Answer:
El trabajo de frontera ejercido por el gas es 10,297.875 joules.
Explanation:
Supongamos que el gas se comporta idealmente y que el proceso es isotérmico y adiabático. El trabajo efectuado por el gas en el proceso equivale al área bajo la curva de la gráfica PV. Puesto que es una línea recta y que tanto la presión como el volumen son variables positivas, tenemos que esa área es la suma del rectángulo inferior (límite inferior de la presión) y el triángulo superior (entre los límites inferior y superior de la presión). Es decir:
[tex]W = P_{min}\cdot (V_{2}-V_{1})+\frac{1}{2}\cdot (P_{max}-P_{min})\cdot (V_{2}-V_{1})[/tex] (1)
Donde:
[tex]W[/tex] - Trabajo de frontera ejercido por el gas, medida en joules.
[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Volúmenes del gas antes y después de la expansión, medidas en metros cúbicos.
[tex]P_{min}[/tex] - Límite inferior de la presión del gas, medida en pascales.
[tex]P_{max}[/tex] - Límite superior de la presión del gas, medida en pascales.
Si sabemos que [tex]V_{1} = 0.06\,m^{3}[/tex], [tex]V_{2} = 0.13\,m^{3}[/tex], [tex]P_{min} = 125,432\,Pa[/tex] y [tex]P_{max} = 168,793\,Pa[/tex], entonces el trabajo de frontera ejercido por el gas es:
[tex]W = (125,432\,Pa)\cdot (0.13\,m^{3}-0.06\,m^{3})+\frac{1}{2}\cdot (168,793\,m-125,432\,m) \cdot (0.13\,m^{3}-0.06\,m^{3})[/tex]
[tex]W = 10,297.875\,J[/tex]
El trabajo de frontera ejercido por el gas es 10,297.875 joules.
Mathew throws a ball straight up in the air. it rises for a period of time and then begins to drop. At which points in the balls journey will be gravity be the greatest force acting on the ball
Answer: The greatest force of gravity on the ball will occur at the point when the ball is near to hit the ground ... hope this helps have a good day :P
why do we not consider the electromagnetic force when we describe the force between the earth and the sun?
Electromagnetic force can not be considered because it very less as compared to gravitational force.
There are four types of Nuclear Forces :
1. Gravitational force
2. Electromagnetic force
3. Strong nuclear force
4. Weak nuclear force
The force acting between the sun and the earth is gravitational force it is about
3.54 ×10²²N
Earth rotates around the sun because it is contained in the sun's gravitational field.
Due to the centripetal force Earth doesn't falls into the sun.
Electromagnetic force can not be considered because it very less as compared to gravitational force. The electromagnetic force between two charged bodies is nearly equal to 10⁵N.
Electromagnetic force can not be considered because it very less as compared to gravitational force.
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if 1 m = 100 cm , then how many cm^2 are there in a m^2 ?? please hel[p
Answer:if 1 m = 100 cm then there should be 200 cm in m^2
Explanation:
Simple pendulum takes 0.01 s to reach the maximum displacement from the equilibrium position, so it's frequency equals.......
a)20 HZ
b)25 HZ
c)50 HZ
c)100 HZ
Simple pendulum takes 0.01 s to reach the maximum displacement from the equilibrium position, so it's frequency equals to 100 Hz
The time period of a simple pendulum: It is defined as the time taken by the pendulum to finish one full oscillation and is denoted by “T”. The amplitude of simple pendulum: It is defined as the distance travelled by the pendulum from the equilibrium position to one side.
The frequency of the pendulum defines how many times the pendulum moves back and forth in a specific period of time
time period of simple pendulum = 1 / frequency
given
time period = 0.01 second
frequency = 1 / 0.01
= 100 Hz
correct answer will be d) 100 HZ
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write down the kirchhoff's laws in the adjoining circuit , find the current in each resistors
Answer:
You need a motor and a battery
Explanation:
The currents through 4 ohm, 2 ohm and 5 ohm are 0.42 amp, 0.15 amp and 0.26 amp respectively.
What are the Kirchhoff's laws ?
Kirchhoff's Current Law states that no charge is lost and that the total current flowing into a junction or node equals the charge leaving it.
Kirchhoff's Voltage Law states that for any closed network, the voltage around a loop is equal to the total of all voltage drops in that loop and is equal to zero.
Let the current through 4 ohm is I₁.
the current through 2 ohm is I₂.
Applying Kirchhoff's Current Law, current through 5 ohm is I = I₁ + I₂.
Applying Kirchhoff's Voltage Law in two loops we get:
4 I₁ + 5(I₁+I₂) = 3
2I₂ + 5(I₁+I₂) = 1
Hence, I₁ = 0.42 amp
I₂ = - 0.15 amp
I₃ = 0.26 amp
Hence, the currents through 4 ohm, 2 ohm and 5 ohm are 0.42 amp, 0.15 amp and 0.26 amp respectively.
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what yall notice about this please help me out
Answer:
There is a blue sky with clouds
There is cracks in the ground and the ground is brown
There is stuff in the dictance
Explanation:
Hope this helps
PLEASE HELP ME
What is the net force acting on the car?
A rocket is launched and the observer measuring the rocket altitude is standing at a distance of 200 feet from the launcher. When the rocket reaches apogee, the observer measures an angle of 45° using an altitude scope. Calculate the altitude of the rocket to the nearest foot.
Answer:
100 foot
Explanation:
Using the formula for calculating range to calculate the speed first as shown;
Range R = U²sin 2theta/gU is the speed
theta is the observe angle
g is the acceleration due to gravity.
200 = U²sin 2(45)/9.8
Usin90 = 200 * 9.8
U² = 1960
U = √1960
U = 44.27 m/s
Get the required altitude
Altitude H = u²/2g
H = 44.27²/2(9.8)
H = 1,959.8329/19.6
H = 99.99 feet
Hence the altitude of the rocket to the nearest foot is 100 foot
If you weight 100 pounds on earth how much would you weight on the moon
Answer:
You should weigh about 16 pounds.
Explanation:
To find this, you need to find your weight and divide it by 6 to find your weight on the moon.
the armature windings of a dc motor have a resistance of 6.5 ω . the motor is connected to a 120-v line, and when the motor reaches full speed against its normal load, the back emf is 108 v. the load is increased so it causes the motor to run at half speed.
The current in the load will be 10 A.
given,
resistance of the load (R) = 6.5Ω
voltage in the load (V) = 120V
Back emf in the load (E)= 108V
now to measure current we will use the formula
current (I)= {voltage (V)-Back emf (E)}/resistance(R)
now as the speed is halved so will the back electro motive force will be halved.
i.e.
back electro motive force will be = 108V/2
=54V
now,
substituting this value in the formula we get,
current (I) = (120-54)/6.5
current (I) = 66/6.5
current (I) =10A
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In his experiments on "cathode rays" during which he discovered the electron, J. J. Thomson showed that the same beam deflections resulted with tubes having cathodes made of different materials and containing various gases before evacuation.(d) Could Thomson observe any deflection of the beam due to gravitation? Do a calculation to argue for your answer. Note: To obtain a visibly glowing patch on the fluorescent screen, the potential difference between the slits and the cathode must be 100V or more.
Thomson could not observe any deflection of the beam due to gravitation.
What is a cathode ray?Cathode ray refers to the beam of electrons that moves from the negatively charged cathode to the positive anode at the other end of the pipe.
J. J. Thomson demonstrated that the identical beam deflections produced tubes with cathodes composed of varied materials and containing a range of gases prior to evacuation.
The gravitational force is approximately 10⁴⁰.
The movement of the light beam shouldn't be impacted because it is 10²⁸ times weaker than the electric force.
Thomson was unable to detect any gravitational beam deflection.
Thus, Thomson could not observe any deflection of the beam due to gravitation.
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A 500N load is lifted to a height in 50 sec. Calculate the power.
Answer:
25000w
Explanation:
P=Fxs
p=500×50
p=25000 watts
Is Ampère's law valid for all closed paths surrounding a conductor? Why is it not useful for calculating →B for all such paths?
Although not always practical, Ampere's law is true for all closed channels around a conductor. The integral can be calculated along many different paths, although it is difficult to do so. Think about a path that is circular but not coaxial with a long, straight wire that carries current. If the current in a conductor has enough symmetry to allow the line integral to be reduced to the size of B times an integral, Ampere's law can be used to calculate B.
What applications does Ampere's Law have?Ampere's Law is applied to,
Find the magnetic induction caused by a long wire carrying current.Analyze the magnetic field a toroid has.Find the magnetic field produced by a long conducting cylinder that carries current.Ascertain the conductor's magnetic field.To learn more about Ampere's Law, visit:
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sheila uses a 45N force on her bowling ball across a 15M lane. what work did she do on the bowling ball? show your work.
Answer:
Work=force × distance
=45×15
=675
Explanation:
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which names are correct for fm−→−? select each correct answer. fp−→− ray f p fd−→− ray f d mf−→− ray m f fw−→−
The correct answer is FM-MF
The crust of the Earth does not naturally contain ferromium. The heaviest element that can be created by neutron bombardment of lighter elements, fermium was first discovered in December 1952 by American scientists from the Argonne National Laboratory near Chicago, Illinois, the Los Alamos National Laboratory in Los Alamos, New Mexico, and The University of California Laboratory in Berkeley. Fermium is the heaviest element that can be synthesized in macroscopic quantities.
We can prove this by saying that if a point's distance from another point is six units, then the distance between those two points will also be six units. Therefore, FM-MF
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Explain how inductive reasoning is inferior to deductive reasoning in establishing theory.
Answer: Created Fall 2016 by Ronald Wilson Inductive reasoning uses a set of specific observations to reach an overarching conclusion; it is the opposite of deductive reasoning. So, a few particular premises create a pattern which gives way to a broad idea that is likely true.
Explanation: No explanation lol. But I hope this is helpful!
suppose the aqueous humor in a person’s eye exerts a force of 0.295 n on the 1.15-cm2 area of the cornea.
The aqueous humor in a person’s eye exerts a force of 0.295 n on the 1.15-cm2 area of the cornea pressure is this in mm Hg P = 2565.2173 N/m².
Equation :Pressure is defined as the normal force F per unit area A over which the force is applied or
P = F / A
Where :
P is the pressure acting on the cornea .
F if the force acting on the cornea .
A is the area of the cornea .
Givens : F = 0.3 N and A = 0.00011 m
Plugging known information to get :
P = F / A
P = 0.295 / 0.000115
P = 2565.2173
P =2565.2173 N/m²
What produces aqueous humor?The ciliary body produces aqueous humor, which travels along the uveoscleral pathway from the posterior chamber through the pupil and anterior chamber (shown by dashed lines and arrowheads) to the ciliary muscle and suprachoroidal region.
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I understand the question you are looking for :
suppose the aqueous humor in a person’s eye exerts a force of 0.295 n on the 1.15-cm2 area of the cornea. What pressure is this in mm Hg?
would a larger, more heavier ball fall faster than a smaller and lighter ball?
Answer:
the larger-heavier would fall faster than lighter-smaller ball
Explanation:
A linearly polarized microwave of wavelength 1.50cm is directed along the positive x axis. The electric field vector has a maximum value of 175V/m and vibrates in the x y plane. Assuming the magnetic field component of the wave can be written in the form B=Bmax sin (k x-Ω t) give values for (f) If this wave were directed at normal incidence onto a perfectly reflecting sheet, what radiation pressure would it exert?
The radiation pressure directed at normal incidence onto a perfectly reflecting sheet is 2.71 × 10⁻⁷ Pa
What radiation pressure would it exert?Given that a linearly polarized microwave of wavelength 1.50cm is directed along the positive x axis and the electric field vector has a maximum value of 175V/m and vibrates in the x y plane.
Also, assuming the magnetic field component of the wave can be written in the form B=Bmax sin (k x-Ω t) give values for (f) If this wave were directed at normal incidence onto a perfectly reflecting sheet, we desire to find the radiation pressure.
The radiation pressure for total reflection is given by P = 2I/c where
I = intensity = E²/2μ₀c where E = maximum electric field = 175 V/m, μ₀ = permeability of free space = 4π × 10⁻⁷ T/m and c = speed of light = 3 × 10⁸ m/sSo, P = 2I/c
P = 2E²/2μ₀c ÷ c
P = E²/μ₀c²
Substituting the values of the variables intot he equation, we have
P = E²/μ₀c²
P = (175 V/m)²/[4π × 10⁻⁷ T/m × (3 × 10⁸ m/s)²]
P = 30625 V²/m²/[4π × 10⁻⁷ T/m × 9 × 10¹⁶ m²/s²]
P = 30625 V²/m²/[36π × 10⁹ Tm/s²]
P = 30625 V²/m²/113.097 × 10⁹ Tm/s²
P = 270.78 × 10⁻⁹ Pa
P = 2.7078 × 10⁻⁷ Pa
P ≅ 2.71 × 10⁻⁷ Pa
So, the radiation pressure directed at normal incidence onto a perfectly reflecting sheet is 2.71 × 10⁻⁷ Pa
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please help me swear will mark brainiest
Answer:
I think that is d option 1.0Nm anticlockwise
is parabolic mirror used for finiding out the intensity of light?
The correct answer is :
Mirror Parabola A mirror with a parabola-shaped reflecting surface is referred to as a parabolic mirror. The mirror is employed in search lights because it can precisely converge a broad parallel beam of light at the focus. Therefore, even at a great distance, the intensity of the reflected beam is unaffected.
Parabolic In search lights, concave mirrors with sizable apertures are utilized. For searching items in search light, we require a powerful parallel beam of light. The reflected rays from an object placed at a concave mirror's focus are parallel to one another and likewise parallel to the main axis. So, we position a light source at the concave mirror's focal point.
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Suppose we have a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna, what is the velocity of the well fed shark?
Answer:
2.64 m/s
Explanation:
Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna
Momentum = MV
Momentum of the yellow shark before collision = 600 × 3 = 1800 kgm/s
Momentum of the tun final before collision = 100 × 0.5 = 50 kgm/s
Total momentum before collision = 1800 + 50 = 1850 kgm/s
Let's assume that they move together after collision. Then,
1850 = ( 600 + 100 ) V
1850 = 700V
V = 1850 / 700
V = 2.64285 m/s
Therefore, the momentum of the shark after collision is 2.64 m/ s approximately
The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n=1.33 is the index of refraction of water.(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc)
It is proved that the speed of the light measured in the laboratory frame is , u =c/n(1+ nv/c)/(1+ v/nc) .
Given ,
The motion of a transparent medium influences the speed of light .
The water moves with speed v in a horizontal pipe .
Assume that the light travels in the same direction as the water moves .
The speed of the light with respect to the water is c/n
Where n = 1.33 is the refractive index of water .
Let us assume ,
u' be the speed of light in water , in the frame moving with the water .
u' is related to the refractive index of water ,n as :
u'=c/n
where , c is the speed of light .
let , u be the speed of light in water in the lab frame .
Now , u and u' are related as : u = (u'+ v )/(1+ u'v/c^2)
Here v is the speed of water in the horizontal pipe .
we know the value of u' , so by substituting the value , we will get ,
u= (c/n+ v)/(1+cv/nc^2)
u= c/n(1+ nv/c)/(1+v/nc)
Hence , it is proved that the speed of the light measured in the laboratory frame is , u =c/n(1+ nv/c)/(1+ v/nc) .
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will anyone give me a free amazon card
how would you derive relation between momentum and newton's second law
Answer:
F = Δ(mv)/Δt = ma
Explanation:
Newton's second law states that the force exerted on an object is equal to the rate of change of its momentum.
Let F represent force and p momentum and p = mv where m = mass and v = velocity
This implies that F = Δp/Δt = Δ(mv)/Δt if the mass is constant, we have
F = mΔv/Δt and we know that acceleration, a = Δv/Δt
So, F = ma which is Newton's second Law.
Show FORMULA AND SOLUTION:
What acceleration will result when a 12N net force applied to a 3kg object? A 6kg object?
Answer:
2m/s²
Explanation:
F = 12 N
m = 3kg
a = ?
F = ma
a = F/m = 12/3 = 4 m/s²
when m = 6kg
a = F/m = 12/6
a = 2 m/s²
Omar and ahmed are lifting weights in the gym. Each lifts a weight of 200N. Omar lifts the weight to a height of 2.0m, whereas ahmed lifts it to a height of 2.1m. Who does more work in lifting the weight.
Answer:
Ahmed did more work in lifting the weight than Omar.
Explanation:
Work done can be determined as the product of a force and the distance moved in the direction of the force.
Work done = F x s
In the given question, since the weight is lifted through a height, then the work done equals potential energy.
i.e Work done = Potential energy
= mgh
But,
Force = weight = mg
So that,
Work done = Fh
where F is the weight and h is the height.
i. Work done by Omar = Fh
= 200 x 2.0
= 400
Work done by Omar is 400 Nm.
ii. Work done by Ahmed = Fh
= 200 x 2.1
= 420
Work done by Ahmed is 420 Nm.
Thus, Ahmed did more work in lifting the weight than Omar.
Under which condition is the resultanat vector equal to zero?
Kangaroos are strong jumpers and can leap over an object 1.8 m high. Calculate a kangaroo's minimum vertical speed when it leaves the ground.
The kangaroo's minimum vertical speed when it leaves the ground is 5.94 m/s
Data obtained from the questionThe following data were obtained from the question given above:
Maximum height (h) = 1.8 mFinal velocity (v) = 0 m/s (at maximum height) Acceleration due to gravity (g) = 9.8 m/s²Initial velocity (u) =?How to determine the initial velocity of the kangarooWe can determine the inintial velocity of the kangaroo as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = u² – (2 × 9.8 × 1.8)
0 = u² – 35.28
Collect like terms
u² = 0 + 35.28
u² = 35.28
Take the square root of both side
u = √35.28
u = 5.94 m/s
From the above calculation, we can conclude that the minimum vertical speed of the kangaroo when it leaves the ground is 5.94 m/s
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An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same direction with a speed of 0.800 c relative to the mother ship. As measured on the Earth, the spaceship is 0.200 ly from the Earth when the landing craft is launched.(b) What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens?
The distance to the Earth at the moment of the landing craft's launch as measured by the aliens is 0.16 light years.
To find the answer, we have to know about the Lorentz transformation.
What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens?It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the distance to the Earth at the moment of the landing craft's launch as measured by the aliens.
It is given that; the spaceship is 0.200 ly from the Earth when the landing craft is launched.By Lorentz transformation equation, the value of the distance to the Earth at the moment of the landing craft's launch as measured by the aliens will be,[tex]L=L_0\sqrt{1-\frac{V^2}{c^2} } =0.2ly\sqrt{1-\frac{(0.6c)^2}{c^2} }=0.2ly\sqrt{1-0.36}= 0.16ly[/tex]
Thus, we can conclude that, the distance to the Earth at the moment of the landing craft's launch as measured by the aliens is 0.16 light years.
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