Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.
Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:
[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]
Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:
[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]
Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:
[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]
Finalmente, reemplazamos los valores para obtener:
[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]
Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.
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https://brainly.com/question/21902769https://brainly.com/question/24322350Which of the following best describes an achievement test that is given at the end of a learning segment to evaluate mastery of objectives?
Answer
An Achievement test is an assessment of developed knowledge or skill. ... Achievement tests are developed to measure skills and knowledge learned in a given grade level, usually through planned instruction, such as training or classroom instruction. Achievement tests are often contrasted with aptitude tests.
Explanation:
What are the two factors that affect the frictional force between objects
*graph is below*
1. What is Peter’s total distance traveled? What is Peter's displacement?
2. Is there a time when Peter is not moving? If so, when?
The total distance covered is 24 Km and Peter was not moving between the points marked 10 mins and 30 mins on the graph. His displacement according to the graph is zero.
The distance time graph shows the distance covered plotted on the vertical axis against the time taken plotted on the horizontal axis. Using this graph, the total distance covered can easily be obtained.
The total distance covered is 12 km + 12 km since equal distance was covered to and fro. Hence the total distance covered is 24 km. Perter was not moving between the points marked 10 mins and 30 mins on the graph.
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A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance s < a from the axis.
The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.
The magnetic field in the gap at a distance s < a can be computed by using the formula:
[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]
where;
Magnetic flux density = Bdistance = d[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]
where;
[tex]\mathbf{J_d}[/tex] = drift current density[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]
[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]
Making the magnetic flux density the subject, we have:
[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]
[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]
Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]
[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]
Recall that distance in question is said to be (s);
∴
[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
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HELP ASAPPPPPPPPPPPPPP
Answer:
F=15N
Explanation:
F=m.a
m=1500g ÷1,000 = 1.5kg
a= 10m/s/s
F =1.5 × 10
F = 15 Newton
The tires of a car make 77 revolutions as the car reduces its speed uniformly from 95.0 km/h to 65.0 km/h. The tires have a diameter of 0.90 m.
If the car continues to decelerate at this rate, how far does it go? Find the total distance.
Answer:
Explanation:
95.0 km/hr = 26.39 m/s
65 km/hr = 18.06 m/s
Circumference of a tire is 0.9π m
77 revolutions is a distance of
77(0.9π) = 69.3π m
v² = u² + 2as
a = (v² - u²) / 2s
a = (18.06² - 26.39²) / (2(69.3π))
a = -0.85 m/s²
s = (v² - u²) / 2a
s = (0² - 26.39²) / 2(-0.85)
s = 409 m
What are the customary units for real power? volt-amperes reactive (VAR) volt-amperes (VA) watts (W)
Answer:
watts is for real power
volt amperes reactive (VA) for reactive power
volt amperes (VA) for apparent power
An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg. If this force serves as the action force, what is the reaction force in the action–reaction pair?
Answer:
Equal reaction from the pair in every action there's an equal and opposite reaction
The item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.
What is gravitational force?All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.
As we know, the gravitational force is given by:
[tex]\rm F = \dfrac{Gm_1m_2}{r^2}[/tex]
Where G is the gravitational constant.
m1 and m2 are masses.
r is the distance between the masses.
It is given that:
An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.
As we know,
An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg and the item will keep moving at a consistent speed.
Thus, the item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.
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How much heat must be removed from 1.61 kg of water at 0 ∘C to make ice cubes at 0 ∘C?
Answer:
Explanation:
All
2. Jason rode his bike 10 miles and only rode for 2 hours how fast was he going?
Answer:
5 miles per hour.
Explanation:
10miles / 2hours = 5 mph
Two objects are being lifted by a machine. One object has a mass of 2 kg, and is lifted at a speed of 2
m/s. The other has a mass of 4 kg and is lifted at a rate of 3 m/s.
a. Which object has more kinetic energy while it is being lifted?
Answer:
Kinetic energy = (1/2) (mass) (speed²)
First object: (1/2) (2 kg) (2 m/s)² = 4 joules .
Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .
The second object had more kinetic energy than the first one had.
Explanation:
Answer:
Kinetic energy = (1/2) (mass) (speed²)
First object: (1/2) (2 kg) (2 m/s)² = 4 joules .
Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .
The second object had more kinetic energy than the first one had.
A pumpkin is launched in the air and travels at a horizontal velocity of 25 meters per second for 5 seconds. How far does it travel horizontally?
Answer:
30.3 meters, 172 degrees
Explanation:
To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.
Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West
Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North
The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.
Applying the Pythagorean theorem leads to the magnitude of the resultant (R).
R2 = (30.0 m)2 + (4.0 m)2 = 916 m2
R = Sqrt(916 m2)
R = 30.3 meters
The angle theta in the diagram above can be found using the tangent function.
tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)
tangent(theta) = 0.1333
theta = invtan(0.1333)
theta = 7.59 degrees
This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.
The current in the long wire is decreasing. What is the direction of the current induced in the conducting loop below the wire
Answer:
anlatamadım herkeze ben sorulara bakamiyorum cumku analmaiyirum yardim edin
bu sorunu nasil cozebilirim
lutfsn...
Heat is transferred from molecules with more kinetic energy to molecules with _________kinetic energy.
Answer:
less
Explanation:
heat travels from hotter to colder object. hotter object has more kinetic energy.
You have a 25 W and a 100 W bulb at home. When you connect one or the other, which bulb carries the greater current
Answer:
100 w bulb has more current.
Explanation:
P=V^2/R.
when velocity is constant power is inversly proportional to resistence, so 25 W will have an hogher resistance tjan a 100 w bulb.
P=VI
when v is constant, power is directly proportional to I. hence, 100 w bulbs will carry more current.
When you connect one or the other, the bulb that's connected carries more current than the one that's not connected.
When you connect BOTH of them, the 100W bulb carries more current than the 25W one.
PLZ HURRY
When you hold a racquet and swing your arm toward the ball, there are two kinds of resistance working against your muscles—the ______ and the ______.
A.
racket, air
B.
air, ball
C.
ball, net
D.
air, racket
Answer:
It should just be A
Explanation:
I dont see the difference between these 2 but ill choose A and update you
Answer:
A: racket, air
Explanation:
Name two environmental factors, one natural and one human-made, that could account for the trend or pattern in bird (quail, wren) and rodent (mice, rabbits) populations before 1997. Help!
Factors such as more crops, fewer predators, invasive species, and changes in the environment explain a decrease or increase in the birds and rodent populations.
Before 1997, both rodents and birds populations increased due to different factors. Moreover, these factors can be classified as natural or human-made factors. Here are the factors that mainly affected these animals:
Bird population:
Increase:
Decrease in predators that increase birds chances to survive (natural factor)
Increase in crops that improve food access for birds (human factor)
Decrease:
Invasive species or increase of predators (natural factor)
Destruction of natural habitats due to an increase in industry and extraction (human factor)
Rodent population:
Increase:
More adaptability that increased rodents chances to survive (natural factor)Increase in crops that improve food access for rodents (human factor)Decrease:
Invasive species or increase of predators (natural factor)Increase in rabbit consumption and use of rodent control products (human factor)Learn more in: https://brainly.com/question/18123593
Suppose that you release a small ball from rest at a depth of 0.590 m below the surface in a pool of water. If the density of the ball is 0.370 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball)
Answer:
Explanation:
The work of the buoyancy force will convert to gravity potential energy
Fresh Water has a density of 1000 kg/m³
The ball has a density of 370 kg/m³
assume the ball is 1 m³
weight of the ball is 370g
The buoyancy force is 1000g
Assume the buoyancy force drops suddenly to zero when the center of the ball clears the water level.
The Work done on the ball is
W = Fd = 1000g(0.590)
the change in potential energy is
PE = mgh = 370g(0.590 + y)
where y is the height above the water level.
1000g(0.590) = 370g(0.590 + y)
1000(0.590) = 370(0.590 + y)
y = (0.590)(1000 - 370) / 370
y = 1.004594... ≈ 1.00 m
In an oscillating LC circuit, when 86.6% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor
We have that for the Question "(a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor"
Answer:
a) maximum charge = [tex]0.366Q_{max[/tex]b) maximum current = [tex]0.931I_{max}[/tex]
From the question we are told
In an oscillating LC circuit, when 86.6% of the total energy is stored in the inductor's magnetic field
A) When 86.6\% energy is stored in inductor
[tex]\%[/tex]of energy stored in electric field = [tex]1 - 0.866 = 13.4\%[/tex]
[tex]\frac{V_E}{V} = \frac{\frac{q^2}{2c}}{\frac{Q^2}{2c}} = 0.134\\\\\frac{q}{Q} = \sqrt0.134\\\\\frac{q}{Q} = 0.366\\\\q = 0.366Q_{max[/tex]
B)
[tex]\frac{V_B}{V} = \frac{\frac{Li^2}{2}}{\frac{LI^2}{2}} = 0.866\\\\\frac{i}{I} = \sqrt0.866\\\\\frac{i}{I} = 0.931\\\\i = 0.931I_{max[/tex]
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5 What is the maximum speed at which a car round a curve of 25m radius on a level road if the coefficient of static friction between the tires and the road is 0.80?
Hi there!
On a level road:
∑F = Ff (Force due to friction)
The net force is the centripetal force, so:
mv²/r = Ff
Rewrite the force due to friction:
mv²/r = μmg
Cancel out the mass:
v²/r = μg
Solve for v:
v = √rμg
v = √(25)(9.81)(0.8) = 14.01 m/s
How much force must be applied to push a 1.35 kg book across the desk at constant speed if the coefficient of sliding friction is 0.30?
The magnitude of the force that must be applied to push the book across the desk is 3.97 N.
The given parameters;
mass, m = 1.35 kgcoefficient of friction, μ = 0.3The acceleration of the book across the desk is calculated as follows;
a = μg
where;
g is acceleration due to gravitya = 0.3 x 9.8
a = 2.94 m/s²
The magnitude of the force that must be applied is calculated as follows;
F = ma
F = 1.35 x 2.94
F = 3.97 N
Thus, the magnitude of the force that must be applied to push the book across the desk is 3.97 N.
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A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.380 kg. Calculate (a) its moment of inertia about its center, and (b) the applied torque needed to accelerate it from rest to 1750 rpm in 5.00 s. Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 55.0 s.
Hi there!
(A)
A grinding wheel is the same as a disk, having moment of inertia of:
[tex]I = \frac{1}{2}MR^2[/tex]
Plug in the given mass and radius (REMEMBER TO CONVERT) to find the moment of inertia:
[tex]I = \frac{1}{2}(0.380)(0.085)^2 = 0.00137 kgm^2[/tex]
(B)
We can use the rotational equivalent of Newton's Second Law to calculate the needed torque:
Στ = Iα = τ₁ - τ₂
Begin by solving for the angular acceleration. Convert rpm to rad/sec:
[tex]\frac{1750r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 183.26 rad/sec[/tex]
Now, we can use the following equation:
ωf = wi + αt (wi = 0 rad/sec, from rest)
183.26/5 = α = 36.65 rad/sec²
τ = Iα = 0.0503 Nm
Since there is a counter-acting torque on the system, we must begin by finding that acceleration:
[tex]\frac{1500r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 157.08 rad/sec[/tex]
ωf = wi + αt
-157.08/55 = α = -2.856 rad/sec²
τ₂ = Iα = 0.0039 Nm
Now, calculate the appropriate torque using the above equation:
[tex]\Sigma\tau = \tau_1 - \tau_2[/tex]
[tex]\Sigma\tau + \tau_2 = \tau_1[/tex]
[tex]0.0503 + 0.0039 = \large\boxed{0.054 Nm}[/tex]
10. On Earth, where is hydrogen not found?
A. Natural gas well
B. Water
C. Atmosphere
o D. Minę
Answer:
i think its D or B
Explanation:
i just think
On Earth,in atmosphere the hydrogen is found in the less quantity. Option C is correct.
What is hydrogen?The chemical element hydrogen has the atomic number 1 and the symbol H. The smallest element is hydrogen.
Hydrogen is found in large numbers on Earth in combination with other elements, such as water and hydrocarbons, yet it is only 0.00005 percent present in Earth's atmosphere.
On Earth,in atmosphere the hydrogen is found in the less quantity.
Option C is correct.
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The force of earth’s gravity is 10N downward. What us the acceleration of a 15kg backpack if lifted with a a 15N force?
Answer:
F-F(gr) = ma
a= {F-F(gr)}/m =
=(15-10)/15=0.33 m/s² (upward)
What’s Newton’s second law? Explain and mention some examples in daily life
Answer:
Newton's second law states that .
The rate of change of linear momentum is directly proportional to the force applied.Formulically
F=maF=Force
m=mass
a=acceleration
The best example is hitting a tennis ball.
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Good morning dear...
Have a beautiful and joyful day ahead.
During a picnic, you and two of your friends decide to have a three-way-tug-of-war, with three ropes in the middle tied into a knot. Roberta pulls to the west with 277 N of force; Michael pulls to the south with 603 N. a) With what force should you pull to keep the knot from moving
Answer:
Explanation:
I'm not 10% sure but i think that if you were to do:
603 + 277 = 880
1000 - 880 = 120
So to keep it from moving, i think you would need to pull with a force of 120N
The force with which it should be pulled to keep the knot from moving is 663.58 N.
What is Newton's third law ?Newton's third law states that, for every action, there is an equal and opposite reaction.
Here,
Force exerted by Roberta, F₁ = 277 N
Force exerted by Michael, F₂ = 603 N
Since, Roberta is pulling to the west and Michael is pulling towards the south, the angle between the forces applied by them is 90°.
The force needed to keep the knot from moving is the resultant force of these two forces.
Therefore, the resultant force,
Fₙ = √F₁² + F₂²
Fₙ = √(277)² + (603)²
Fₙ = 663.58 N
Hence,
The force with which it should be pulled to keep the knot from moving is 663.58 N.
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Which processes result in the release of carbon? Select three options.
Animals break down food molecules to obtain energy.
The remains of producers are broken down by decomposers.
The remains of consumers are broken down by soil decomposers.
Producers take in carbon dioxide.
Producers make sugars and starches.
Answer:
The remains of producers are broken down by decomposers.
The remains of consumers are broken down by soil decomposers.
Animals break down food molecules to obtain energy.
Explanation:
Answer:
It is A. B. and C.
Explanation:
Hope this helps
3. What can you infer about the position of the galaxies 100 million years before this telescope photo was taken
The galaxies are in constant motion so it can be inferred that their position will not be the same as a hundred million years ago.
A galaxy is a term to refer to the set of stars, gas clouds, planets, cosmic dust, dark matter, and energy gravitationally united in a more or less defined structure in the universe.
Galaxies are in constant motion because the elements that make them up are not static at any time; The galaxies rotate around the center of the galaxy. If they were still, the gravitational attraction would cause them to immediately fall towards the center of the galaxy: it is the same that would happen to the Earth and the other planets if they stopped rotating around the Sun, they would fall towards the Sun.
According to the above, it can be inferred that the galaxies after a while have moved from the place where they were last seen.
Note: This question is incomplete because there is some missing information. However, I can answer based on my general prior knowledge.
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Please solve this This is my exam question please be fast
Answer:
OK sure why not!!!!!!!!!!!!