1.013 x 10⁵ ) / ( 8820 ) =

h = 11.48 m ≈ 11.50 m

Hope this helps!

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Why does lifting one end of the track lead to constant acceleration?

Rainbow [258]

We put a motion detector at one end of the track and put a cart on the track. ... Next, we put a motorized fan on the cart and let it push the cart down the track. ... This is what I would expect based on the velocity graph, since acceleration equals the slope of the velocity graph, which remainsconstant in time.

7 0

2 years ago

Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley

kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = $\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2$

T = $4mx^2 + 2my^2 -2mxy$

also the general potential energy can be expressed as

U = $-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant$

The Lagrangian of the problem can now be setup as

$L =4mx^2 +2my^2 -2mxy +2mgy + constant$

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange equation = $4x-y =0\\-2y+x +g = 0$

solving the equations simultaneously

x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0

1 year ago

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

8 0

2 years ago

Mr. Smith is designing a race where velocity will be measured. Which course would allow velocity to accurately get a winner?

liraira [26]

I’m not completely sure but most likely is is the 10 mile bike ride, I hope I can help! (:

6 0

2 years ago

Read 2 more answers

A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going

tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2$

For the sound

Distance = Speed × Time

$\text{Distance}=340\times (2.5-t)$

Here, distance traveled by the stone and sound is equal

$\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0$

$t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds$

The time taken to fall down is 2.4158 seconds

$h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m$

Height of the buidling is 28.6260196842 m

7 0

2 years ago