Can you think of reasons why the charge on each ball decreases over time and where the charges might go
Answer:
By the principle of corona discharge.
Explanation:
The charge on each ball will decreases over time due to the electrical discharge in air.
According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g
Answer:
w = 1,066 rad / s
Explanation:
For this exercise we use Newton's second law
F = m a
the centripetal acceleration is
a = w² r
indicate that the force is the mass of the body times the acceleration
F = m 0.58g = m 0.58 9.8
F = 5.684 m
we substitute
5.684 m = m w² r
w = [tex]\sqrt{5.684/r}[/tex]
To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m
w = [tex]\sqrt{ 5.684/5}[/tex]
w = 1,066 rad / s
Physical quantities expresed only by their magnitude is
Answer:
Scalar quantity is the Physical quantity expresed only by their magnitude.
Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.
Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
Explanation:
Let us assume that
[tex]q_{1} = q_{2} = +q[/tex]
[tex]q_{3} = -q[/tex]
As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).
Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r
Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]
where,
k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]
Substitute the values into above formula as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)
Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex] ... (2)
As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.
Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.
Answer:
v = 804.23 m/s
Explanation:
Given that,
The maximum distance covered by a cannon, d = 33 km = 33000 m
We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,
[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]
So, the initial velocity of the shell is 804.23 m/s.
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath. The temperature of the water bath is monitored, and it is determined from this data that 133.0 kJ of heat flows into the system during the reaction. The position of the piston is also monitored, and it is determined from the data that the piston does 241.0 kJ of work on the system during the reaction.
a. Does the temperature of the water bath go up or down?
b. Does the piston move in or out?
c. Does heat flow into or out of the gaseous mixture?
d. How much heat flows?
A 49.5-turn circular coil of radius 5.10 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.535 T. If the coil carries a current of 26.5 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm
Explanation:
Given;
number of turns of the circular coil, N = 49.5 turns
radius of the coil, r = 5.10 cm = 0.051 m
magnitude of the magnetic field, B = 0.535 T
current in the coil, I = 26.5 mA = 0.0265 A
The magnitude of the maximum possible torque exerted on the coil is calculated as;
τ = NIAB
where;
A is the area of the coil
A = πr² = π(0.051)² = 0.00817 m²
Substitute the given values and solve for the maximum torque
τ = (49.5) x (0.0265) x (0.00817) x (0.535)
τ = 0.00573 Nm
τ = 5.73 x 10⁻³ Nm
What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?
Answer:
The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.
i.Name two commonly used thermometric liquids.
ii.State two advantages each of the thermometric liquids mentioned above
Answer:
mercury and alcohol
ii) used to test temperatures
i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.
ii) It does not wet (cling to the sides of) the tube.
Alcohol:
i) Alcohol has greater value of temperature coefficient of expansion than mercury.
ii) it's freezing point is below –100°C.
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 309 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars
Answer:
The velocity is 2661.5 m/s.
Explanation:
Radius, horizontal distance, d = 309 km
height, h = 25 km
acceleration due to gravity on moon, g =3.71 m/s^2
Let the time taken is t and the horizontal velocity is u.
horizontal distance = horizontal velocity x time
309 x 1000 = u t .... (1)
Use second equation of motion in vertical direction.
[tex]h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s[/tex]
So, put in (1)
309 x 1000 = u x 116.1
u = 2661.5 m/s
A flat coil of wire is placed in a uniform magnetic field that is in the y-direction.
The magnetic flux through the coil is maximum if the coil is:_________.
(a) in the XY plane
(b) in either the XY or the YZ plane
(c) in the XZ plane
(d) in any orientation, because it is constant.
Answer:
The correct answer is c
Explanation:
Flow is defined by
Ф = B . A
bold letters indicate vectors.
The magnetic field is directed to the y axis, The area of the coil is represented by a vector normal to the plane of the coil, so to have a flux
i.i = j.j = k.k = 1
and the tori scalar products are zero
a) If the coil must be in the xy plane so that its normal vector is in the Z axis, so there is no flux
b) if the coil is in the plane yz the normal veto is in the x axis, so the flux is zero
C) If the coil is in XZ, the normal vector points in the y direction, usually the scalar product is one and there is a flux in this configuration
The correct answer is c
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answer:
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Explanation:
We are given that
[tex]PV^{1.4}=C[/tex]
Where C=Constant
[tex]\frac{dP}{dt}=-7KPa/minute[/tex]
V=420 cubic cm and P=99KPa
We have to find the rate at which the volume increasing at this instant.
Differentiate w.r.t t
[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]
Substitute the values
[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]
[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]
[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Answer:
[tex]\dot V=2786.52~cm^3/min[/tex]
Explanation:
Given:
initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]
initial volume during the process, [tex]V_1=420~cm^3[/tex]
The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.
Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]
Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:
[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]
[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]
[tex]\dot V=2786.52~cm^3/min[/tex]
[tex]\because P\propto\frac{1}{V}[/tex]
[tex]\therefore[/tex] The rate of change in volume will be increasing.
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity of the light from this bulb at a distance of 0.400 m from the bulb
Answer: [tex]29.85\ W/m^2[/tex]
Explanation:
Given
Power [tex]P=60\ W[/tex]
Distance from the light source [tex]r=0.4\ m[/tex]
Intensity is given by
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Inserting values
[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]
Answer:
29.85 W/ m^2
Explanation:
A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?
Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.
the speed of the pulse depends on what?
Answer:
The pulse speed depends on the properties of the medium and not on the amplitude or pulse length of the pulse.
Explanation:
Hope this helps.
which has higher eneergy electron r proton
Answer:
proton have higher energy than electron
Explanation:
tag me brainliest
Answer:
proton
Explanation:
proton is higher energy than the electron
A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance d by a rope supplying a force F at an angle of elevation q. The surface has a frictional force acting during this motion. How much work was done by friction during this motion? The student calculates the value to be –Fd sinq. How does this value compare to the correct value?
a. It is the correct value.
b. It is too high.
c. It is too low.
d. The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
Which graph would be created by a pendulum with the greatest amplitude?
Answer:
Graph (c) would be created by a pendulum with the greatest amplitude.
Explanation:
The amplitude of a wave is the greatest displacement covered by an object. It refers to the maximum amount of displacement of a particle on the medium from its rest position. It is the distance from rest to crest.
Out of three graphs, the amplitude is greatest in graph 3 as the distance from rest is crest in this case is maximum. Hence, the correct option is (c).
why do you like the full moon ?
Answer:
The Moon brings perspective. Observing the Moon, and I mean really looking – sitting comfortably, or lying down on a patch of grass and letting her light fill your eyes, it's easy to be reminded of how ancient and everlasting the celestial bodies are. When I do this, it always puts my life into perspective.Answer:
because it look more impressive than empty dark sky .
A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
[tex]d_2=1.5*10^-3m[/tex]
Explanation:
From the question we are told that:
Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]
Space 1 [tex]d_1=2.3*10^{-3}[/tex]
Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]
Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by
[tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]
[tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]
[tex]d_2=1.5*10^-3m[/tex]
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (Hint: At what angle do field lines intersect equipotential lines?) Draw sufficient field lines that you can "see" the electric field.
Answer:
The angle between the electric field lines and the equipotential surface is 90 degree.
Explanation:
The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.
The electric field lines are always perpendicular to the equipotential surface.
As
[tex]dV = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
For equipotential surface, dV = 0 so
[tex]0 = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
The dot product of two non zero vectors is zero, if they are perpendicular to each other.
How does the theory of relativity explain the gravity exerted by massive objects?
A. More massive objects create stronger forces of gravity.
B. More massive objects create shallower curves of space-time.
C. More massive objects pull objects from farther away.
D. More massive objects create larger curves of space-time.
(D)
Explanation:
The more massive an object is, the greater is the curvature that they produce on the space-time around it.
The theory of relativity explain the gravity exerted by massive objects is
more massive objects create larger curves of space-time (option-d).
Do bigger objects exert more gravity?The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.
Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.
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1. What is the total distance the car moves until it stops?
a. 250 m
b. 450 m
c. 300 m
d. 600 m.
Which is the most difficult subject?
Answer:
Quantum Mechanics
Explanation:
Well, that's what I think personally.
Topic: Chapter 19: Some wiggle room
A hummingbird flaps its wings up to 70 times per second, producing a 70 Hz
hum as it flies by. If the speed of sound is 340 m/s, how far does the sound
travel between wing flaps?
= 4.86 m
= 58.9 m
= 0.206 m
= 23,800 m
Answer:
4.86 m
Explanation:
Given that,
The frequency produced by a humming bird, f = 70 Hz
The speed of sound, v = 340 m/s
We need to find how far does the sound travel between wing flaps. Let the distance is equal to its wavelength. So,
[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]
So, the sound travel 4.86 m between wings flaps.
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll.
What will be the car’s speed as it coasts into the gas station on the other side of the valley?
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J
Answer:
a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy is -18279.78 J
c) heat lost by the gas is zero or 0
d) the work done on the gas is -18279.78 J
Explanation:
Given the data in the question;
P[tex]_i[/tex] = 1 atm = 101325 pascal
P[tex]_f[/tex] = ?
V[tex]_i[/tex] = 0.1550 m³
V[tex]_f[/tex] = 0.742 m³
we know that for an adiabatic process γ = 1.4
P[tex]_i[/tex]V[tex]_i^Y[/tex] = P[tex]_f[/tex]V[tex]_f^Y[/tex]
P[tex]_f[/tex] = P[tex]_i[/tex][tex]([/tex] V[tex]_i[/tex] / V[tex]_f[/tex] [tex])^Y[/tex]
we substitute
P[tex]_f[/tex] = 1 × [tex]([/tex] 0.1550 / 0.742 [tex])^{1.4[/tex]
= [tex]([/tex] 0.2088948787 [tex])^{1.4[/tex]
= 0.11166 atm
a) the initial and final temperatures
Initial temperature
T[tex]_i[/tex] = P[tex]_i[/tex]V[tex]_i[/tex] / nR
given that n = 3.10 mol
= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )
= 15705.375 / 25.7734
T[tex]_i[/tex] = 609.4 K
Final temperature
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )
= 8394.95 / 25.7734
= 325.7 K
Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy
ΔE[tex]_{int[/tex] = nC[tex]_v[/tex]ΔT
here, C[tex]_v[/tex] = ( 5/2 )R
ΔE[tex]_{int[/tex] = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J
c) the heat lost by the gas
Since its an adiabatic process,
Q = 0
Therefore, heat lost by the gas is zero or 0
d) the work done on the gas
W = ΔE[tex]_{int[/tex] - Q
= -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J
a) The Initial Temperature and Final Temperature of gas are 601.68 K and 321.61 K respectively.
b) The change in internal energy is -18279.78 J.
c) The heat lost by the gas is zero.
d) The work done on the gas is -18279.78 J.
Given data:
The moles of sample is, n = 3.10 mol.
The initial volume of sample is, [tex]V_{1}=0.1550 \;\rm m^{3}[/tex].
The final volume of sample is, [tex]V_{2}=0.742 \;\rm m^{3}[/tex].
The initial pressure of the sample is, [tex]P_{1}=1.00 \;\rm atm[/tex].
(a)
We know that the relation between the pressure and volume for an adiabatic process is as follows,
[tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex]
Here, [tex]\gamma[/tex] is a adiabatic index. And for air, its value is 1.41.
Solving as,
[tex]P_{2}=P_{1} \times\dfrac{V_{1}^{\gamma}}{V_{2}^{\gamma}}\\\\\\P_{2}=1.00 \times\dfrac{0.1550^{1.41}}{0.742^{1.41}}\\\\\\P_{2} = 0.11166 \;\rm atm[/tex]
Now, calculate the final temperature using the ideal gas equation as,
[tex]P_{2}V_{2}=nRT_{2}\\\\T_{2}= \dfrac{P_{2} \times V_{2}}{nR}\\\\T_{2}= \dfrac{0.11166 \times 10^{5}\times 0.742}{3.10 \times 8.31}\\\\T_{2}=321.61 \;\rm K[/tex]
Similarly, calculate the initial temperature as,
[tex]P_{1}V_{1}=nRT_{1}\\\\T_{1}= \dfrac{P_{1} \times V_{1}}{nR}\\\\T_{1}= \dfrac{1 \times 10^{5}\times 0.1550}{3.10 \times 8.31}\\\\T_{1}=601.68 \;\rm K[/tex]
Thus, we can conclude that the initial and final temperature of the gas is 601.68 K and 321.61 K respectively.
(b)
The change in internal energy is given as,
ΔE = nCΔT
here, C = ( 5/2 )R
ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J.
c)
The heat lost by the gas . Since its an adiabatic process, so there will be no heat interaction.
Q = 0
Therefore, heat lost by the gas is zero or 0
d)
The work done on the gas
W = ΔE - Q
W = -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J.
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Harmonics a.are components of a complex waveform. b.have frequencies that are integer multiples of the frequency of the complex waveform. c.are pure tones. d.have sinusoidal waveforms. e.all of the above
Answer:
b.have frequencies that are integer multiples of the frequency of the complex waveform
Explanation:
Please correct me if I am wrong
The working substance of a certain Carnot engine is 1.90 of an ideal
monatomic gas. During the isothermal expansion portion of this engine's
cycle, the volume of the gas doubles, while during the adiabatic expansion
the volume increases by a factor of 5.7. The work output of the engine is
930 in each cycle.
Compute the temperatures of the two reservoirs between which this engine
operates.
Answer:
Explanation:
The energy for an isothermal expansion can be computed as:
[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)
However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:
[tex]V_b = 2V_a[/tex]
Equation (1) can be written as:
[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]
Also, in a Carnot engine, the efficiency can be computed as:
[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]
[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]
In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]
relating the above two equations together, we have:
[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]
Making the work done (W) the subject:
[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]
From equation (1):
[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]
[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]
If we consider the adiabatic expansion as well:
[tex]PV^y[/tex] = constant
i.e.
[tex]P_bV_b^y = P_cV_c^y[/tex]
From ideal gas PV = nRT
we can have:
[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]
[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]
From the question, let us recall aw we are being informed that:
If the volumes changes by a factor = 5.7
Then, it implies that:
[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]
∴
[tex]T_H = T_L (5.7)^{y-1}[/tex]
In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]
As such:
[tex]T_H = T_L (5.7)^{1.6-1}[/tex]
[tex]T_H = T_L (5.7)^{0.67}[/tex]
Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]
[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]
From in the question:
W = 930 J and the moles = 1.90
using 8.314 as constant
Then:
[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]
[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]
[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]
[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]
From [tex]T_H = T_L (5.7)^{0.67}[/tex]
[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]
[tex]\mathbf{T_H \simeq 125K}[/tex]
