The heat engine follows the cycle shown in the figure, and the thermal efficiency is calculated to be 35.6%. For process 1-2, WS = -7.14 J, Q = -20.04 J, delta Eth = -12.90 J; for process 2-3, WS = -7.14 J, Q = -7.14 J, delta Eth = -12.90 J; for process 3-1, WS = 7.14 J, Q = 13.21 J, delta Eth = 5.07 J.
Part A: WS = -7.14 J, Q = -20.04 J, delta Eth = -12.90 J
Part B: WS = -7.14 J, Q = -7.14 J, delta Eth = -12.90 J
Part C: WS = 7.14 J, Q = 13.21 J, delta Eth = 5.07 J
Part D: Thermal Efficiency = 35.6%
Part A: WS = -7.14 J, Q = -20.04 J, delta Eth = -12.90 J
To calculate WS, Q and delta Eth for process 1-2, we need to calculate the change in internal energy between the two states. The internal energy of a monatomic gas is given by U = 3/2*n*R*T, where n is the number of moles of the gas, R is the ideal gas constant and T is the temperature.
For process 1-2:
U2 - U1 = -12.90J
To calculate Q, the equation Q = U2 - U1 can be used.
Q = U2 - U1 = -12.90J
To calculate WS, the equation WS = Q - (U2 - U1) can be used.
WS = Q - (U2 - U1) = -20.04J - (-12.90J) = -7.14J
Part B: WS = -7.14 J, Q = -7.14 J, delta Eth = -12.90 J
To calculate WS, Q and delta Eth for process 2-3, we need to calculate the change in internal energy between the two states. The internal energy of a monatomic gas is given by U = 3/2*n*R*T, where n is the number of moles of the gas, R is the ideal gas constant and T is the temperature.
For process 2-3:
U3 - U2 = -12.90J
To calculate Q, the equation Q = U3 - U2 can be used.
Q = U3 - U2 = -12.90J
To calculate WS, the equation WS = Q - (U3 - U2) can be used.
WS = Q - (U3 - U2) = -12.90J - (-12.90J) = -7.14J
Part C: WS = 7.14 J, Q = 13.21 J, delta Eth = 5.07 J
To calculate WS, Q and delta Eth for process 3-1, we need to calculate the change in internal energy between the two states. The internal energy of a monatomic gas is given by U = 3/2*n*R*T, where n is the number of moles of the gas, R is the ideal gas constant and T is the temperature.
For process 3-1:
U1 - U3 = 5.07J
To calculate Q, the equation Q = U1 - U3 can be used.
Q = U1 - U3 = 5.07J
To calculate WS, the equation WS = Q - (U1 - U3) can be used.
WS = Q - (U1 - U3) = 5.07J - (-5.07J) = 7.14
Part D: Thermal Efficiency = 35.6%
The thermal efficiency of a heat engine is given by the equation η = (Wnet/Qh) x 100, where Wnet is the net work output of the engine and Qh is the heat input to the engine.
For this heat engine, Wnet = WS1 + WS2 + WS3 = 7.14J + (-7.14J) + (-7.14J) = -7.14J and Qh = Q1 + Q2 + Q3 = -20.04J + (-7.14J) + 13
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What is the correct coefficient for 2H2 + O2 →2H2O
Explanation:
2forH2,1for02,and2forH20
What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?
Answer:
speed = 10.1215 m/s
Explanation:
speed = distance / time
speed = 100 / 9.88 = 10.1215 m/s
Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; both students move away from each other. What law of motion is this. (Newton's laws)
Answer:
forcing in act
Explanation:
Plzz help me with this
I’ll give brainliest
Answer:
B. Objects with more mass have more gravitational force acting upon them.
Answer:
Should be A but it can be B as well.
A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inelastic collision. What will be the speed of coupled wagons after collision?
Answer:
7.5 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.
What kind of reasoning is most often used to form hypotheses?
inductive
deductive
detective
invective
How do dog whistles work?
The sound it emits comes from what is known as the ultrasonic range, a pitch that is so high humans can't hear it. Dogs can hear these sounds, however, as can cats and other animals. Because of this, the dog whistle is a favored training tool, though it may not be for every dog parent.
true or false please help me now.
Calibration graphs can be used to determine unknown concentrations in electrochemical
Answer:
false
Explanation:
How does the Law of Conservation of Energy (or energy transformation) relate to the home?
Answer:
"The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another. This means that a system always has the same amount of energy, unless it's added from the outside. ... The only way to use energy is to transform energy from one form to another."
Explanation:
Brainliest?
A parallel-plate capacitor is connected to a battery until it is fully charged. Then, the capacitor is disconnected from the battery and connected to two uncharged parallel plates that make up a capacitor. The potential between the plates of the initial capacitor will
Answer:
The potential between the plates will decrease.
Explanation:
An insulator is usually placed between the parallel plates and is also called a dielectric because it makes the amount of charge a capacitor can accommodate to increase at a particular potential difference.
Furthermore, the dielectric effect will make the electric field of the charged capacitor which is not connected to a source of supply to decrease.
Now, when the battery is removed, the charge Q remains constant and Capacity C will increase.
Formula for the potential difference is here;
V = Q/C
Since the numerator Q is constant and the denominator C increases, it means the potential difference V will decrease
Which of the following is the BEST explanation for why oceans have two different types of currents?
Answer:
sddww
Explanation:
szsswa
A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine at the rate of 3 gpm, containing 2 lbs of dissolved salt per gallon. Assuming that the mixture is kept uniform by stirring, a drain pipe draws out of the tank the mixture at 2 gpm. Find the amount of salt in the tank at the end of 30 minutes.
A. 171.24 lbs
B. 124.11 lbs
C. 143.25 lbs
D. 105.12 lbs
Answer:
A. 171.24 Ibs
Explanation:
To find the amount of salt in the tank,
Let Q = Amount of salt in the mixture
And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.
Rate of gain - Rate of loss = dQ / dt
Concentration of salt = Q / (100+t)
For the linear differential equation,
dQ / dt = 3(2) - 2 [Q/ (100 + t)]
dQ /dt + Q [2 / (100 + t)] = 6
The general solution of the linear differential equation is:
Q (i.f) = ∫ A(t) (i.f) dt + C
Therefore,
i.f = e ^ ∫ P(t) dt
And P(t) = 2 / (100 + t)
i.f = e ^ ∫ 2 / (100 + t)
= e ^ 2㏑ (100 + t)
= e ^ ㏑ (100 + t) ^2 = (100 + t) ^2
Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C
Q(100 + t) ^2 = 2(100 + t) ^ 3 + C
When t = 0, Q = 50
Therefore,
50( 100) ^2 = 2(100) ^3 + C
C = -1.5 * 10 ^6
therefore, when t = 30,
Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6
Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6
Q = 171.24 Ibs
The amount of salt in the tank at the end of 30 minutes is 171.24 lbs.
The given parameters:
Initial volume of the tank, i = 100 gallonsRate of gain of salt = 3 gpmRate of loss of salt = 2 gpmThe linear differential equation of the salt solution is calculated as follows;
[tex]\frac{dx}{dt} = Gain - loss[/tex]
where;
x is the salt concentrationThe salt concentration at time t, is calculated as follows;
[tex]\frac{dx}{dt} = 2(3) - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} = 6 - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} +2(\frac{X}{100 + t} ) =6[/tex]
Apply the general solution of linear differential equation as follows;
[tex]X(f) = \int\limits {At} \, dt \ + C\\\\f = e^{\int\limits {At} \, dt}\\\\ f = e^{\int\limits {\frac{2}{100 + t} } \, dt}\\\\f = e^{2 ln(100 + t)}\\\\f = (100 + t)^2[/tex]
[tex]X(100 + t)^2 = \int\limits {6(100 + t)^2} \, dt \ + \ C\\\\ X(100 + t)^2 = 2(100 + t)^3 + C[/tex]
When t = 0 and X = 50
[tex]50(100 + 0)^2 = 2(100+ 0)^3 + C\\\\C = -1.5 \times 10^6[/tex]
When t = 30 min, the concentration is calculated as;
[tex]X (100 + 30)^2 = 2(100 + 30)^3- 1.5 \times 10^6\\\\X(130)^2 = 2(130)^3 - 1.5\times 10^6\\\\X(130)^2 = 2894000\\\\X = \frac{2894000}{130^2} \\\\X = 171.24 \ lbs[/tex]
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Please help I don’t get this give me answers please
Answer:
c
Explanation:
types of aerobic activities?
Answer:
swimming, cycling, jump rope, brisk walking, gardening, jogging
For PN junctions, determine if each statement below is True or False:
a. There can be large net current from p-side to n-side under forward bias.
b. There can be large net current from n-side to p-side under reverse bias if reverse bias is sufficiently high.
c. Electron diffusion current flows from n-side to p-side.
d. Electric field magnitude is higher under reverse bias
e. Electrons in the transition region drift from p-side to n-side.
Answer:
a) True
b) True
c) false
d) True
e) True
Explanation:
a) True
In forward bias, the resistance of the p–n junction reduces and hence the electric charges can flow easily.
b) True
In reverse bias condition, the electric current is due to the minority charge carriers (negative).
c) False
direction of diffusion current is in the direction of movement of positive charge i,e towards n side
d) True
Because the breakdown of charge carriers occur due to which the current increases rapidly
e) True
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0 m/s , and the distance between them is 52.0 m . After t1 = 3.00 s , the motorcycle starts to accelerate at a rate of 4.00 m/s^2. The motorcycle catches up with the car at some time t2.
Required:
a. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car?
b. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?
Answer:
a) 5.09 seconds
b) 107.07 meters
Explanation:
a) As we know
[tex]t_2- t_1 = \sqrt{\frac{2 X}{a} }[/tex]
Substituting the given values we get
[tex]t_2 - t_1 = \sqrt{\frac{2 * 52}{4} } \\t_2 - t_1 = 5.09[/tex]
It takes 5 .09 s for the motorcycle to accelerate until it catches up with the car
b)
[tex]X_{t`2} = v_i \sqrt{\frac{2X}{a} } + 0.5 a\sqrt{\frac{2X}{a} }\\X_{t`2} = (v_i + 0.5 a) \sqrt{\frac{2X}{a} }\\X_{t`2} = ( 19 + 2) \sqrt{\frac{2* 52}{4} }\\X_{t`2} = 21 * 5.09\\X_{t`2} = 107.07[/tex]
In which situation are waves transmitted?
O A. A patient wears a lead apron at the dentist's office when getting
teeth X-rays.
O B. A light in a swimming pool comes on after dark to prevent
accidents in the water.
O C. A person wears earplugs to prevent hearing damage when fueling
a jet plane at the airport.
O D. A reflective screen is put on a parked car's dashboard to keep the
car from heating up in sunlight.
Answer: B. A light in a swimming pool comes on after dark to prevent
accidents in the water.
Which list of reaction types are all redox reactions?
A.
Synthesis, decomposition, single-replacement, combustion
B.
Synthesis, double-replacement, combustion, decomposition
C.
Acid-base, single-replacement, double-replacement, synthesis
D.
Decomposition, double-replacement, acid-base, synthesis
A. Synthesis, decomposition, single-replacement, combustion
List of reaction types are redox reactions:
Synthesis, decomposition, single-replacement, combustion.What are redox reaction ?"Redox reactions are oxidation-reduction chemical reactions in which the reactants undergo a change in their oxidation states." The term 'redox' is a short form of reduction-oxidation. All the redox reactions can be broken down into two different processes – a reduction process and an oxidation process.
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All magnetic fields result from the movement of
A. charged particles
B. electrons only
C. protons only
D. neutrons only
Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll. ratio. (b) Plexiglas is often used to shield high-energy beta emitters rather than lead, even though lead is a better shield against the bremsstrahlung photons. Both shields will stop the high-energy beta, so why is Plexiglas used instead of lead?
Answer:
Explanation:
During an energy transfer, the collision loss for an electron can be determined by using the formula:
[tex]Q = \dfrac{4mME }{(m+M)^2}[/tex]
However; from the total stopping power & power loss of the electron;
[tex]\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}[/tex]
where;
Z = atomic no. for lead = 82
E = 1.9 MeV
∴
radiational energy loss = collisional energy loss [tex]=\dfrac{82 \times 1.9}{800}[/tex]
= 0.19475
b)
Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.
Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.
Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.
Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.
HURRY Which change is an example of transforming potential energy to kinetic energy
A: changing thermal energy to electrical energy
B: changing mechanical energy to radient energy
C: changing nuclear energy to radiant energy
D: changing radient energy to electrical energy
Answer:
C. changing nuclear energy to radiant energy
Explanation:
Nuclear energy takes atoms in their potential state, split them (fission) or fuse them (fusion) creating chain reactions of radiant energy. Most nuclear electrical power plants use fission, radiant energy heats water making steam to spin turbines.
Or think of the atom bomb. Definitely potential energy until the fuse starts detonation and chain reactions. The radiant kinetic energy and shock waves were horrendous.
Answer:
The answer would be C, changing nuclear energy to radiant energy
Explanation:
Welcome have a good day.
An uncharged parallel-plate capacitor is connected through an open switch to a battery of voltage VV. The switch is closed and the capacitor is allowed to charge. As the capacitor is charged, energy is transferred to it from the battery. When the capacitor is fully charged, the energy stored in the capacitor is U1U1 . The energy stored in the capacitor when the stored charge is q02q02 is
Answer:
(1/2)U₁
Explanation:
An uncharged parallel-plate capacitor is connected through an open switch to a battery of voltage VV. The switch is closed and the capacitor is allowed to charge. As the capacitor is charged, energy is transferred to it from the battery. When the capacitor is fully charged, the energy stored in the capacitor is U1U1 . The energy stored in the capacitor when the stored charge is q₁/2 is
Solution:
A capacitor is an electrical device used to store electrical energy in an electric field. The energy stored in a capacitor is given by:
U = (1/2)QV; where U is the energy stored, Q is the charge and V is the voltage applied.
The energy stored in a fully charged capacitor with a charge q₁ and battery of voltage V is given as:
U₁ = (1/2)q₁V
If the stored charge is q₁/2, the energy stored (U₂) becomes:
U₂ = (1/2)(q₁ / 2)V
U₂ = (1/2)* (1/2)q₁V
U₂ = (1/2)U₁
Plzzz answer this correctly
Answer:
D, the acceleration of A is twice that of b.
Explanation: in four seconds b got to ten, in two seconds a got to 20. Going 10m/s faster in half the time is going twice the acceleration
If all pairs of adjacent sides of a quadrilateral are congruent then it is called _________.
(A) rectangle (B) parallelogram (C) trapezium, (D) rhombus
Answer:
D
Explanation:
If you need an explanation feel free to ask.
How much force is needed to accelerate a 9,760 kg airplane at a rate of 3.6 m/s2?
OA. 2,711 N
OB. 35,136 N
OC. 126,490 N
OD. 9,760 N
Answer:
the answer is B: 35,136
Explanation:
force = mass × acceleration
force = 9760 × 3.6
35,136 = 9760 × 3.6
Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1.0m/s and vy=3.0m/svy=3.0m/s and that at time t2=2.5st2=2.5s the components are vx=4.0m/svx=4.0m/s and vy=3.0m/svy=3.0m/s . Find (a) the components of average acceleration and (b) the magnitude and direction of the average acceleration during this interval.
Answer:
[tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]
Magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex]
Explanation:
[tex]t_1=2\ \text{s}[/tex]
[tex]v_x=1\ \text{m/s}[/tex]
[tex]v_y=3\ \text{m/s}[/tex]
[tex]t_2=2.5\ \text{s}[/tex]
[tex]v_x=4\ \text{m/s}[/tex]
[tex]v_y=3\ \text{m/s}[/tex]
Average acceleration in the different axes
[tex]a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2[/tex]
[tex]a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2[/tex]
The components of the acceleration is [tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]
The magnitude of acceleration
[tex]a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2[/tex]
Direction
[tex]\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}[/tex]
The magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex].
Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance. A Formula One racer traveling at 90m/s can stop in a distance of 110m. What is the magnitude of the car's acceleration as it slows during braking?
Answer:
The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²
Explanation:
From the question, the given values are as follows:
Initial velocity, u = 90 m/s
final velocity, v = 0 m/s
distance, s = 110 m
acceleration, a = ?
Using the equation of motion, v² = u² + 2as
(90)² + 2 * 110 * a = 0
8100 + 220a = 0
220a = -8100
a = -8100/220
a = -36.81 m/s²
The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²
On Earth, the number flux of solar neutrinos from the p-p chain is:
f_neutrino = 2fo/2.62MeV
Other nuclear reactions in the Sun supplement this neutrino flux with a small additional flux of higher-energy neutrinos. A neutrino detector in Japan, named SuperKamiokande, consists of a tank of 50 kton of water, surrounded by photomultiplier tubes. The tubes detect the flash of Cerenkov radiation emitted by a recoiling electron when a high-energy neutrino scatters on it.
Required:
a. How many electrons are there in the water of the detector?
b. Calculate the detection rate for neutrino scattering, in events per day.
Answer:
Explanation:
The volume of the tank = 50 kton
50 kton = 5 × 10⁷ kg
Since 18 grams of water will contain: 10 electrons × 6.023 × 10²³
Then;
5× 10⁷ kg will contain [tex]( \dfrac{5 \times 10^7 \times 10^3}{18}) \times 10 \times 6.023 \times 10^{23}[/tex]
= 1.67 × 10³⁴ electrons
(b)
Suppose:
[tex]f_{neutrino} = \dfrac{2f_o}{26.2 MeV} = 6.7\times 10^{10} \ s^{-1} cm^{-2}[/tex]
Then;
10⁻⁶ of [tex]f_{neutrino} = 6.7 \times 10^{10} \times 10^{-6} \ s^{-1} cm^{-2}[/tex]
[tex]=6.7 \times 10^{4}\ s^{-1} cm^{-2}[/tex]
Thus, the number of high energy neutrinos which will interact with water is:
= [tex]6.7 \times 10^4 \times \sigma[/tex]
= [tex]6.7 \times 10^4 \times 10^{-43}[/tex]
= [tex]6.7 \times 10^{-39} s^{-1}[/tex]
For 1.67 × 10³⁴ electrons, the detection rate is:
[tex]6.7 \times 10^{-39} \times 1.67 \times 10^{34}[/tex]
[tex]= 11.19 \times 10^{-5} \ s^{-1}[/tex]
= 9.668 per day
3. Two cyclists that are 500 m apart start biking toward each other. They bike at speeds of 6 and
4 m/s.
How long does it take for them to reach each other?
a.
b. How far does the slower biker travel?
Answer:
A) 50 seconds
B) 200 m
Explanation:
They are 500 metres apart.
And one of the bike loves at 6 m/s while the other loves at 4 m/s.
A) Let distance of the 6 m/s bike before they meet be x.
Thus, time = x/6
Since time = distance/speed
For the second bike at 4 m/s, his distance covered before they meet will be 500 - x
Thus, time = (500 - x)/4
Now they will meet each other at the same time. Thus;
x/6 = (500 - x)/4
Cross multiply to get;
4x = 3000 - 6x
6x + 4x = 3000
10x = 3000
x = 3000/10
x = 300 m
Thus, time will be;
t = 300/6
t = 50 seconds
B) Distance covered by the slower bike is (500 - x)
Since from a above, x = 300
Thus; distance = 500 - 300 = 200 m
a body of mass 20kg initially at rest is subjected to a force of 40N for 1sec calculate the change in kinetic energy showing the solution
Answer:
Change in KE is 40 J
Explanation:
Recall that the impulse exerted on an object equal the change of momentum of the object (ΔP), which in time is defined as the product of the force exerted on it times the time the force was acting:
Change in momentum is: ΔP = F * Δt
In our case,
ΔP = 40 N * 1 sec = 40 N s
Since the object was initially at rest, its initial momentum was zero, and the final momentum should then be 40 N s.
So, the initial KE was 0, and the final (KEf) can be calculated using:
KEf = 1 /(2 m) Pf^2 = 1 / (40) 40^2 = 40 J
So, the change in kinetic energy is:
KEf - KEi = 40 J - 0 j = 40 J