Answer:
iIn a nuclear plant, the heat source is from the nuclear reaction whereas in a thermal power plant it is from the combustion of coal. The difference is in the inlet steam parameters to the turbine in a nuclear plant. Thermal power plants use steam at superheated conditions. ... The nuclear plant uses a 'wet steam turbine'.
Explanation:
The heat source in a nuclear power plant is the nuclear reaction, whereas the heat source in a thermal power plant is coal combustion. The difference is in the turbine's input steam characteristics.
What is a power plant?Power plant is an industrial structure that generates electricity. The majority of power plants are linked to the electrical grid.
Nuclear power bare a form of thermal power plant. You have a reactor where fission takes place and heat is generated, a heat exchanger that transports this heat to where it is needed.
Thermal power plant equipment converts this heat into electric energy, usually via a steam turbine.
The reactor, heat exchanger, and thermal conversion technology all have different designs and technologies, but the overall architecture is quite similar to other types of thermal power plants.
The heat source in a nuclear power plant is the nuclear reaction, whereas the heat source in a thermal power plant is coal combustion. The difference is in the turbine's input steam characteristics.
Hence, the two types of power plants differ in difference is in the turbine's input steam characteristics.
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A motorcycle moves according to the velocity-time graph
shown in Figure 3.28. Find the average acceleration of
the motorcycle during each of the following segments of
the motion: (a) A, (b) B, and (c) C.
Answer:
a) 1 m/s²
b) -1 m/s²
c) 0 m/s² (constant speed, not accelerating)
Explanation:
A
delta speed = 10 - 0
delta time = 10 - 0
delta speed / delta time = 10/10
delta speed / delta time = 1
B
ds = 5 - 10
dt = 15 - 10
ds / dt = -5 / 5
ds / dt = -1
C
ds = 5 - 5
dt = 25 - 15
ds / dt = 0 / 10
ds / dt = 0
The average acceleration of A, B, and C will be 1 m/s², -1 m/s², and 0 m/s².
What is acceleration?An object is considered to have been accelerated if its velocity changes. Depending on whether an object is moving faster, slower, or in a different direction, its velocity may change. Examples of acceleration include a falling apple, the moon orbiting the earth, and a car that has stopped at a stop sign. These examples demonstrate how acceleration occurs whenever a moving object modifies its direction, speed, or both.
There are different types of acceleration :
Uniform accelerationNon-Uniform AccelerationAverage accelerationAverage acceleration is defined as the average change in velocity with respect to the average change in time.
SI unit is m/s² and it is vector quantity.
According to the question,
For A, Acceleration= (v₁-v₀)/(t₁-t₀)
⇒10-0/10-0
= 1 m/s².
For B, Acceleration= (v₃-v₂)/(t₃-t₂)
⇒5-10/15-10
= -1 m/s²
For C, Acceleration will be constant because change in velocity is constant, this is the case of uniform motion, so its acceleration is also going to be constant (as per the definition of acceleration change in velocity ratio change in time).
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The octopus’s tentacle keeps _ right after it is bitten off ? a. Moving b. Breathing c. Growing
Answer:
The answer is A
Explanation:
The octopus’s tentacle keeps moving right after it is bitten off
a. Block on a smooth incline.
A block of mass m= 3.8 kg on a smooth inclined plane of angle 36° is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 3 kg hanging vertically. Take the positive direction up the incline and use 9.81 m/s2
for g.
What is the acceleration of each block to 1 decimal place?
Answer:
Explanation:
F = ma
3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a
a = 1.10566...
a = 1.1 m/s²
the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.
Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.
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/24322350What are some physical factors we deal with for sleeping?
Answer:
we deal our body movement during sleeping
what were your preparetion before going the different physical fitness test?
Answer:
Avoid heavy strenuous exercise for the 24 hours prior to testing. Do not exercise at all on the day of testing to ensure you are well rested. Wear appropriate clothing for the conditions (e.g. shorts/track pants and t-shirt/singlet/sports top) and non-slip athletic footwear with laces securely fastened
What is First Aid.
I mark u brainliest answer
Answer:
First aid refers to the emergency or immediate care you should provide when a person is injured or ill until full medical treatment is available.
Explanation:
Một khối khí hidro bị nén đến thể tích bằng 1/2 lúc đầu khi nhiệt độ không đổi. Nếu vận tốc trung bình của phân tử hidro lúc đầu là V thì vận tốc trung bình sau khi nén là bao nhiêu ?
answer: what language is this???
Which is NOT a function of the
cell wall?
A. Protects cell from bursting
B. Provides support for plant cells
C. Protects cell from harsh internal
environments
D. Absorbs sunlight to give energy to the cell
Answer:
The answer is D
Explanation:
The chloroplast absorbs sunlight for energy not the cell wall
I will mark brainlist
A wave is disturbance that transfers energy and matter.
true
false
Answer:
False
Explanation:
A wave is a disturbance that transfers energy from one place to another without transferring matter.
Answer:
I'm pretty sure it true sorry if I'm wrong
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
Answer:
Explanation:
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
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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A 0.015 kg marble sliding to the right at 0.225 m/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 0.180 m/s. After the collision, the first marble moves to the left at 0.180 m/s. Find the velocity of the second marble after the collision.
Explanation:
[tex]if \: the \: masses \: of \: the \:two \: marbles\: equal \\ then \: each \: velocity \: sharing \: with \: other \: velocity[/tex]
A person is pushing a box with 5 N to the right while another person pushes with 10 N to the left, what is the net force on the box?
Answer:
Fnet = 5 Newton
Explanation:
Fnet = 10 N - 5 N
Fnet = 5 N
A 500 kg cart is rolling to the right at 1.3 m/s. a 60 kg man is standing on the right end of the cart. what is the speed of the cart if tha man suddenly starts running to the left with a speed 10.0 m/s relative to the cart
Answer:
P1 = 1.3 (500 + 60) = 728 kg-m total momentum to right at start
P2 = (v2 - 10) 60 + 500 v2
total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart
728 = 560 v2 - 600
v2 = 1328 / 560 = 2.37 m/s new speed of cart
Check:
After: p2 for cart = 500 * 2.37 = 1186
p1 for man = (2.37 - 10) * 60 = -458
P2 = p1 + p2 = 728 total momentum unchanged
2) A ray of light in air is approaching the boundary with water at an
angle of 52 degrees. Determine the angle of refraction of the light
ray. (Refractive index of air = 1, water = 1.33)
=
Answer:
Explanation:
ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface
n₁sinθ₁ = n₂sinθ₂
1 sin52 = 1.33sinθ₂
θ₂ = arcsin(sin52 / 1.33)
θ₂ = 36°
as measured from the perpendicular to the surface
A cylindrical rod formed from silicon is 16.8 cm long and has a mass of 2.17 kg. The density of silicon is 2.33 g/cm3 . What is the diameter of the cylinder
Answer:
We are given the length, the mass and the density of the cylinder. First let us calculate for the volume by dividing the mass by the density.
volume = mass /density
where mass = 2.17 kg = 2170 g, therefore:
volume = 2170 g / (2.33 g/cm^3)
volume = 931.33 cm^3
We know that the volume of a cylinder has the formula:
volume = π r^2 h
since h = 16.8 cm, therefore calculating for radius:
931.33 cm^3 = π r^2 (16.8 cm)
r^2 = 17.646 cm^2
r = 4.2 cm
Hence the diameter (d) is:
d = 2 r
d = 8.4 cm
Explanation:
The diameter of the cylindrical rod is approximately 0.382 cm.
To find the diameter of the cylindrical rod, we can use the formula for the volume of a cylinder and then solve for the diameter.
The formula for the volume of a cylinder is:
V = π[tex]r^{2}[/tex]h,
where V is the volume, r is the radius, and h is the height (or length) of the cylinder.
In this case, we know the length of the cylinder (h) is 16.8 cm. We need to find the radius (r) in order to calculate the diameter.
The mass of the cylinder can be related to its volume and density using the formula:
m = ρV,
where m is the mass, ρ is the density, and V is the volume.
Rearranging this formula, we can solve for V:
V = m / ρ.
Now we have two equations:
V = π[tex]r^{2}[/tex]h,
V = m / ρ.
Setting these two equations equal to each other, we can solve for r:
π[tex]r^{2}[/tex]h = m / ρ.
Substituting the given values:
π[tex]r^{2}[/tex] * 16.8 cm = 2.17 kg / (2.33 g/[tex]cm^3[/tex]).
Let's solve this equation for r:
[tex]r^{2}[/tex] = (2.17 kg / (2.33 g/[tex]cm^3[/tex])) / (π * 16.8 cm).
[tex]r^{2}[/tex] ≈ 0.036775 [tex]cm^2[/tex].
Taking the square root of both sides:
r ≈ 0.191 cm.
Finally, we can find the diameter (d) by multiplying the radius by 2:
d ≈ 2 * 0.191 cm.
d ≈ 0.382 cm.
Therefore, the diameter of the cylindrical rod is approximately 0.382 cm.
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A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.
Answer:
1.5 m/s
Explanation:
Conservation of momentum means the momentum of the system before the collision is the same as after.
The before, after momentum of each ball is ...
5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)
10 kg ball: (10 kg)(0 m/s), (10 kg)(v)
The sum of the "before" products is the same as the sum of the "after" products:
(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v
(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides
v = (15 kg·m/s)/(10 kg) = 1.5 m/s
The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.
i need help with these please
Answer:
Explanation:
An airplane accelerates from a speed of 33 m/s at the constant rate of 3.0 m/s2 over a distance of 500 m. What the final velocity?
Answer:
Explanation:
v² = u² + 2as
v² = 33² + 2(3.0)(500)
v² = 4089
v = 63.9452...
v = 64 m/s
Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points
The potential difference between the initial and final point is 3.0 V.
The given parameters:
distance between the plates, d = 0.2 mvoltage across the plates, V = 12 Vposition of the electron from negative plate, x₁ = 0.1 mposition of the electron from the positive plate, x₂ = 0.0 5mThe potential difference between the initial and final point is calculated as follows;
[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]
where:
[tex]d_2[/tex] is the distance of the electron between the positive and negative plate
[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]
[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]
Thus, the potential difference between the initial and final point is 3.0 V.
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Three point charges are placed on the y-axis: a chargeqaty=a,a charge-2qat the origin, and a chargeqaty= -a.Such an arrangement is called an electricquadrupole.(a) Find the magnitude and direction of the electric field at points on thepositive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x >> a.Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.
5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
A wooden barrel full of water has a flat circular top of radius 25.0 cm with a small hole in it. A tube of height 8.00 m and inner radius 0.582 cm is suspended above the barrel with its lower end inserted snugly in the hole. Water is poured into the upper end of the tube until it is full. The density of water is 1.00 × 103 kg/m3.
What is the force with which the water in the barrel pushes up on the top of the barrel?
Answer:
that is all i know
Explanation:
radius= 25.0cm
height= 8m
inner radius= 0.582cm
density= 1.00 × 103kgm³f= m× awhat is the ratio of the average distances that oxygen will diffuse in a given time in air and water
HELPP
When two forces are in opposite directions, and they are the exact same magnitude, the forces will _______.
a. subtract from each other
b. cancel out
c. go on infinitely
d. eventually reach equilibrium
Answer:
i think the correct answer is B. cancel out
HELP ASAP!!!!! Choose all the answers that apply. Technology A)influences science
B)helps scientists observe fast phenomena
C)is the same as science
D) influences history
E)helps scientists observe slow phenomena
question below................
Answer:
do your work why you in class and you would know
Explanation:
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months of training, he realizes that one important component in running a successful marathon is carbo-loading, the consumption of a sufficient quantity of carbohydrates prior to the race that the body can store as glycogen to burn during the race. The typical energy requirement for runners is 1 kcal/km per kilogram of body weight, and each mole of oxygen intake allows for the release of 120 kcal of energy by oxidizing (burning) glycogen.
(a) If the professor finishes the marathon in 4:45:00 h, what is the professor's oxygen intake rate, in liters per minute, during the race if he metabolizes all of the carbo-loaded glycogen during the race and the ambient temperature is 21.5°C? 2.28 Read the problem statement again carefully. Is the air at standard temperature and pressure during the marathon? How would this affect the volume of 1 mol of oxygen? L/min
(b) The human body has an efficiency of 25.0%. Only 25.0% of the energy released from oxidizing glycogen is used as macroscopic mechanical energy, and the remaining 75.0% is used for body processes such as pumping blood and respiration, and then leaves the body through the skin via radiation, evaporative cooling, and other processes. What is the average mechanical power (in W) generated by the professor during the run? 197.561 What is the total energy required by the professor during the run? How efficient is the human body, and how long did the race last? W
(c) What is the change in entropy (in J/K) of the professor's body if his core temperature has risen to 38.3°C during the run and his skin temperature is at 36.0°C during the marathon? J/K
(d) What is the change of the entropy (in J/K) of the air surrounding the professor during the race if the ambient temperature remains constant at 21.5°C? J/K
