answer.
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
Ulleksa [173]
2 years ago
4

The level of toluene (a flammable hydrocarbon) in a storage tank may fluctuate between 10 and 400 cm from the top of the tank. s

ince it is impossible to see inside the tank, an open-end manometer with water or mercury as the manometer fluid is to be used to determine the toluene level. one leg of the manometer is attached to the tank 500 cm from the top. a nitrogen blanket at atmospheric pressure is maintained over the tank contents. felder, richard m.; rousseau, ronald w.; bullard, lisa g.. elementary principles of chemical processes, 4th edition (page 81). wiley. kindle edition.
Physics
1 answer:
Alika [10]2 years ago
8 0

Complete Question  

The complete question is shown on the first and second uploaded image  

Answer:

When water is used the reading is  R =  2281.6 \  cm

When mercury is used the reading is R =  23.83 \ cm  

The best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Explanation:

From the question we are told that  

    The length of the leg of the manometer to the top of the tank is  d =  500cm

   The toluene level where in the tank where the height of the manometer fluid level in the open arm is equal to the height where the manometer is connected to the tank is  h =150 cm  

   The manometer reading is  R  

Generally at the point where the height of the open arm is equal to the height of the of the point connected to the tank ,  

   The pressure at the height of the both arms of the manometer corresponding to the base of the tank are equal  

     i.e   P_1 = P_2

Here P_1 is the pressure of the manometer at the point corresponding to the base of the tank and this is mathematically represented as  

       P_{atm} + P_1 =  P_{atm} + P_t

Here P_t is the pressure due to the toluene level in the tank and in the arm of the manometer connected to the tank and this is mathematically represented as  

         P_t  =  \rho_t  * g  * h_i

Here  

      \rho_t is the density of toluene with value  \rho_t =  867 kg/m^3

       

 h_i is the height of the connected arm above the point equivalent to the base of the tank , this mathematically represented as

          h_i =  d - h + R

  and  P_2 is the the pressure at the open arm of the manometer at the point equivalent to the base of the base of the tank and this is mathematically represented as

         P_2 =  \rho_f * g *  h_f

Here  

       \rho_f is the density of the fluid in use , if it is water the density is  

       \rho_w =  1000 \  kg /m^3

and  if it is  mercury the density is  

        \rho_m =  13600 \  kg /m^3

h_f is the height of the  fluid in the open arm of the manometer from the point equivalent to the base of the tank which is equivalent the manometer reading R

So when the fluid is water we have

      P_{atm} +  \rho_t* g *(d - h + R) =  P_{atm} + \rho_f * g *  h_f

=>   \rho_t* (d - h + R) =   \rho_w *  h_f        

=>   867  (500 - 150 + R) =    1000 *  R

=>    R =  2281.6 \  cm

So when the fluid is mercury we have      

  \rho_t* (d - h + R) =   \rho_m *  h_f          

=>   867  (500 - 150 + R) =   13600  *  R  

=>   R =  23.83 \ cm  

The difference in the mercury reading for mercury due to the fact that they have different densities as we have seen in this calculation

So the best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

       

You might be interested in
Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
Anit [1.1K]
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 =  625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
8 0
2 years ago
Read 2 more answers
A 2-column table with 4 rows. The first column labeled substance has entries calcium chloride, calcium bromide, calcium carbonat
yanalaym [24]

Answer:

SAMPLE C

Explanation:

5 0
2 years ago
Read 2 more answers
This is a physical property of all visible light determined by the light's frequency and visible to the human eye.
motikmotik
Color <span>is a physical property of all visible light determined by the light's frequency and visible to the human eye.</span>
6 0
2 years ago
A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a
Contact [7]

Answer:

0.83 ω

Explanation:

mass of flywheel, m = M

initial angular velocity of the flywheel, ω = ωo

mass of another flywheel, m' = M/5

radius of both the flywheels = R

let the final angular velocity of the system is ω'

Moment of inertia of the first flywheel , I = 0.5 MR²

Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'

0.5 x MR² x ωo = 0.6 MR² x ω'

ω' = 0.83 ω

Thus, the final angular velocity of the system of flywheels is 0.83 ω.

5 0
2 years ago
Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of
mrs_skeptik [129]

Answer:

B.

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

A_1V_1=A_2V_2

In the problem A_2 is 2A_1, then

A_1V_1=2A_1V_2

V_2 = \frac{V_1}{2}

<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>

For the particular case of Bernoulli we have to

P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2

P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)

For the previous definition we can now replace,

P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)

\Delta P =  \frac{3}{8} \rho V_1^2

<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0
2 years ago
Other questions:
  • A tall cylinder contains 30 cm of water. oil is carefully poured into the cylinder, where it floats on top of the water, until t
    8·2 answers
  • To warm 2.0 l of tea (d = 1.01 g/ml; sp. heat = a cook places a 500 g block of stone at a temperature of 200f into the teapot. a
    8·1 answer
  • About three billion years ago, single-celled organisms called cyanobacteria lived in Earth’s oceans. They thrived on the ocean’s
    12·2 answers
  • If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a hei
    5·2 answers
  • If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f
    14·2 answers
  • A sound technician is testing the sound acoustics in a theatre for an upcoming music concert. As he moves towards the speakers,
    5·2 answers
  • A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h
    10·1 answer
  • A majorette in the Rose Bowl Parade tosses a baton into the air with an initial angular velocity of 2.5 rev/s. If the baton unde
    11·1 answer
  • The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by w
    7·1 answer
  • An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!