Answer:
magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
Explanation:
Given the data in the question;
Speed of carousel N = 24 rpm
From the diagram below, selected path direction defines the Axis of slip.
Hence, The Coriolis is acting along the axis of transmission
Now, we determine the angular speed ω of the carousel.
ω = 2πN / 60
we substitute in the value of N
ω = (2π × 24) / 60
ω = 2.5133 rad/s
Next, we convert the given velocity from mph to ft/s
we know that; 1 mph = 1.4667 ft/s
so
[tex]V_{slip[/tex] = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s
Now, we determine the magnitude of the Coriolis acceleration
[tex]a_c[/tex] = 2( [tex]V_{slip[/tex] × ω )
we substitute
[tex]a_c[/tex] = 2( 8.8002 ft/s × 2.5133 rad/s )
[tex]a_c[/tex] = 44.235 ft/s²
Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
If acceleration is zero what statement about velocity is
true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing
Answer: A
Velocity is zero because the acceleration isn't affected, and velocity is the rate of change, so it can't be any other options.
Answer:
B)Velocity is constantExplanation:
If an object moves with a velocity and there is no acceleration, then the velocity remains constant. His velocity after five second will be equal to his initial velocity.#keeplearning dude:)1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
A 0.5kg ball of clay originally moving at 6 m/s strikes a wall and comes to rest in 0.25s, what is the magnitude of the impulse given to the ball of clay?
A) 0.75 kg m/s
B) 1.5 kg m/s
C) 3.0 kg m/s
D) 12 kg m/s
Answer:
C I did USA testprep
Explanation:
6. If an object accelerates at 3m/s/s, how long does it take for the object to travel at a speed of 12 m/s.
Answer:
4 seconds
Explanation:
Assuming that the object started from rest,
v = at
--> t = v/a = (12 m/s) / (3 m/s^2)
= 4 seconds
You and a friend are playing with a Coke can that you froze so it's solid to demonstrate some ideas of Rotational Physics. First, though, you want to calculate the Rotational Kinetic Energy of the can as it rolls down a sidewalk without slipping. This means it has both linear kinetic energy and rotational kinetic energy. [The freezing only matters because if there is liquid inside, the calculation for the Moment of inertia becomes more complicated]. A Coke can can be modeled as a solid cylinder rotating about its axis through the center of the cylinder. This can has a mass of 0.33 kg and a radius of 3.20 cm. You'll need to look up the equation for the Moment of Inertia in your textbook. It is rotating with a linear velocity of 6.00 meters / second in the counter-clockwise (or positive) direction. You can use this to determine the angular velocity of the can (since it is rolling without slipping). What is the Total Kinetic Energy of the Coke can
Answer:
K_{total} = 8.91 J
Explanation:
In this exercise you are asked to find the kinetic energy of the can of coca-cola
K_total = K_ {Translation} + K_ {rotation}
the translational kinetic energy is
K_ {translation} = ½ m v²
the kinetic energy of rotation is
K_ {rotation} = ½ I w²
The moment of inertia of a cylinder is
I = ½ m r²
we substitute
K_ {total} = ½ m v² + ½ (½ m r²) w²
angular and linear velocity are related
v = w r
we substitute
K_ {total} = ½ m v² + ¼ m r² v² / r²
K_ {total} = m v² (½ + ¼)
K_ {total} = ¾ m v²
let's calculate
K_ {total} = ¾ 0.33 6.00²
K_{total} = 8.91 J
The law of conservation of angular momentum states that if no external force acts on an object, then its angular momentum does not change. true or false
Answer:
the answer is false.
Explanation:
i took the test and it is false trust me!!!!!!!!!
Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10-m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope (5-m diameter) on Palomar Mountain in California
Answer:
Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Explanation:
If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.
Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Can someone take there time and answer this :)
Answer: I think B.)
Explanation:
What type of force is F-?
Answer:
Frictional Force
Explanation:
i think so
Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kgm2. The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable. (Hint: use what you know about relative velocity to help solve the problem
Answer:
[tex]0.3165\ \text{rad/s}[/tex]
Explanation:
m = Mass of person = 65 kg
d = Diameter of round table = 6.5 m
r = Radius = [tex]\dfrac{d}{2}=3.25\ \text{m}[/tex]
v = Velocity of person running = 3.8 m/s
[tex]I_t[/tex] = Moment of inertia of turntable = [tex]1850\ \text{kg m}^2[/tex]
Moment of inertia of the system is
[tex]I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2[/tex]
As the angular momentum of the system is conserved we have
[tex]L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}[/tex]
The angular velocity of the turntable is [tex]0.3165\ \text{rad/s}[/tex].
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Carbon-14 is the typical radioisotope used to date materials; however, it has a limitation to 40,000 years. A scientist who wants to date materials older than 40,000 years would most likely use which radioisotope?
Answer:
the decay of uranium ending in lead, of potassium (40K) that becomes argon, the decay of rubidium
Explanation:
For the radioactive dating process, a material is needed that has a known average life time and that we can know the amount of material at a given moment,
In the case of carbon 14 (14C), living beings stop capturing it from the air and plants when they die, so knowing the amount they currently have, it is possible to calculate the time in which they stopped absorbing, but the life time average is 5730 years, the maximum time that can be used is up to about 10 average visa times
To analyze extra samples have high half-life times
* the decay of uranium ending in lead
* the decay of potassium (40K) that becomes argon T1 / 2 = 1,251 10⁹ years
* the decay of rubidium (87Ru) which becomes strontium T1 / 2 = 4.92 10¹⁰ years
Below are the ingredients of photosynthesis. Organize them to show what is used and what is produced
during this process.
carbon dioxide
water
oxygen
glucose
sunlight
_____plus_____in the presence of_____result in_____plus_____
Answer:
carbon dioxide plus water in the presence of energy from light, results in glucose plus oxygen
The process is like carbon dioxide plus water in the presence of sunlight result in glucose plus oxygen.
What is photosynthesis?Photosynthesis is a process that plants and other organisms use to convert light energy into chemical energy that can then be released to fuel the organism's activities via cellular respiration.
Photosynthesis is important because it is the process by which plants convert carbon dioxide into oxygen using sunlight. Animals require oxygen to survive, and plants serve as nature's air filter, filtering out harmful carbon dioxide.
Photosynthesis is the conversion of light energy by phototrophs into chemical energy, which is then used to power cellular activities. Sugars, which are formed from water and carbon dioxide, are used to store chemical energy.
The process of photosynthesis includes:
[tex]6CO_2+6H_2O \frac{Sunlight}{chlorophyll} > C_6H_1_2O_6+6O_2[/tex]
Thus, this is the basic process of photosynthesis.
For more details regarding photosynthesis, visit:
https://brainly.com/question/1388366
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Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 132 W/m2 . You may want to review (Pages 897 - 898) . Part A What is the angle from vertical of the axis of the second polarizing filter
Answer:
θ = 32.4º
Explanation:
For this exercise let's use Malus's law
I = Io cos² θ
in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical
I₀ = 370/2 = 185 W / m²
this is the radiation that affects the second polarizer, let's apply the expression of Maluz
θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])
θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])
θ = cos⁻¹ (0.844697)
θ = 32.4º
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Answer:
[tex]523269.9\ \text{N/m}[/tex]
Explanation:
q = Charge
r = Distance
[tex]q_1=25\ \text{C}[/tex]
[tex]r_1=3000\ \text{m}[/tex]
[tex]q_2=40\ \text{C}[/tex]
[tex]r_2=850\ \text{m}[/tex]
The electric field is given by
[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]
The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]
A sound wave travels with a velocity of 1.5 m/s and has a frequency of 500 Hz. What is its wavelength?
Which macronutrient is made up of carbon and hydrogen elements joined together in long groups called hydrocarbons?
Proteins Magnesium Potassium Fats
Answer:
FATS
Explanation:
Fats are made up of carbon and hydrogen elements joined together in long groups called hydrocarbons. The simplest unit of fat is the fatty acid, of which there are two types: saturated and unsaturated.
The macronutrient which is made up of carbon and hydrogen elements joined together in long groups called hydrocarbons is: D. Fats.
A macronutrient refers to a very large molecule (structure) that comprises covalently bonded organic atoms (hydrocarbons) and smaller molecular structures (monomers).
In Science, macronutrients are grouped into four (4) main categories and these are;
Nucleic acid.Carbohydrates.Proteins.Fats (Lipids).Fats (Lipids) are macronutrients that are typically made up of both carbon and hydrogen elements, which are joined together in long groups referred to as hydrocarbons.
Find more information: https://brainly.com/question/14681125
g A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 39.0kg and diameter 78.0cm. The power is off for 34.0s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.At what rate is the flywheel spinning when the power comes back on?
Answer:
[tex]10.54\ \text{rad/s}[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
[tex]\omega_f[/tex] = Final angular velocity
t = Time = 34 s
[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]
[tex]\alpha[/tex] = Angulr acceleration
From the kinematic equations of angular motion we have
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]
The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].
What is surface tension
Answer:
Surface tension is, the surface where the water meets the air, water molecules cling even more tightly to each other.
If F = force, which equation illustrates the Law of Conservation of Momentum?
A) F1 = F2
B) F1 = - F2
C) - F1 = -F 2
D) F1 + - F2 = F3
Answer:
b
Explanation:
f1=-f2 that could be thank u
what is force,momentum,and velocity.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Momentum is force or speed of movement.
Velocity defines the path of the motion of the frame or the object
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV.
What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface?
Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.
Answer:
Explanation:
energy of photon having wavelength of 400 nm = 1237.5/400 eV
= 3.1 eV.
Maximum kinetic energy of photoelectrons = 1.1 eV .
Threshold energy Ф = 3.1 - 1.1 = 2 eV .
energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .
Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy
= 4 - 2 = 2 eV .
Required kinetic energy K₀= 2 eV.
If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity or the boy skater
his mass is 50 kg?
Question: Two people stand facing each other at a roller-skating rink then push off each other. If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity of the boy skater if his mass is 50 kg?
Answer:
3 m/s
Explanation:
Applying,
The Law of conservation of momentum
Momentum of the girl skater = momentum of the boy skater
MV = mv...................... Equation 1
Where M = mass of the girl skater, V = velocity of the girl skater, m = mass of the boy skater, v = velocity of the boy skater
From the question, we were asked to calculate v
v = MV/m.................. Equation 1
Given: M = 30 kg, V = 5 m/s, m = 50 kg
Substitute these values into equation 1
v = (30×5)/50
v = 3 m/s
Hence the velocity of the the boy skater is 3m/s
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 15.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons.
Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:
[tex]W = mg[/tex]
where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,
[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]
Now, we will calculate the value of g on the neutron star:
[tex]g' = \frac{GM}{R^2}[/tex]
where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,
[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]
Therefore, the weight on the surface of the neutron star will be:
[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]
W' = 1.66 x 10¹⁴ N
A truck is traveling on a level road. The driver suddenly applies the brakes, causing the truck to decelerate by an amount g/2. This causes a box in the rear of the truck to slide forward. If the coefficient of sliding friction between the box and the truckbed is 2/5, find the acceleration of the box relative to the truck and relative to the road.
Answer:
Truck [tex]\dfrac{g}{10}[/tex]
Road [tex]-\dfrac{g}{10}[/tex]
Explanation:
[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]
[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]
Frictional force is given by
[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]
Net acceleration is given by
[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]
The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.
Model the Earth's atmosphere as 79% N2, 19% O2, and 2% Argon, all of which are in thermal equilibrium at 280 K. At what height is the density of O half its value at sea level
Answer:
[tex]9.495 \times 10^3\ m[/tex]
Explanation:
From the given information:
Using the equation of Barometric formula as related to density, we have:
[tex]\rho (z) = \rho (0) e^{(-\dfrac{z}{H})} \ \ \ \ --- (1)[/tex]
Here;
[tex]p(z) =[/tex] the gas density at altitude z
[tex]\rho(0) =[/tex] the gas density at sea level
H = height of the scale
[tex]H = \dfrac{RT}{M_ag } \ \ \ --- (2)[/tex]
Also;
R represent the gas constant
temperature (T) a= 280 K
g = gravity
[tex]M_a =[/tex] molaar mass of gas; here, the gas is Oxygen:
∴
[tex]M_a =[/tex] 15.99 g/mol
= 15.99 × 10⁻³ kg/mol
[tex]H = \dfrac{8.3144 \times 280}{15.99 \times 10^{-3} \times 9.8 }[/tex]
[tex]H =14856.43 \ m[/tex]
Now we need to figure out how far above sea level the density of oxygen drops to half of what it is at sea level.
This implies that we have to calculate z;
i.e. [tex]\rho(z) =\dfrac{\rho(0) }{(2)}[/tex]
By using the value of H and [tex]\rho(z)[/tex] from (1), we have:
[tex]\dfrac{\rho(0) }{(2)} = \rho (0) e^{(-\dfrac{z}{14856.43})}[/tex]
∴
[tex]\dfrac{1}{2} = e^{(-\dfrac{z}{14856.43})} \\ \\ e^{(-\dfrac{z}{14856.43})} =\dfrac{1}{2}[/tex]
By rearrangement and taking the logarithm of the above equation; we have:
[tex]- z = 14856.43 \times \mathtt{In}\dfrac{1}{2} \\ \\ -z = 14856.43 \times (-0.6391) \\ \\ z = 9495 \ m \\ \\ z = 9.495 \times 10^3\ m[/tex]
As a result, the oxygen density at [tex]9.495 \times 10^3\ m[/tex] is half of what it is at sea level.
Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.
Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
Determine the Thrust developed
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : calculate the area of the duct
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
next : calculate the velocity of propeller
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
Finally determine the thrust developed
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN