Wo thin uniformly charged rods, each with length L and total charge +Q, are parallel and separated by a distance a. The first rod has one end at the origin and its other end on the positive y-axis. The second rod has its lower end on the positive x-axis
Determine the x-component of the differential force dF2 exerted on a small portion of the second rod, with length dy2 and position y2, by the first rod. (This requires integrating over differential portions of the first rod, parameterized by dy1. )
Answers
The x-component of differential force is, dF2x = kQ²/L² [1/(y2-a) - 1/(y2+a-L)].
Let's consider a small segment of the second rod, with length dy2 and position y2. We want to find the x-component of the differential force dF2 exerted on this segment by the first rod.
The x-component of the electric field vector at the position of the segment is given by the product of the total electric field and the cosine of the angle between the electric field vector and the x-axis.
The total electric field at the position of the segment is given by the integral of the electric field due to the first rod over all its elements dl, which are parameterized by dy1:
E = \(\int k\lambda \dfrac{1}{((y_2-a)^2+y_1^2)^{1/2}} cos\theta dx_1\)
where λ1 is the linear charge density of the first rod, θ is the angle between the line connecting the element dl of the first rod and the position of the segment, and dx1 is an element of length along the first rod.
Using the geometry of the problem, we can express cosθ in terms of y1, y2, a, and L:
cosθ = (y1(L-y2))/[(y2-a)²+y1²]^(1/2)L
Substituting this expression into the integral,
E = \(k\lambda_1 L\int dy_1 \dfrac{(y_1(L-y_2))}{[(y_2-a)^2+y_1^2]^{3/2}}\)
Integrating this expression over the length of the segment, we get the x-component of the differential force dF2:
dF2x = \(\int Edq 2 cos\theta\)
where dq2 is the charge on the segment:
dq2 = λ2dy2 = Q/L dy2
Substituting the expressions for E and cosθ, and performing the integration, we get:
dF2x = \(\dfrac{kQ^2}{L^2} \int dy_1 \dfrac{y_1(L-y_2)}{[(y_2-a)^2+y_1^2]^{3/2}}\)
This integral can be evaluated by making the substitution u = y2-a, which gives:
dF2x = kQ²/L² [1/(y2-a) - 1/(y2+a-L)]
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Related Questions
Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in front of A. If the speed of sound in this room is 340m/s will the observer hear a loud sound or a quiet sound?
Answers
Answer:
The observer hears a loud sound
Explanation:
In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.
You first calculate the distance between the observer and the loudspeakers.
The distances are given by:
d1: distance to loudspeaker A = 2.10m
d2: distance to loudspeaker B
\(d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m\)
Next, you calculate the wavelength of the sound waves by using the following formula:
\(\lambda=\frac{v_s}{f}\)
vs: speed of sound = 343 m/s
f: frequency of the waves = 400Hz
λ: wavelength
\(\lambda=\frac{343m/s}{400Hz}=0.8575m\)
Next, you calculate the path difference between the distance from the observer to the loudspeakers:
\(\Delta d=3.827m-2.10m=1.727m\)
You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:
\(\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2\)
Then, there will be a constructive interference, and the sound who the observer hears is loud.
How much work is done by 15 kW engine during 3.0 min?
Answers
W = 3.15 · 10⁶ J work is done by 15 kW engine during 3.0 min.
We are aware that power is determined using the equation P = W/t, where P equals power (W), W equals work (J), and t equals time (s)
Using this formula, we can create a formula for work by adding t to both sides of the equation.
W = Pt
Converting 15 kW to W yields 15,000 W.
3.0 minutes is 180 seconds, therefore let's convert that.
Solve for W by substituting these numbers into the work formula.
W = Pt
W = (15,000 W) · (180 s) (180 s)
W = 2700000 J
W = 2.7 · 10⁶ J
This is the amount of work a 15 kW engine can accomplish in three minutes.
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Please ANSWER!!!!
BRAINLIEST will be named!!!!!!!!!!!!!!
Answers
Based on radiometric dating of Apollo rock samples, the rocks have been detected to be about 4.5 Billion years old.
Hope this helps!
If an unbalanced force acts on an object, what will happen?
Answers
Answer:
An unbalanced force acting on an object causes it to accelerate. There are two points to note about the acceleration of an object when an unbalanced force acts on it. The bigger the unbalanced force acting on the object the bigger the acceleration of the object.Explanation:
An element in an electrical heating unit is applied to a 232-volt power supply. The current flow through the element is 19 amps. What is the resistance of the element?
Answers
The resistance of the element in an electrical heating unit when applied to a 232-volt power supply with a current flow of 19 amps is approximately 12.21 ohms.
From Ohm's Law, the relationship between voltage, current and resistance is given byV = IR, where V is voltage, I is current, and R is resistance. Substituting the given values in the equation, V = IR232 = 19R
Rearranging the equation, we have R = V/I = 232/19
The resistance of the element in an electrical heating unit when applied to a 232-volt power supply with a current flow of 19 amps is approximately 12.21 ohms.
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The set up time of a flip-flop is 10 ps, the hold-time is 50 ps, the de lay of the flip-flop is 10 ps. The delay of logic circuits in feedback is 70 ps. What is the maximum frequency of the clock that can be used to operate the state-machine correctly? A) 7.1 GHz B) 7.7 GHz C) 11.1 GHz D) 14.2 GHz
Answers
Option A is correct. The maximum frequency of the clock that can be used to operate the state machine correctly is 7.1 GHz.
To determine the maximum clock frequency, need to consider the setup time, hold time, and delay of the flip-flop, as well as the delay of logic circuits in feedback. The maximum frequency can be calculated using the formula:
Maximum Frequency = 1 / (2 * (Setup Time + Hold-Time + Delay Flip-Flop + Delay Feedback))
Substituting the given values:
Maximum Frequency = 1 / (2 * (10 ps + 50 ps + 10 ps + 70 ps))
Maximum Frequency = 1 / (2 * 140 ps) = 1 / 280 ps
Converting the frequency to GHz:
Maximum Frequency = \(1 / (280 ps * 10^{-12} s/ps * 10^9 Hz/GHz)\)
Maximum Frequency ≈ 7.1 GHz
Therefore, the maximum frequency of the clock that can be used to operate the state machine correctly is approximately 7.1 GHz.
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A 2.0 kg block is released from rest at the top of a rough plane inclined at 37° to the horizontal, as shown
below. The block slides down the incline with an acceleration of 4.0 m/s. What is the approximate magnitude
of the force of friction on the block as it slides down the incline?
Answers
Approximate magnitude of the force of friction on the block as it slides down the incline is calculated as 4N.
What is force of friction?
Force of friction is the force resisting the relative motion of solid surfaces, fluid layers and material elements sliding against each other. There are many types of friction like static friction, sliding friction, rolling friction, and fluid friction. Static friction is the strongest which is followed by sliding friction and rolling friction is the weakest. Fluid friction takes place in fluids which are liquids or gases.
Given mass is 2 kg and
acceleration is 4.0 m/s
m *g *sin 37° - F = m*a
F= m *g *sin 37° - m*a
=( 2* 9.81 * sin 37°- 2 *4)
=3.8 N
approximately equal to 4N
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Se colocan tres objetos, muy cerca uno del otro, dos al mismo tiempo. Cuando se juntan los objetos A y B, se repelen. Cuando se acercan los objetos B y C, se repelen. De los siguientes enunciados, ¿cuál es el verdadero? a) Los objetos A y C tienen cargas del mismo signo. b) Los objetos A y C poseen cargas de signos opuestos. c) Los tres objetos tienen cargas de distinto signo. d) Uno de los objetos es neutro. e) Es necesario llevar a cabo experimentos adicionales para determinar los signos de las cargas.
Answers
Answer:
Los objetos A y C tienen cargas del mismo signo (opcion a)
Explanation:
Hay dos tipos de cargas : cargas positivas y cargas negativas.
La Ley de Coulomb dice que la fuerza electrostática entre dos cargas puntuales es proporcional al producto de las cargas e inversamente proporcional al cuadrado de la distancia que las separa, y tiene la dirección de la línea que las une y se cumple que:
La fuerza ejercida sobre una carga apunta hacia la otra cuando las dos tienen distinto signo (fuerza atractiva). El sentido de la fuerza se dirige hacia el lado opuesto de la carga cuando ambas tienen el mismo signo (fuerza repulsiva).Es decir que las cargas de igual signo se repelen, mientras que las de diferente signo se atraen.
Entonces, si se juntan los objetos A y B y se repelen significa que la carga es del mismo signo.
Cuando se acercan los objetos B y C, se repelen. Entonces significa que posee carga de igual signo.
Por lo que podes concluir que los objetos A y C tienen cargas del mismo signo (opcion a)
what is latent heat of vaporization?
Answers
The amount of energy required for a unit mass of a substance to undergo a phase change from liquid to gas.
If a lever lifts a load four times the effort applied and effort distance is 5 times the load distance, calculate its efficiency
Answers
Answer:
If effort distance was 4 times, efficiency would be 100%.
Since it takes 5 times for effort distance, efficiency drops to output/input
output is 1*F
input is (1/4*F)*5
so: F/1/5*F/4 = 4F/5F = .8 or 80%
The efficiency of the lever is 80%.
To calculate the efficiency of the lever, we can use the formula for mechanical advantage and efficiency.
Mechanical Advantage (MA) is the ratio of the load (L) to the effort (E) in a lever system:
MA = Load / Effort
Given that the load is four times the effort applied:
Load = 4 * Effort
Also, the effort distance (dEffort) is five times the load distance (dLoad):
dEffort = 5 * dLoad
Now, we can write the formula for efficiency (η) of a lever system:
Efficiency (η) = (Mechanical Advantage / Ideal Mechanical Advantage) * 100%
The Ideal Mechanical Advantage (IMA) is the ratio of the effort distance to the load distance:
IMA = dEffort / dLoad
Substitute the given values into the IMA equation:
IMA = (5 * dLoad) / dLoad
IMA = 5
Now, we can calculate the Mechanical Advantage (MA) using the relationship between the load and effort:
MA = Load / Effort
MA = (4 * Effort) / Effort
MA = 4
Finally, we can calculate the efficiency (η):
Efficiency (η) = (Mechanical Advantage / Ideal Mechanical Advantage) * 100%
η = (4 / 5) * 100%
η = 0.8 * 100%
η = 80%
The efficiency of the lever is 80%.
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A uniform flat disk of radius R and mass 2M is pivoted at point P. A point mass of 1/2 M is attached to the edge of the disk. Calculate the total moment of inertia IT of the disk with the point mass with respect to point P, in terms of M and R.
Answers
The total moment of inertia IT of the disk with the point mass with respect to point P is 3/4 MR^2, in terms of M and R.
The total moment of inertia IT of the disk with the point mass with respect to point P can be calculated by using the parallel axis theorem, which states that IT = ICM + md^2, where ICM is the moment of inertia of the disk about its center of mass, m is the mass of the point mass, and d is the distance between the point mass and the pivot point P.
The moment of inertia of the disk about its center of mass ICM can be found from the formula for a uniform disk: ICM = 1/2 MR^2.
The distance between the point mass and the pivot point P is equal to the radius R of the disk.
Substituting these values into the parallel axis theorem, we get:
IT = ICM + md^2
= 1/2 MR^2 + (1/2 M)(R^2)
= 3/4 MR^2
Hence, the total moment of inertia IT of the disk with the point mass with respect to point P is 3/4 MR^2, in terms of M and R.
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4. A 70.0 kg boy and a 45.0 kg girl use an elastic rope while engaged in a tug-of-war on an icy,
frictionless surface. If the acceleration of the girl toward the boy is 2.25 m/s², find the
acceleration of the boy toward the girl. (3M law : m2 a2 = - ml al)
Given:
ml =
m2 =
a2=
Unknown:
al=?
Equation:
m2 a2= - ml al
Answers
Answer:
\(a_1 = 1.446m/s^2\)
Explanation:
Given
\(m_1 = 70.0kg\) -- Mass of the boy
\(m_2 = 45.0kg\) -- Mass of the girl
\(a_2 = 2.25m/s^2\) -- Acceleration of the girl towards the boy
Required
Determine the acceleration of the boy towards the girl (\(a_1\))
From the question, we understand that the surface is frictionless. This implies that the system is internal and the relationship between the given and required parameters is:
\(m_1 * a_1 = m_2 * a_2\)
Substitute values for \(m_1, m_2\) and \(a_2\)
\(70.0 * a_1 = 45.0 * 2.25\)
Make \(a_1\) the subject
\(\frac{70.0 * a_1}{70.0} = \frac{45.0 * 2.25}{70.0}\)
\(a_1 = \frac{45.0 * 2.25}{70.0}\)
\(a_1 = \frac{101.25}{70.0}\)
\(a_1 = 1.446m/s^2\)
Hence, the acceleration of the boy towards the girl is 1.446m/s^2
A periodic wave having a frequency of 18 hertz and a speed of 343 meters per second has a wavelength of ____ m. (**\.\*)
Answers
Answer:19.1m
Explanation:
A ball is thrown up into the air. When it falls half-way back, what kind of energy does it have?
Answers
Answer:
it's gravitational potential energy would have converted to kinetic energy half way back when it's falls
A bush pilot wants to fly her plane to a lake that is 250.0 km [N30.0°E] from her starting point. The plane has an air speed of 210.0 km/h, and a wind is blowing from the west at 40.0 km/h. (a) In what direction should she head the plane to fly directly to the lake? (b) If she uses the heading determined in (a), what will be her velocity relative to the ground? (c) How long will it take her to reach her destination?
Answers
(a) The plane must fly at an angle of N 35.5° E to immediately approach the lake.
(b) The velocity of the plane is 245.6 km/h.
(c) The time taken is 1.02 hours.
Lake displacement from the starting position, D = 250 km N 30° E
As the x-axis, let's choose the line connecting the plane's beginning point with the lake, which is located in the direction N 30° E.
D = 250 ^i km
The plane's airspeed,
v(p) = 210 km/h
From the east comes the wind. Therefore, using the coordinate system described above, the wind speed,
v(w) = 40 [ cos ( − 30° ) ^i + sin ( − 30°) ] km/h
v(w) ≈ ( 36.64 ^i − 20 ^j ) km/h
(a) To eliminate the perpendicular component of the wind velocity, move the plane in the direction θ with respect to the x-axis.
v(p) = 210 ( cos(θ) ^i + sin(θ) ^j ) km/h
In order for the y-component of the plane velocity to cancel out the y-component of the wind velocity, the correct angle must be calculated.
Therefore,
210 sin(θ) km/h = 20 km/h
sin(θ) = 20 / 210
θ ≈ 5.5°
Therefore, the aircraft should fly 5.5 degrees off of the x-axis. This means that the plane must fly at an angle of N 35.5° E to immediately approach the lake.
(b) The velocity relative to the ground is:
v = v(w) + v(p)
v = 210 ( cos 5.5° ^i + sin 5.5° ^j ) km/h + ( 36.64 ^i − 20 ^j ) km/h
v ≈ ( 209 ^i + 20 ^j ) km/h + ( 36.64 ^i − 20 ^j ) km/h
v = 245.6 ^i km/h
The plane is moving at a speed of 245.6 km/h relative to the ground.
(c) From the starting point, the time taken to reach the lake is:
time = distance/ speed
T = 250 km / 245.6 km/h
T ≈ 1.02 h
Therefore, from the starting position, it will take around an hour to get to the lake.
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Types of Spectra 5) Stars like our Sun have low-density, gaseous atmospheres surrounding their hot, dense cores. If you were looking at the spectra of light coming from the Sun (or any star), which of the three types of spectrum would be observed? Explain your reasoning.
Answers
The spectrum observed from the Sun (or any star) would exhibit an absorption spectrum. This is because the outer gaseous atmosphere of the star absorbs specific wavelengths of light, resulting in dark absorption lines in the spectrum.
In the cooler, lower-density outer atmosphere, where white light from the star travels, some atoms or molecules in the atmosphere absorb photons with particular energy. In the spectrum, these absorptions show up as black lines at specific wavelengths. The specific set of absorption lines that each element or molecule generates results in a distinctive pattern that can be used to identify the elements that are present in the star's atmosphere.
The absorption spectrum offers insightful data on the chemical make-up and physical characteristics of the star. Astronomers can ascertain the elements present, their abundances, and other characteristics like the temperature, pressure, and velocity of the star's atmosphere by examining the absorption lines.
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Give an example of periods (periodic table)
Answers
Example: Row 1, which includes Hydrogen and Helium.
Answer:
helium, hydrogen, nitrogen, oxygen,iron,uranium,
Explanation:
there are 118 elements but here are some
Increasing the amplitude by a factor of ______ doubles the energy of a simple harmonic oscillator
Answers
Increasing the amplitude by a factor of four times doubles the energy of a simple harmonic oscillator.
The amplitude of a periodic variable is a measure of its change in a single period. The amplitude of a non-periodic signal is its magnitude compared with a reference value. There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.
Simple harmonic oscillator is an oscillator that is neither driven nor damped. It consists of a mass m, which experiences a single force F, which pulls the mass in the direction of the point x = 0 and depends only on the position x of the mass and a constant k.
A specific example of a simple harmonic oscillator is the vibration of a mass attached to a vertical spring, the other end of which is fixed in a ceiling.
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Please answer for this in class 12 physics
Answers
Answer:
w1:w2 = 16:1
Explanation:
As we know that ration of amplitude is equal to
\(\frac{a_1}{a_2} = \sqrt{\frac{w_1}{w_2} }\)
here w1 and w2 are the widths of the slit.
\(\frac{I _{min}}{I_{max}} = \frac{(a_1-a_2)^2}{(a_1+a_2)^2} \\\frac{9}{25} = \frac{(1-\frac{a_2}{a_1} )^2}{(1+\frac{a_2}{a_1} )^2} \\\frac{3}{5} = \frac{(1-\frac{a_2}{a_1} )}{(1+\frac{a_2}{a_1} )}\\8 \frac{a_2}{a_1} = 2\\\frac{a_1}{a_2} = 4\)
Again
\(\frac{a_1}{a_2} = \sqrt{\frac{w_1}{w_2} }\)
\(4 = \sqrt{\frac{w_1}{w_2} }\\\frac{w_1}{w_2} = 16\)
w1:w2 = 16:1
Consider a beaker with two holes X and Y near the base, such that hole X is ABOVE hole Y. The holes are closed with water-resistant tape and the beaker is filled with water.
Both the holes are then uncovered at the same time. If water from hole X comes out in a stream that touches the ground at a distance of 5 cm away from the beaker, how far from the beaker will the stream from hole Y touch the ground?
1. exactly 5 cm
2. less than 5 cm
3. more than 5 cm
4. (It will be more or less than 5 cm depending on the distance between the two holes.)
Answers
As the height of the hole increases, the velocity of the stream increases, results in a higher distance from the beaker. Therefore, the distance of the stream from the hole Y will be less than that from X thus, less than 5 cm is correct.
What is stream speed ?The speed of stream can be determined using the height and acceleration due to gravity g. We can use the equation for velocity using the parameters g and h.
v = √2gh
Therefore, as the height h increases, v increases.
Similarly the distance s = vt
therefore, the distance increases with v.
Here, the hole x is above the hole Y. Then the stream from X will have the greater speed and it covers greater distance (5 cm )from the beaker. Stream from Y slow compared to that in X hence covers a distance less than 5 cm. Hence, option 2 is correct.
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In a heat engine, 2.00 mol of a monoatomic gas are carried through the cycle ABCDA. The segment AB represents an isothermal expansion, the segment BC is an adiabatic expansion, the segment CD is an isobaric compression, and DA is a constant volume process. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm.
a) What is the pressure at B?
b) What is the temperature at C?
c) Find the total work done by the gas in one cycle.
Answers
(a)The pressure at B is 0.1248 atm.
(b)The temperature at C is 727.1 K.
(c)The total work done by the gas in one cycle is -1979J
General calculation:
We can use the First Law of Thermodynamics to analyze the heat engine cycle:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. For a complete cycle, ΔU = 0, so:
Q = W
We can also use the ideal gas law to relate the pressure, volume, and temperature of the gas:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature.
(a)How to find the pressure at B segment?
To find the pressure at B, we can use the fact that the segment AB is an isothermal expansion. This means that the temperature remains constant, so:
PV = nRT
PB = (nRT)/(2V) = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/(2V) = (0.0821 L·atm/mol)(600 K)/V
Since the pressure at A is 5.00 atm, we can use the fact that the temperature is constant to find the volume at A:
PV = nRT
VA = (nRT)/P = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/5.00 atm = 197.76 L
Since the volume at B is twice the volume at A, we have:
VB = 2VA = 395.52 L
Substituting into the expression for PB, we get:
PB = (0.0821 L·atm/mol)(600 K)/395.52 L = 0.1248 atm
Therefore, the pressure at B is 0.1248 atm.
(b) How to find the temperature at segment C?To find the temperature at C, we can use the fact that the segment BC is an adiabatic expansion. This means that no heat is added or removed from the system, so:
\(PV^\gamma\)= constant
where γ is the ratio of specific heats (for a monoatomic gas, γ = 5/3). We can use the fact that the volume at C is equal to the volume at A to find the pressure at C:
\(PAV^\gamma = PCV^\gamma\)
PC = \(PA(V/A)^\gamma\) = 5.00 atm\((1/2)^(^5^/^3^)\) = 1.556 atm
Since the segment BC is adiabatic, the temperature changes but no heat is added or removed from the system. Using the ideal gas law, we can relate the pressure, volume, and temperature:
PV = nRT
TC = (PCVC)/(nR) = (1.556 atm)(197.76 L)/(2.00 mol)(0.0821 L·atm/mol·K) = 727.1 K
Therefore, the temperature at C is 727.1 K.
(c) How to find the total work done by the gas in one cycle?The total work done by the gas in one cycle is the sum of the work done in each segment of the cycle:
W = WAB + WBC + WCD + WDA
For segment AB, the work done is:
WAB = -QAB = -∫PdV = -nRT∫(1/V)dV = -nRT ln(VB/VA) = -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2) = -602 J
For segment BC, the work done is:
WBC = -QBC = -∫PdV = -nγRT∫(1/V)dV = -nγRT
We know that VB = 2VA and VC = 2VD, so we can express the ratio VB/VC in terms of VA/VD:
VB/VC = (2VA)/(2VD) = VA/VD
Substituting into the expression for WBC, we get:
WBC = -nγRT ln(VA/VD)
For segment CD, the work done is:
WCD = -QCD + PCDΔV = -nCpΔT + PCDΔV
where Cp is the specific heat at constant pressure, ΔT is the change in temperature, and ΔV is the change in volume. We know that the segment CD is isobaric, so ΔV = VB - VA = (2VA) - VA = VA. We can also use the ideal gas law to relate the pressure, volume, and temperature:
PV = nRTPC = (nRT)/VDSubstituting into the expression for WCD, we get:
WCD = -nCpΔT + (nRT/VD)VA = -nCp(TC - TD) + (nRT/VD)VA
For segment DA, the work done is:
WDA = -QDA + ΔU = -nCvΔT
where Cv is the specific heat at constant volume. We know that the segment DA is isovolumetric, so ΔV = 0. Using the First Law of Thermodynamics, we know that ΔU = 0 for a complete cycle, so:
QDA = -WDA = nCvΔT
Substituting into the expression for WDA, we get:
WDA = -nCvΔT
Adding up the work done in each segment, we get:
W = WAB + WBC + WCD + WDA
= -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(Cp)(TC - TD) + (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(Cv)(TC - TA)
We know that Cp and Cv for a monoatomic gas are related by Cp = Cv + R, so we can express Cp in terms of Cv:
Cp = Cv + R = (3/2)R + R = (5/2)R
Substituting and simplifying, we get:
W = (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(727.1 K)+ (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(600 K)
W = -966.2 J - 4957 J - 7476 J + 5154 J
= -1979 J
Therefore, the total work done by the gas in one cycle is -1979 J
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What are the benefits of getting the Hubble into space over an
Earth-based telescope?
Answers
The Hubble Space Telescope has several benefits over Earth-based telescopes.
The following are the benefits of getting the Hubble into space over an Earth-based telescope:
Higher resolution images:Because the Hubble is located in space, it is free of the atmosphere's distortions and can produce higher resolution images than Earth-based telescopes. This gives scientists a better view of the universe, allowing them to see more clearly and in greater detail.
Exploration of distant objectAnother benefit of having the Hubble in space is that it can explore distant objects that Earth-based telescopes can't see. This is due to the fact that Earth's atmosphere absorbs certain types of light, making it difficult to detect distant objects
. This is especially true for ultraviolet and infrared light, which is critical to understanding the universe.A wider field of view:The Hubble also has a wider field of view than Earth-based telescopes, which means it can see more of the sky at once.
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Lisette decided to take this game more seriously and put both hands on the rope, applying a 150 N force to the left. Alexis still struggled with her 100 N force to the right.
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Answer:
orce or acceleration b. Ms. Sunderland finally decided to take this seriously and put both hands on the rope and applied a 1000 N force to the left, while Jared and James still struggled with their 300 N force to the right. What is ... 150 more force. 10. During 5th period Dylan resisted the forces applied by her classmates in a ...
Explanation:
what would be the importance of a topographic map in a rescue situation?
Answers
Answer:
Topographic maps give the user the ability to view a three-dimensional landscape on a two-dimensional map. One who is able to read a topo map can identify the elevation and location of valleys, peaks, ridges, and other land features.
Explanation:hope this helps
categorize we evaluate the current with an equation developed in this section, so we categorize this example as a substitution correct: your answer is correct. problem. combine imax
Answers
Example categorized as correct substitution problem using equation from section to evaluate current.
This means that the given problem involves replacing variables in an equation with given values and solving for the unknown variable. The solution obtained using this method is deemed correct according to the equation developed in the section. The use of equations in problem-solving is a common practice in various fields, including mathematics, physics, and engineering. By categorizing problems, it becomes easier to identify the appropriate methods to use in solving them, which can improve problem-solving efficiency and accuracy. Example categorized as correct substitution problem using equation from section to evaluate current. Imax is the maximum current and the answer obtained is deemed correct according to the equation developed in the section.
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What is required for the heat to flow from one object to another?
Uniform temperature
Uniform density
Temperature difference
Density difference
Answers
Answer:
Explanation:
The answer is Temperature difference
Answer:
C
Explanation:
is the answer ,..........................
Which is the best definition
of electric charge?
A. It is a force of attraction and repulsion by magnetic
materials.
B. It is the flow of electrons that have a negative charge to
protons that have a positive charge.
C. It is a force of attraction based on an object's mass.p oh
Answers
The correct answer is None of these above.
While some electrons are bonded in an atom, they are all similarly attached to protons. The atom's electrons that are farthest from the nucleus can be extracted. Because there are less electrons in the atom when some electrons are taken out, there are more protons than electrons. The electrically neutral substance becomes positively charged after the elimination of electrons.
The body can also get electrons from an outside source, which is the reverse of the first scenario. In this instance, the body's electron count rises and it acquires a negative charge.
An electric charge is the difference between an excess or shortage of electrons in a body.
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1) A U-tube, in which both ends are open to the atmosphere, is partially filled with water. Oil which does not mix with water is poured into one side of the tube until it stands a distance d=12.3 mm above the water level on the other
side which has meanwhile risen a distancea=67.5 mm from its original level Find the density of the oil
2) The mercury column in a barometer has a measured height h of 740.35 mm. the temperature is-5.0 C
which temperature the density of mercury is 1.3608 104 kg/m3. the free-fall acceleration at the site of the
barometer is 9.7835 m/s2. What is the atmospheric pressure?
Answers
Answer:
1) 916 kg/m³
2) 9.8566×10⁴ Pa
Explanation:
1) Originally, the water is at the same level on both sides. When the oil is added, one side goes down 67.5 mm and the other side goes up 67.5 mm. The top of the oil is 12.3 mm above the water, so the total height of the oil relative to the low end of the water is:
12.3 mm + 67.5 mm + 67.5 mm = 147.3 mm
And the height of the water relative to the low end is:
67.5 mm + 67.5 mm = 135 mm
The pressure at the bottom of the oil equals the pressure of the water at the same elevation.
P = P
ρgh = ρgh
ρh = ρh
ρ (147.3 mm) = (1000 kg/m³) (135 mm)
ρ = 916 kg/m³
2) P = ρgh
P = (1.3608×10⁴ kg/m³) (9.7835 m/s²) (0.74035 m)
P = 9.8566×10⁴ Pa
which equation according to ohms law
Answers
Answer: Ohm's law states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit. The formula for Ohm's law is V=IR.
A student uses a 320 Hz tuning fork during a lab investigation into the speed of sound. If the air temperature in the classroom is 20
degrees Celsius, what is the wavelength of the sound waves produced by the tuning fork?
Answers
Answer:
Hecht.
Explanation: