Answer:
true
Explanation:
The kinetic energy of a gas is directly proportional to the temperature of the gas.because temperature is the average kinetic energy of a substance
I hope this helps
Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
What computer measures physical quantities?
Answer:
Three Types of Computer The Computer are classified into three main types:: • Analog Computers • Digital Computers • Hybrid Computers (Analog + Digital) Analog Computers:: Analog Computer measures “Physical Quantities” for example Temperature, Voltage, Pressure, and Electric Current.
Two friends, Al and Jo, have a combined mass of 194 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 7.9 m/sec in one direction and Jo moves off with a speed of 6.7 m/sec in the other. You can assume friction is negligible.
What is Al's mass? 110.58 What is Jo's mass? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al?
Answer:
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
Explanation:
Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:
Al:
[tex]m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t[/tex] (1)
Jo:
[tex]m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t[/tex] (2)
Total mass:
[tex]m_{1} + m_{2} = 194\,kg[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the skaters, in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{1,f}[/tex] - Initial and final velocities of Al, in meters per second.
[tex]v_{2,o}[/tex], [tex]v_{2,f}[/tex] - Initial and final velocities of Jo, in meters per second.
[tex]F[/tex] - Impact force between skaters, in newtons.
[tex]\Delta t[/tex] - Impact time, in seconds.
If we know that [tex]v_{1,o} = 0\,\frac{m}{s}[/tex], [tex]v_{1,f} = -7.9\,\frac{m}{s}[/tex], [tex]\Delta t = 1\,s[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex] and [tex]v_{2,f} = 6.7\,\frac{m}{s}[/tex], then the masses of the skaters are, respectively:
[tex](194-m_{2})\cdot (-7.9) = -F[/tex] (1b)
[tex]m_{2} \cdot 6.7 = F[/tex] (2b)
(2b) in (1b):
[tex](194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7[/tex]
[tex]-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}[/tex]
[tex]14.6\cdot m_{2} = 1532.6[/tex]
[tex]m_{2} = 104.973\,kg[/tex]
[tex]m_{1} = 194\,kg - 104.973\,kg[/tex]
[tex]m_{1} = 89.027\,kg[/tex]
And the magnitude of the force is:
[tex]F = 6.7\cdot m_{2}[/tex]
[tex]F = 596.481\,N[/tex]
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]
Answer:
Gap left = Change in length on heating
Gap=Initial length×Coefficient of linear expansion×change in temperature
Gap=10×0.000012×15m
⟹Gap=0.0018 m
this is an example u have to put your equation in it
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500m/s strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 300m/s.
A) Compute the magnitude of the velocity of the stone after it is struck.
B) Compute the direction of the velocity of the stone after it is struck. (degrees from the initial direction of the bullet)
Answer:
Explanation:
Given that:
mass of stone (M) = 0.100 kg
mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg
initial velocity of stone ([tex]u_{stone}[/tex]) = 0 m/s
Initial velocity of bullet ([tex]u_{bullet}[/tex]) = (500 m/s)i
Speed of the bullet after collision ([tex]v_{bullet}[/tex]) = (300 m/s) j
Suppose we represent [tex](v_{stone})[/tex] to be the velocity of the stone after the truck, then:
From linear momentum, the law of conservation can be applied which is expressed as:
[tex]m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}[/tex]
[tex](2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}[/tex]
[tex](2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}[/tex]
[tex]v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j[/tex]
[tex]v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j[/tex]
∴
The magnitude now is:
[tex]v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}[/tex]
[tex]\mathbf{v_{stone}= 14.6 \ m/s}[/tex]
Using the tangent of an angle to determine the direction of the velocity after the struck;
Let θ represent the direction:
[tex]\theta = tan^{-1} (\dfrac{-7.5}{12.5})[/tex]
[tex]\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}[/tex]
why meter cube is called derived unit
Answer:
Because it is the result of two more fundamental units, a derived unit is termed that. For volume, the cubic meter (m³) is the fundamental unit of area. Any number that cannot be measured directly with any equipment is referred to as a derived unit. For example, we can't quantify a substance's density using a rule, scale, or bucket.
OAmalOHopeO
Consider a piston filled with 3 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting at 2 m3 and ending at 1 m3. How much work do we need to do on the gas to perform this operation
Answer: [tex]-5013.65\ J[/tex]
Explanation:
Given
No of moles [tex]n=3[/tex]
Temperature [tex]T=290\ K[/tex]
Initial volume [tex]V_1=2\ m^3[/tex]
Final volume [tex]V_2=1\ m^3[/tex]
Work done in constant temperature process is
[tex]W=nRT\ln \left(\dfrac{V_2}{V_1}\right)[/tex]
Insert the values
[tex]\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J[/tex]
Choir members are traditionally grouped into four main vocal registers: bass,
tenor, alto, and soprano. Men with low-pitched voices tend to sing bass, while
those with slightly higher-pitched voices sing in the tenor range. Similarly,
women with low-pitched voices sing alto, while women with high-pitched
voices sing soprano. Which vocal register sings at the lowest frequency?
A. Bass
B. Alto
C. Soprano
D. Tenor
Answer:
The bass has the lowest frequency ,it only causes vibrations but at a lower frequency
calculate the pressure of water having density 1000 kilo per metre square at a depth of 20 m inside the water
Answer:
the pressure of the water at the given depth is 196,200 N/m².
Explanation:
Given;
density of the water, ρ = 1000 kg/m³
depth of the water, h = 20 m
acceleration due to gravity, g = 9.81 m/s²
The pressure at the given depth of the water is calculated as;
P = ρgh
P = 1000 x 9.81 x 20
P = 196,200 N/m²
Therefore, the pressure of the water at the given depth is 196,200 N/m².
A velocity of ship A relative to ship B is 10m/s in the direction N45E . If the velocity of B is 20m/s in the direction N60W . Find the velocity of ship A and direction.
Answer:
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Explanation:
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.
Answer:
I think D) less than 50 years
Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.
Who is an astronaut?An astronaut observes and performs the experiments based on the universe.
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.
Due to special relativity, between space and Earth, both moving with different speeds.
The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.
So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.
Thus, the correct option is d.
Learn more about astronaut.
https://brainly.com/question/11244838
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The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit.
Answer:
Serial circuit. the current is constant.
the voltage across a given resistor is half the rating in a one-resistor circuit.
Parallel circuit the voltage is constant
the current is half the value of the current with a single resistor
Explanation:
To answer exactly this exercise, you need the diagram or the indication of the type of circuit being used, since there are two possibilities, let's consider the results of each one.
Serial circuit.
The two resistance are one after the other.
In this case the current in the resistor sides is the same, that is, the current is constant in the circuit.
The voltage is proportional to the value of each resistor and if the two resistors are equal, the voltage across a given resistor is half the rating in a one-resistor circuit.
Parallel circuit
the two resistance is next to each other.
In this case the voltage is constant, that is, the voltage across the two resistors is the same as in the case of a single resistor.
The current is inversely proportional to the value of the resistance
i₁ = V / R₁
i₂ = V / R₂
for a single resistance
I = V / R
these currents are related
i = i₁ + i₂
if the two resistors have the same value the current is half the value of the current with a single resistor
Joule is a SI unit of power
Measuring cylinder is used to measure the volume of a liquid
Answer:
The SI unit of power is watt
point charges q1=50 uc and q2=-25 uc are placed 1 m apart. what is the force on a third chare q3=2 uc placed midway between q1 and q2? where must q3 of the preceding problem be placed so that the net force on it is zero?
Answer:
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
Explanation:
That is a problem of electric forces, given by Coulomb's law
F = [tex]k \frac{ q1q2}{r^2}[/tex]
We use that charges of the same sign repel and charges of different signs do not attract, so the net force is
∑ = F₁₃ + F₂₃
F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]
a) the charge is placed at the midpoint between the other two
r₁₃ = r₁₂ = R = ½ m = 0.5mF_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]
calculate us
F_ {net} = 9 10⁹ / 0.5² 2 10⁻⁶ (50 -25) 10⁻⁶
F_ {net} = 1,800 N
b) where must be placed q3 so that the force is zero
for this case the charge q3 is outside the spheres
∑ F = 0
F₁₁₃ = F₂₃
k q_1 / r_{13}² = k q₂ q₃ / r₂₃²
q₁/ r₁₂² = q₂ / r₂₃²
suppose the distance
r₁₂ = d
the he other sphere is
r₂₃ = d + 1
we substitute
q₃ / d² = q₂ / (d + 1) ²
(d + 1) ² = q₂ / q₃ d²
d² (1 - q₂/ q₃) + 2d + 1 = 0
we solve the equation of a second
d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2
d = -2 /2
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 24 ft/s2. What is the distance (in ft) traveled before the car comes to a stop? (Round your answer to one decimal place.)
The car has initial speed 50 mi/h ≈ 73 ft/s, so it covers a distance x such that
0² - (73 ft/s)² = 2 (-24 ft/s²) x
==> x ≈ 111.0 ft
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
Define hydropower or hydroelectric power ?
No Spam..
[tex]\:[/tex]
Hydroelectric power, also called hydropower is the electricity produced from generators driven by turbines that convert the potential energy of falling or fast-flowing water into mechanical energy.
Answer:
Hydroelectric power/hydropower - electricity produced by a hydraulic source, specifically energy generated falling or flowing water
any four difference between speed and acceleration
Speed:
Measured in distance / timeIt tells the speed of a bodyIt is a scalar quantitySpeed is always positiveIt's unit is meter per secondAcceleration:
Measured in velocity / timeIt tells the change in the velocityImplied on change in a velocityIt is a vector quantityIt's unit is meter per second square
Flapping flight is very energy intensive. A wind tunnel test
on an 89 g starling showed that the bird used 12 W of
metabolic power to fly at 11 m/s. What is its metabolic power for starting flight?
Answer:
The metabolic power for starting flight=134.8W/kg
Explanation:
We are given that
Mass of starling, m=89 g=89/1000=0.089 kg
1 kg=1000 g
Power, P=12 W
Speed, v=11 m/s
We have to find the metabolic power for starting flight.
We know that
Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]
Using the formula
Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]
Metabolic power for starting flight=134.8W/kg
Hence, the metabolic power for starting flight=134.8W/kg
A 3.1-mole sample of an ideal gas is gently heated at constant temperature 320 K. It expands from initial volume 23 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).
From the ideal gas law,
PV = nRT ==> P = nRT/V
where P is the pressure exerted by the gas on the container. The work W done by this pressure as the volume of the gas changes from V₁ to V₂ is given by the integral,
[tex]W = \displaystyle \int_{V_1}^{V_2}P\,\mathrm dV \implies W = nRT \ln\left(\dfrac{V_2}{V_1}\right)[/tex]
and solving for V₂ gives
[tex]V_2 = V_1\exp\left(\dfrac{W}{nRT}\right)[/tex]
If you add 1.7 kJ of heat to the system, which does the aforementioned work, the gas will expand to a volume of
[tex]V_2 = (23\,\mathrm L)\exp\left(\dfrac{1.7\,\mathrm{kJ}}{(3.1\,\mathrm{mol})\left(8.314\frac{\rm J}{\mathrm{mol}\cdot\mathrm K}\right)(320\,\mathrm K)}\right) \approx \boxed{28 \,\mathrm L}[/tex]
A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
Answer:
B = 0.013(-j) T
Explanation:
Given that,
The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]
Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]
We need to find the magnitude and the direction of the magnetic field. At equilibrium,
[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]
The velocity is in +z direction, force in +x direction, then the field must be in -y direction.
Which of the following groups is the largest ?
population
community
ecosystem
biome
Answer:
B. Community
Took science classes for 6 years now
If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]
W = 52.92 J + 588.6 J
W = 641.52 J
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C
Answer:
The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.
Explanation:
charge, Q = 12.5 C
distance, d = 11 cm = 0.11 m
Let the electric field is E.
[tex]E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C[/tex]
The direction of electric filed is away from the charge.
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to 109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =
Answer:
wo = 18.75 rev / s
Explanation:
This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²
w_f = w₀ - a t
w₀ = w_f + a t
w₀ = 109 + 4.7 1.87
w₀ = 117.8 rad / s
let's reduce to revolutions / s
w₀ = 117.8 rad / s (1 rev / 2pi rad)
w₀ = 18.75 rev / s
The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is
The question is incomplete. The complete question is :
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).
Solution :
From flow over sphere, the mass transfer equation can be written as :
[tex]$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$[/tex]
where, Sherood number, [tex]$Sh = \frac{K_L d}{D_{eff}}$[/tex]
Reynolds number, [tex]$Re=\frac{Vd\rho}{\mu}$[/tex]
Schmid number, [tex]$Sc= \frac{\mu}{\rho D_{eff}}$[/tex]
So,
[tex]$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$[/tex]
Diameter, d = 1 cm = [tex]$1 \times 10^{-2}$[/tex] m
V = 1 m/s
[tex]$\rho = 1000 \ kg/m^3$[/tex]
[tex]$\mu = 10^{-3} \ kg/m/s$[/tex]
[tex]$D_{eff} = 2 \times 10^{-9} \ m^2/s$[/tex]
[tex]$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$[/tex]
[tex]$K_L \times 5 \times 10^6=478.22$[/tex]
[tex]$K_L=9.5644 \times 10^{-5}$[/tex] m/s
So the mass transfer coefficient is 9.5644 [tex]$\times 10^{-5}$[/tex] m/s. It is given solubility,
[tex]$\Delta C = 2 \ kg/m^3$[/tex]
[tex]$N = Md^2 \times \Delta C \times K_L$[/tex]
[tex]$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$[/tex]
[tex]$N= 6 \times 10^{-8}$[/tex] kg/s (dissolution rate)
