Answer:
Do we have to choose out of those three?
Explanation:
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process. Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m3 to 0.026 m3 while doing work on a piston.
This question is incomplete, the complete question is;
The entropy of an ?-ideal gas changes in the following way as a function of temperature and volume:
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.
Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m³ to 0.026 m³ while doing work on a piston.
1) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy? ΔS = ? J/K
2) For this adiabatic expansion, what is the final temperature? T[tex]_f[/tex] = ? K
Answer:
1) the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) the final temperature is 158.66 K
Explanation:
Given the data in the question;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
P[tex]_i[/tex] = 100 kPa = 100000 Pa
V[tex]_i[/tex] = 0.01 m³
V[tex]_f[/tex] = 0.026 m³
T[tex]_i[/tex] = 300 K
1) the change in entropy due to the volume change alone
from the question; ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
so change in entropy due to the volume change alone is;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex])
we know that, from ideal gas law; PV = nRT
so, nR = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] ---- let this be equation 1
∴ ΔS = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] × ln(V[tex]_f[/tex]/V[tex]_i[/tex])
we substitute
ΔS = [( 100000 Pa × 0.01 m³) / 300 K ] × ln(0.026m³ / 0.01m³ )
ΔS = 3.185 J/K
Therefore, the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) Final temperature
we know that, in an adiabatic expansion;
[tex]PV^Y[/tex] = K
where Y = 5/3
so
[tex]P_i[/tex][tex]V_i^{(5/3)[/tex] = [tex]P_f[/tex][tex]V_f^{(5/3)[/tex]
[tex]P_f[/tex] = [tex]P_i[/tex][tex]( \frac{V_i}{V_f})^{(5/3)[/tex]
we substitute
[tex]P_f[/tex] = ( 100000 Pa) [tex]( \frac{0.01 m^3}{0.026 m^3})^{(5/3)[/tex]
[tex]P_f[/tex] = 20341.255 Pa
Also from ideal gas law;
PV = nRT
T = PV / nR
so
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
but from equation 1; nR = PV/T
so
T[tex]_f[/tex] = (P[tex]_f[/tex]V[tex]_f[/tex]) / (P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] )
T[tex]_f[/tex] = ( P[tex]_f[/tex]V[tex]_f[/tex]T[tex]_i[/tex] / P[tex]_i[/tex]V[tex]_i[/tex] )
we substitute
T[tex]_f[/tex] = ( 20341.255 Pa × 0.026 m³ × 300 K) / 100000 Pa × 0.01 m³ )
T[tex]_f[/tex] = 158.66 K
Therefore, the final temperature is 158.66 K
a 0.1 kg object oscillates as a simple harmonic motion along x axis with a frequency f=3.185 hz. At a position x1 , the object has a kinetic energy of o.7 j and a potential energy 0.3 J.The amplitude of oscillation A is:
Answer:
The total energy must be .7 J + .3 J = 1 J for a particle at the endpoint or midpoint of motion.
Also, omega = (k / m)^1/2
f = omega / (2 * pi)
omega^2 = 4 pi^2 * f^2 = k / m
k = 4 * pi^2 * f^2 * m = 40.05
Max KE or PE = 1/2 k A^2
A^2 = 2 * E / k = 2 * 1 / k = .0499
A = .223 meters
Electron spin: Radio astronomers can detect clouds of hydrogen too cool to radiate optical wavelengths of light by means of the 21 cm spectral line corresponding with the flipping of the electron in a hydrogen atom from having its spin parallel to the proton spin to having it antiparallel. From this wavelength, and thus E between states, find the magnetic field experienced by the electron in a hydrogen atom
Answer:
the magnetic field experienced by the electron is 0.0511 T
Explanation:
Given the data in the question;
Wavelength λ = 21 cm = 0.21 m
we know that Bohr magneton μ[tex]_B[/tex] is 9.27 × 10⁻²⁴ J/T
Plank's constant h is 6.626 × 10⁻³⁴ J.s
speed of light c = 3 × 10⁸ m/s
protein spin causes magnetic field in hydrogen atom.
so
Initial potential energy = -μ[tex]_B[/tex]B × cos0°
= -μ[tex]_B[/tex]B × 1
= -μ[tex]_B[/tex]B
Final potential energy = -μ[tex]_B[/tex]B × cos180°
= -μ[tex]_B[/tex]B × -1
= μ[tex]_B[/tex]B
so change in energy will be;
ΔE = μ[tex]_B[/tex]B - ( -μ[tex]_B[/tex]B )
ΔE = 2μ[tex]_B[/tex]B
now, difference in energy levels will be;
ΔE = hc/λ
2μ[tex]_B[/tex]B = hc/λ
2μ[tex]_B[/tex]Bλ = hc
B = hc / 2μ[tex]_B[/tex]λ
so we substitute
B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]
B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]
B = 510556326.09
B = 0.0511 T
Therefore, the magnetic field experienced by the electron is 0.0511 T
The volumes of two bodies are measured to be
V₁ = (10.2 ± 0.02) cm³ and V₂ = (6.4 ± 0.01) cm³. Calculate sum and difference in
volumes with error limits.
Answer:
sum of volumes = (16.6 ± 0.03) cm³
and difference of volumes = (3.8 ± 0.03) cm³
Explanation:
Here,
V₁ = (10.2 ± 0.02) cm³ and V₂ = (6.4 ± 0.01) cm³.
Now,
∆V = ± (∆V₁ + ∆V₂)
= ± (0.02 + 0.01) cm³
= ± 0.03 cm³
V₁ + V₂ = (10.2 + 6.4) cm³ = 16.6 cm³ and
V₁ - V₂ = (10.2 - 6.4) cm³ = 3.8 cm³
Thus, sum of volumes = (16.6 ± 0.03) cm³
and difference of volumes = (3.8 ± 0.03) cm³
-TheUnknownScientist
Answer:
I hope it's helpful .............
Which of the following is NOT a geological event caused by the movement of tectonic plates?
A) Sand Dunes
B) Mountain forming
C) Earthquakes
D) Volcanoes
Which statement describes what some people might consider a benefit of wireless technology?
O Micah interacts with people to play video games.
O Nadia pays a monthly fee to have high-speed internet for her tablet.
O Luca relies on a printed store receipt instead of the emailed version.
O Quinn watches television for hours while her friends spend time together.
Help
Although the use of absorbances at 450 nm provided you with maximum sensitivity, the absorbances at, say, 400 nm or 500 nm are not zero and could have been used throughout this experiment. Would the same value of K be obtained at one of these wavelengths
Answer:
Yes, the value will be the same.
Explanation:
Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.
What turns the drive shaft of the generator?
Help
Answer:
Produces 60-cycle AC electricity; it is usually an off-the-shelf induction generator. High-speed shaft: Drives the generator. Low-speed shaft: Turns the low-speed shaft at about 30-60 rpm. Nacelle: Sits atop the tower and contains the gear box, low- and high-speed shafts, generator, controller, and brake.
Explanation:
A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How many radians did the disk turn while stopping ? C. how many revolutions?
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
[tex]\omega_f = \omega_i + \alpha t[/tex]
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
[tex]0\ rad/s = 3.14\ rad/s + \alpha(3\ s)[/tex]
α = - 1.047 rad/s²
B.
We can use the second equation of motion to find the angular distance:
[tex]\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2[/tex]
θ = 14.1 rad
C.
θ = (14.1 rad)(1 rev/2π rad)
θ = 2.24 rev
what causes the moon to change its appearance over a month?
Answer:
The reason we see different phases of the Moon here on Earth is that we only see the parts of the Moon that are being lit up by the Sun. When the Moon is between Earth and the Sun, the lit side is hidden from us. As it moves around Earth, more and more of the lit side comes into view. Then it begins to disappear again, creating the different phases we see.
Explanation:
3. A ball thrown vertically upward returns to its starting point in 4s. Find its initial speed. [4]
Answer:
9.8 ×4 equal 39.2 m/s This is v intial
Calculate the equivalent resistance for each of the following circuits.
Answer:
9. 4.8 Ω
10. 0.92 Ω
Explanation:
9. Determination of the equivalent resistance.
Resistor 1 (R₁) = 10 Ω
Resistor 2 (R₂) = 20 Ω
Resistor 3 (R₃) = 30 Ω
Resistor 4 (R₄) = 40 Ω
Equivalent Resistance (R) =?
The equivalent resistance can be obtained as follow:
1/R = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
1/R = 1/10 + 1/20 + 1/30 + 1/40
Find the least common multiple (lcm) of 10, 20, 30 and 40. The result is 120. Divide 120 by each of the denominator and multiply the result obtained by the numerator as shown below:
1/R = (12 + 6 + 4 + 3) / 120
1/R = 25 / 120
Invert
R = 120 / 25
R = 4.8 Ω
Thus, the equivalent resistance is 4.8 Ω
10. Determination of the equivalent resistance.
Resistor 1 (R₁) = 2 Ω
Resistor 2 (R₂) = 3 Ω
Resistor 3 (R₃) = 4 Ω
Equivalent Resistance (R) =?
The equivalent resistance can be obtained as follow:
1/R = 1/R₁ + 1/R₂ + 1/R₃
1/R = 1/2 + 1/3 + 1/4
Find the least common multiple (lcm) of 2, 3 and 4. The result is 12. Divide 12 by each of the denominator and multiply the result obtained by the numerator as shown below:
1/R = (6 + 4 + 3) / 12
1/R = 13 / 12
Invert
R = 12 / 13
R = 0.92 Ω
Thus, the equivalent resistance is 0.92 Ω
balance Al +N₂ --> AIN
Answer:
[tex]2Al + N_{2} ==> 2AlN[/tex]
Explanation:
A 1.00 x 109 kg object is raised vertically at a
constant velocity of 4.00 m/s by a crane. What
is the power output of the crane?
Answer:
P = 3.92 10¹⁰ W
Explanation:
The power is data by the expression
P = W / t
the work of a force is
W = F. y
the bold ones represent vectors. In this case the displacement is vertical upwards and the vertical forces upwards, therefore the angle is zero and the cos 0 = 1
W = F y
we substitute
P = F y / t
P = F v
as the body rises at constant speed the acceleration is zero and from the equilibrium condition
F -W = 0
F = mg
we substitute
P = m g v
let's calculate
P 1.00 10⁹ 9.8 4
P = 3.92 10¹⁰ W
What would be a good reason to increase friction between surfaces? to allow objects to slide past each other easily to reduce wear and tear to provide traction so that slipping does not occur to avoid contact that results in increased temperature and overheating
Answer:
B to provide traction so that slipping does not occur
Explanation:
To increase friction between two surfaces, allow the objects to slide past each other thereby force that is called friction.
What is friction?Friction is a type of force acting on an object to hinder its motion . It is negative by direction, since this force, hinder the motion of the object. Friction is also known as an opposing force because it always acts in the opposite direction of a moving or attempting to move body.
The virtue of friction causes a moving body to slow down. Friction is useful at times because it prevents car tires from skidding on the road and also allows us to walk on the pavement without slipping.
To increase friction , make an uneven, rugged, or sticky point of contact. When two or more bodies slide or rub against each other, three things can happen: small irregularities, nooks and crannies on the surfaces can catch on each other.
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PLEASE HELP ME PLEASE! BRAINLEST
Answer:
10kg
Explanation:
Weight is "how much does gravity drag this down".
Mass is "how much matter is there here".
The relation is:
[tex]F_g = mg[/tex]
where [tex]F_g[/tex] is the weight, [tex]m[/tex] is the mass and [tex]g[/tex] is the gravitational acceleration (roughly equal to 10N/kg on Earth).
From the task we know that:
[tex]F_g = 100N\\g = 10\frac{N}{kg}[/tex]
So let's input it into the relation:
[tex]100N = m\cdot 10\frac{N}{kg}\\10N = m \cdot 1\frac{N}{kg}\\10N \cdot \frac{kg}{N} = m\\~\\m = 10kg[/tex]
A 10Ω and a 15Ω resistor are connected in series across a 110V potential difference. (Can you find them) please help
A) what is the total resistance of the circuit?
B) what is the current through each resistor?
C) what is the voltage drop across each resistor
Answer:
(A) The total resistance of the circuit is 25 Ω
(B) The current through each resistor is 4.4 A
(C) For 10Ω: Potential drop = 44 V
For 15Ω: Potential drop = 66 V
Explanation:
Given;
potential difference, V = 110V
resistors in series, = 10Ω and a 15Ω
(A) The total resistance of the circuit is calculated as follows;
Rt = 10Ω + 15Ω = 25Ω
(B) The current through each resistor;
Same current will flow through the two resistors since they are in series.
I = V/Rt
I = 110 / 25
I = 4.4 A
(C) The voltage drop across each resistor;
For 10Ω: Potential drop = IR₁ = 4.4 x 10 = 44 V
For 15Ω: Potential drop = IR₂ = 4.4 x 15 = 66 V
Helppp answer question pic in photo
Research the main categories of mental disorders and their evaluation and assessments. For example, you can use your favorite Internet search engine to look up the following sites:
National Institute of Mental Health
Mental Help.net
American Psychological Association Help Center
Step 2 Report on methods of diagnosis and assessments of the major mental disorders.
Write a one and one-half-page report on mental disorders, their diagnosis, and treatment. Identify the following elements in your paper:
The major categories of mental disorders. ( Hint: Anxiety disorders are one category.)
The commonly used methods of diagnosis.
The treatment of choice (meaning the favorite current treatment) for this disorder.
Answer:
D
Explanation:
It’s D
Where does lymph come from?
A lens with f= 20.0 cm creates a
virtual image at -37.5 cm (in front
of the lens). What is the object
distance?
Answer:
13.04
Explanation:
credit to the comment above
A lens with f= 20.0 cm creates a virtual image at -37.5 cm, then the object distance is approximately 0.0767 cm.
To determine the object distance in this scenario, we can use the lens formula:
[tex]\rm \frac{1}{f} =\frac{1}{v} -\frac{1}{u}[/tex]
Here, we have:
f = 20.0 cm (positive for a converging lens)
v = -37.5 cm (negative because it is a virtual image formed in front of the lens)
Substituting the values into the lens formula:
1÷20.0 = 1 ÷ (-37.5) - 1 ÷ u
Simplifying the equation:
1 ÷ 20.0 + 1÷ 37.5 = 1 ÷ u
Multiplying through by the common denominator:
(37.5 + 20.0) ÷ (20.0 * 37.5) = 1 ÷ u
57.5 ÷ 750 = 1 ÷ u
Dividing both sides by 57.5:
1 ÷ u = 750 ÷ 57.5
u = 57.5 ÷ 750
u ≈ 0.0767 cm
Thus, the object distance is approximately 0.0767 cm.
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According to Kepler's second law at the two shaded areas above are equal then...
A. it takes the same amount of time for planet A to travel between T1 and T2 as it does to travel between T3 and T4
B. Planet A will cause an eclipse on Planet B during those time intervals
C. it takes Planet A more time to travel between T1 and T2 as it does to travel between T3 and T4
D. it takes Planet A less time to travel between T1 and T2 as it does to travel between T3 and T4
Answer:
A. Kepler's second law says that the radius vector sweeps equal areas in equal times.
Please please help me please
A piece of irregularly shaped metal weighs 300N in air. When the metal is completely submerged in water, it weighs 232.5N. Find the volume and specific gravity of the metal.
Answer:
Volume of metal piece = 0.0069 m³ (Approx.)
Explanation:
Given:
Weight of metal in air = 300 N
Weight of metal in water = 232.5 N
Find:
Volume of metal piece
Specific gravity of metal
Computation:
We know that;
Density of water = 1,000 kg/m³
Buoyant force applied on metal piece = Weight of metal in air - Weight of metal in water
Buoyant force applied on metal piece = 300 N - 232.5 N
Buoyant force applied on metal piece = 67.5 N
Buoyant force = Volume of metal x Density of water x Gravitational force
67.5 = Volume of metal x 1,000 x 9.8
Volume of metal piece = 0.0069 m³ (Approx.)
Help answer question in licture
Answer:
D
Explanation:
D is high pitch c is low pitch
Answer:
D creo que esa es la nota mas alta, la segumda
1. If a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of
the spring?
d. What is the net force on the bowling ball rolling lane
Answer:
Friction.
Explanation:
You set a tuning fork into vibration at a frequency of 683 Hz and then drop it off the roof of the Physics building where the acceleration due to gravity is 9.80 m/s2. Determine how far the tuning fork has fallen when waves of frequency 657 Hz reach the release point
Answer:
The distance traveled by the tuning fork is 9.37 m
Explanation:
Given;
source frequency, [tex]f_s[/tex] = 683 Hz
observed frequency, [tex]f_o[/tex] = 657 Hz
The speed at which the tuning fork fell is calculated by applying Doppler effect formula;
[tex]f_o = f_s [\frac{v}{v + v_s} ][/tex]
where;
[tex]v[/tex] is speed of sound in air = 343 m/s
[tex]v_s[/tex] is the speed of the falling tuning fork
[tex]657 = 683[\frac{343}{343 + v_s} ]\\\\\frac{657}{683} = \frac{343}{343 + v_s}\\\\0.962 = \frac{343}{343 + v_s}\\\\0.962(343 + v_s) = 343\\\\343 + v_s = \frac{343}{0.962} \\\\343 + v_s = 356.55\\\\v_s = 356.55 - 343\\\\v_s = 13.55 \ m/s[/tex]
The distance traveled by the tuning fork is calculated by applying kinematic equation as follows;
[tex]v_s^2 = v_o^2 + 2gh[/tex]
where;
[tex]v_o[/tex] is the initial speed of the tuning fork = 0
g is acceleration due to gravity = 9.80 m/s²
[tex]v_s^2 = 0 + 2gh\\\\h = \frac{v_s^2}{2g} \\\\h = \frac{13.55^2 }{2\times 9.8} \\\\h = 9.37 \ m[/tex]
Therefore, the distance traveled by the tuning fork is 9.37 m
Two identical objects, A and B, move along straight, parallel, horizontal tracks. The graph above represents the position as a function of time for the two objects.
(a) At a time of 2 seconds, where the lines intersect, do the displacements of the two objects from their initial positions have the same magnitude? Briefly explain your answer.
(b) At a time of 2 seconds, where the lines intersect, do the velocities of the two objects have the same magnitude? Briefly explain your answer.
(c) At a time of 2 seconds, where the lines intersect, which object, if either, has a net force with a greater magnitude exerted on it? If the net force has the same magnitude for both objects, indicate this explicitly
(d) In a clear, coherent paragraph-length response, explain your response to part (c). Be sure to reference and compare the graphed information for both objects A and B.
Answer:
After a little online search, I've found the graph of this question, the graph can be seen below.
a) The displacement is defined as the distance between the final position and the initial position.
In the graph, the vertical axis represents the distance. We also can see that both of the lines start in position 0, so at any given time, the displacement of the objects is given by the vertical position in the graph.
Thus, at t = 2 seconds, both lines have the same y-value, this means that the displacements have the same magnitude.
b) The velocity is related to the slope of the curve,
We can clearly see that the slope of graph A and the slope of gaph B are different at t = 2 seconds (graph A is steeper) then we can conclude that the velocities do not have the same magnitude.
c) By Newton's second law, we know that F = m*a
Force equals mass times acceleration.
Acceleration is the rate of change of velocity. When the position graph is a straight line, in any point of the line the slope will be the same, thus the object has always the same velocity, thus the object is not accelerated.
If we do not have a straight line (like in graph A) then the velocity is changing, then we have acceleration, then we have a force.
Then object A has a greater net force (because object B has a net force equal to 0)
d) It is already explained in point c.
Which feature of electromagnets makes them more useful than permanent
magnets in many modern technologies?
A. Electromagnets are not dependent on a power supply.
B. The strength of the magnetic field is more consistent in
electromagnets.
C. Electromagnets cannot be turned on and off.
D. The current can be adjusted to control the strength of the
magnetic field.
Answer:
The current can be adjusted to control the strength of the
magnetic field.