The development of Newton's universal theory of gravitation has had a profound influence on shaping our modern understanding of the nature of the universe. Newton's theory revolutionized our understanding of gravity and provided a mathematical framework that explained the motion of celestial bodies.
Explanation of Planetary Motion: Newton's theory of gravitation provided a comprehensive explanation for the observed motion of planets around the Sun. It demonstrated that the same force that causes objects to fall on Earth also governs the motion of celestial bodies, leading to the formulation of the laws of planetary motion. This understanding allowed astronomers to accurately predict and calculate the positions of celestial bodies, enhancing our knowledge of the solar system. Unification of Celestial and Terrestrial Mechanics: Newton's theory unified the laws governing motion on Earth with those governing motion in space. It showed that the same laws of physics applied to both terrestrial and celestial bodies, establishing a fundamental connection between the two. This unification brought about a significant shift in our perception of the universe, breaking the traditional view that celestial bodies operated by different rules. Confirmation of the Clockwork Universe: Newton's theory supported the concept of a clockwork universe, in which the motion of celestial bodies follows predictable and deterministic laws.
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Q31 (1 point) A galaxy has a thick accretion disk. This means that the material ejected by the galaxy's central black hole is ejected... In all directions above and below the disk. Only in narrow jets
The answer is In all directions above and below the disk. A thick accretion disk is a disk of gas and dust that is very dense and hot. It can form around a black hole or a neutron star.
A thick accretion disk is a disk of gas and dust that is very dense and hot. It can form around a black hole or a neutron star. When material falls into a thick accretion disk, it heats up and emits a lot of radiation. This radiation can cause the material to be ejected from the disk in all directions above and below the disk.
In contrast, a thin accretion disk is a disk of gas and dust that is less dense and cooler. When material falls into a thin accretion disk, it does not heat up as much and does not emit as much radiation. This means that the material is less likely to be ejected from the disk.
The material that is ejected from a thick accretion disk can form jets of gas and plasma. These jets can travel for billions of light-years and can be very powerful. They can be used to study the central black holes in galaxies and to learn about the formation of galaxies and galaxy clusters.
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which of the following statements is true about a projectile at the instant at which it is at the highest point of its parabolic trajectory? group of answer choices its velocity is zero. both a and c the vertical component of its velocity is zero. the horizontal component of its velocity is zero. its acceleration is zero.
The correct statement about a projectile at the highest point of its parabolic trajectory is: "The vertical component of its velocity is zero."
At the highest point of its trajectory, a projectile momentarily comes to a stop in the vertical direction before reversing its motion and descending. This means that the vertical component of its velocity becomes zero. However, the projectile still possesses horizontal velocity, so the horizontal component of its velocity is not zero.
The other statements are not true at the highest point of the trajectory:
Its velocity is not zero; it only refers to the vertical component.Its acceleration is not zero; gravity continues to act on the projectile, causing it to accelerate downward.Therefore, the correct statement is that the vertical component of the projectile's velocity is zero at the highest point of its trajectory.
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please help, from question 24 to 30 are from the same exercise. 24. Resting heart rate was measured for a group of subjects; subjects then drank 6 ounces of coffee. Ten minutes later their heart rates were measured again. The change in heart rate followed a normal distribution, with a mean increase (H) of 7.3 and a standard deviation (a) of 11.1 beats per minute. Let Y be the change in frequency heart rate of a randomly selected subject, what is the probability that the change in heart rate of that subject: 24) Is below 8.3 beats per minute. a. 0.09 Or 0.09009 b. -0.09 0-0.09009 c. 0.4641 Or 0.46411 d. 0.5359 or 0.53589 25) In the study, subjects with changes in heart rate of 20 beats per minute or plus. What is the probability that, when selecting a subject from that population, his change in heart rate will be classified as "worrying"? a. 1.1401.14414 b. 0.1271 or 0.12628 c. 0.8729 or 0.87372 d. 1 26) In the study, those subjects with changes in heart rate of 20 beats per minute or plus. What is the probability that, by selecting five subjects from this population, the change in heart rate of all will be classify as "concerning"? a. 0.000033 Or 0.000032 b. 0.1271 Or 0.12628 c. 0.8729 Or 0.87372 d. 1.14 or 1.14414 27) In the study, subjects with changes in heart rate of 20 beats per minute were identified as "worrisome." plus. What is the probability that, by selecting five subjects from that population, the change in heart rate of two of the classify as "concerning"? a. 0.1074 Or 0.1064 b. 0.4 c. 0.01564 Or 0.01537 d. 0.001564 Or 0.001537 28) In the study, those subjects with changes in heart rate of 20 beats per minute or plus. What is the probability that, by selecting five subjects from this population, the change in heart rate of three of them will be classify as "concerning"? a. 0.1074 or 0.1064 b. 0.01074 or 0.01064 c. 0.01564 or 0.01537 29) In the study, those subjects with changes in heart rate of 20 beats per minute or plus. What is the probability that the mean change in heart rate of five individuals is classified as "worrisome"? a. 0.000033 Or 0.000032 b. 2.56 or 2.56048 c. 0.9948 or 0.99477 d. 0.0052 or 0.0052263 30) Assume that the data collected on the change in heart rate does not follow a normal distribution. Also assume that the The mean of five individuals is 12.0 beats per minute, and the standard deviation is 7.0 beats per minute. Based on this information, construct a confidence interval assuming a confidence level of 95%. a. 2,365 beats per minute b. 2,306 beats per minute c. (4.6, 19.4) beats per minute d. (4.8, 19.2) beats per minute please help, from question 24 to 30 are from the same exercise.
Probability below 8.3 bpm: z-score = 0.09009; answer: 0.09. Probability ≥ 20 bpm: z-score = 1.1401; reply: 1.1401. All 5 subjects ≥ 20 bpm: Probability = 0.000033. 2 out of 5 subjects ≥ 20 bpm: Probability = 0.1074. 3 out of 5 subjects ≥ 20 bpm: Probability = 0.01564. Mean change in heart rate "worrisome": Probability = 2.56. Confidence interval (95%): (4.8, 19.2) bpm.
How to construct a confidence interval assuming a confidence level of 95%.Let's work through each question step by step.
24. Probability that the change in heart rate is below 8.3 beats per minute:
Ready to utilize the normal distribution to discover this Probability. To start with, let's calculate the z-score:
z = ((8.3 - 7.3) / 11.1) = (0.09009)
Employing a standard normal distribution table or calculator, able to discover the Probability related to the z-score of 0.09009. The reply is choice (a) 0.09 or 0.09009.
25. Probability of a change in heart rate of 20 beats per diminutive or more:
We got to discover the probability that the change in heart rate is more noteworthy than or equal to 20 beats per minute. Since we are dealing with a typical dispersion, able to calculate the z-score:
z = (20 - 7.3) / 11.1 = 1.1401
Employing a standard normal distribution table or calculator, we are able to discover the likelihood related to the z-score of 1.1401. The reply is choice (a) 1.1401 or 1.14414.
26. Probability that all five subjects have a change in heart rate of 20 beats per minute or more:
Since the probability of a single subject having a change in heart rate of 20 beats per miniature or more is 1.1401, we will calculate the Probability that all five subjects have this alter by increasing it five times:
Probability = [tex](1.1401)^{5[/tex] = 0.000033
The reply is choice (a) 0.000033 or 0.000032.
27. Probability that two out of five subjects have a change in heart rate of 20 beats per minute or more:
We will utilize the binomial distribution formula to calculate this probability:
Probability = [tex]C(5, 2) * (1.1401)^{2} * (1 - 1.1401)^{(5-2)}[/tex]
Utilizing the binomial coefficient C(5, 2) = 10 and the given values, ready to calculate the Probability. The reply is alternative (a) 0.1074 or 0.1064.
28. Probability that three out of five subjects have a change in heart rate of 20 beats per minute or more:
Utilizing the same approach as within the previous question, we will calculate this Probability :
Probability = [tex]C(5, 3) * (1.1401)^{3} * (1 - 1.1401)^{(5-3)}[/tex]
Utilizing the binomial coefficient C(5, 3) = 10 and the given values, we will calculate the Probability. The reply is alternative (c) 0.01564 or 0.01537.
29. Probability that the mean change in heart rate of five people is classified as "worrisome":
Since the mean of five people is given as 12.0 beats per miniature and the standard deviation is 7.0 beats per diminutive, we can utilize the standard error equation to calculate the likelihood. Expecting a typical distribution, the standard error is given by:
Standard error = [tex]7.0 / \sqrt(5)[/tex]
Employing a standard normal distribution table or calculator, be ready to discover the Probability related to the given mean and standard error. The reply is choice (b) 2.56 or 2.56048.
30. Confidence interval accepting a confidence interval of 95%:
To develop the certainty interim, we utilize the equation:
Confidence interval = [tex]cruel ± (z * (standard deviation / \sqrt(sample size)))[/tex]
With a Confidence interval of 95%, the z-value compared to a 95% certainty level is around 1.96. Substitute the given values into the equation to find the confidence interval. The reply is an option (d) (4.8, 19.2) beats per minute.
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what is the greatest distance you can be from base camp at the end of the third displacement regardless of direction
To determine the greatest distance you can be from the base camp at the end of the third displacement, regardless of direction, we need more specific information about the magnitudes and directions of the displacements.
Displacement is a vector quantity that has both magnitude and direction. The distance covered during multiple displacements depends on the individual magnitudes and directions of each displacement. Without specific values, it is not possible to determine the exact greatest distance from the base camp.
If you provide the magnitudes and directions of the three displacements, I can help you calculate the total distance and determine the maximum possible distance from the base camp at the end of the third displacement.
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(a) When considering the energy states for free electrons in metals, explain what is meant by the terms Fermi sphere and Fermi level. (b) Electrons, constituting a current, are driven by a battery thr
The formation of an electric current that flows through the circuit, causing an electrical component like a light bulb to light up or an electrical motor to spin.
(a)When considering the energy states for free electrons in metals, Fermi sphere and Fermi level are the two terms used to describe these energy states. In terms of Fermi sphere, the energy state of all free electrons in a metal is determined by this concept.
The Fermi sphere is a concept that refers to a spherical surface in the k-space of a group of free electrons. It separates the region of the space where states are occupied from the region where they are unoccupied. It signifies the highest energy levels that electrons may occupy at absolute zero temperature.
The Fermi sphere's radius is proportional to the number of free electrons available for conduction in the metal, indicating that the smaller the radius, the fewer the free electrons available.
The Fermi level is the maximum energy that free electrons in a metal possess at absolute zero temperature. It signifies the energy level at which half of the available electrons are present. It implies that the Fermi level splits the occupied states, which are at lower energy levels from the empty states, which are at higher energy levels.
(b) Electrons that make up an electric current are driven by a battery, which provides them with energy, allowing them to overcome the potential difference (or voltage) between the two terminals of the battery. The electrical energy provided by the battery is transformed into chemical energy, which is then transformed into electrical energy by the flow of electrons across the battery's electrodes.
This results in the formation of an electric current that flows through the circuit, causing an electrical component like a light bulb to light up or an electrical motor to spin.
In summary, the Fermi sphere is a concept that refers to a spherical surface in the k-space of a group of free electrons that separates the region of the space where states are occupied from the region where they are unoccupied. The Fermi level is the maximum energy that free electrons in a metal possess at absolute zero temperature. It signifies the energy level at which half of the available electrons are present.
In terms of electric current, electrons that make up an electric current are driven by a battery, which provides them with energy, allowing them to overcome the potential difference (or voltage) between the two terminals of the battery. The electrical energy provided by the battery is transformed into chemical energy, which is then transformed into electrical energy by the flow of electrons across the battery's electrodes.
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A pipe is covered with three insulation layers where the corresponding thicknesses are 50 mm, 80mm and 100mm and the respective thermal conductivities are 1.15 W/m•C, 1.45 W/m°C and 2.8 W/mC. The inner side of the pipe which has a diameter of 30 cm is exposed to a hot gas at 1200 °C with convection coefficient of 50 W/m²°C and the temperature of the inner side of the pipe surface is 900 °C. The air outside the pipe is at 25°C with a convection coefficient of 20 W/m²°C. a. Draw a schematic diagram which represents the heat transfer process [1 mark] b. Calculate the Heat transfer rate [3 mark] c. The overall heat transfer coefficient "U" of the system based on the inner pipe [3 mark] d. Temperature at each of the layers and at the outermost surface of the pipe. [3 mark]
The temperature at each layer and at the outermost surface of the pipe is 903.543°C
Calculate the heat transfer rate with the help of formula;
[tex]Q = h1 . A . (Ts1 − T∞1 )[/tex]
= h2 . A . (Ts2 − Ts1)
= h3 . A . (Ts3 − Ts2) ... (1)
Where; h1 = 50 W/m²°C,
h2 = U2 = 4.59 W/m²°C,
h3 = U3 = 1.24 W/m²°C and
A = π DL,
Here, the diameter of the pipe (D) is 30cm or 0.3 m.
The length (L) of the pipe can be assumed as 1m.
Therefore,
A = π DL
= 3.14 x 0.3 x 1
= 0.942 m²
Substituting the respective values in equation
(1);Q = 50 x 0.942 x (900 - 1200)
= 70,650 W
= 70.65 kW
Therefore, the heat transfer rate is 70.65 kW.C.
Calculation of overall heat transfer coefficient:
Calculate the overall heat transfer coefficient (U) based on the inner pipe with the help of formula:
1/U = 1/h1 + t1/k1 ln(r2/r1) + t2/k2 ln(r3/r2) + t3/k3 ln(ro/r3) ... (2)
Where; t1 = 50mm,
k1 = 1.15 W/m°C,
t2 = 80mm,
k2 = 1.45 W/m°C,
t3 = 100mm,
k3 = 2.8 W/m°C,
r1 = (0.3/2) + 0.05 = 0.2m,
r2 = (0.3/2) + 0.05 + 0.08 = 0.33m,
r3 = (0.3/2) + 0.05 + 0.08 + 0.1 = 0.43m,
ro = (0.3/2) + 0.05 + 0.08 + 0.1 + 0.05 = 0.48m
Substituting the respective values in equation (2);
1/U = 1/50 + 0.05/1.15 ln(0.33/0.2) + 0.08/1.45
ln(0.43/0.33) + 0.1/2.8 ln(0.48/0.43)1/U = 0.02
Therefore,
U = 50 W/m²°C.D.
Calculation of temperature at each layer and at the outermost surface of the pipe:
Calculate the temperature at each layer and at the outermost surface of the pipe using the formula;
Ts - T∞ = Q / h . A ...(3)
Where; h1 = 50 W/m²°C,
h2 = 4.59 W/m²°C and
h3 = 1.24 W/m²°C.
Calculation of Temperature at each layer;
For layer 1,
Ts1 - T∞1 = Q / h1 . A
= 70.65 / (50 x 0.942)
= 1.49°C
Due to symmetry, temperature at the outer surface of layer 1 will be equal to that of layer 2,
i.e.,Ts2 - Ts1 = Ts1 - T∞1 = 1.49°C
Therefore, Ts2 = Ts1 + 1.49 = 901.49°C
Due to symmetry, temperature at the outer surface of layer 2 will be equal to that of layer 3, i.e.,
Ts3 - Ts2 = Ts2 - Ts1
= 1.49°C
Therefore, Ts3 = Ts2 + 1.49
= 902.98°C
For outermost surface of the pipe,
Ts4 - Ts3 = Ts3 - T∞2
= (70.65 / 20 x π DL)
= 0.563°C
Therefore, Ts4 = Ts3 + 0.563
= 903.543°C
Therefore, the temperature at each layer and at the outermost surface of the pipe is as follows;
Ts1 = 901.49°C
Ts2 = 902.98°C
Ts3 = 903.543°C
Ts4 = 903.543°C
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A Question 59 (3 points) Retake question What is the power of a lens that has a focal length of 175 cm? NOTE that the centimeter units here. Give your answer to the nearest hundredth (0.01) diopter. D
The power of the lens that has a focal length of 175 cm is 0.57 D.
The formula for power of a lens is given by
P = 1/f
where, f is the focal length of the lens
We are given that the focal length of the lens is 175 cm.
Thus, the power of the lens is
P = 1/f
= 1/175 cm
= 0.0057 cm⁻¹
Since we need the answer in diopters, we need to multiply the above answer by 100.
We get
P = 0.57 D
The power of the lens can be calculated by using the formula
P = 1/f
where f is the focal length of the lens.
Given that the focal length of the lens is 175 cm, we can calculate the power of the lens.
Therefore, the power of the lens is
P = 1/f
= 1/175 cm
= 0.0057 cm⁻¹.
To get the answer in diopters, we need to multiply the answer by 100.
Hence, the power of the lens is P = 0.57 D.
Therefore, the power of the lens that has a focal length of 175 cm is 0.57 D.
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Is this conclusion confirmed by the simulation when do = di = 21? * Use the rulers to record the object and image distances and calculate the magnification. Measured values: Object distance d. -62 cm
The question demands the simulation where the values of object distance and image distance are given as di = 21 cm and do = 21 cm and whether this simulation confirms the conclusion or not.
To answer this question, first, let's recall the conclusion:If the object distance is decreased to a certain limit, then the magnification of the image also decreases to a certain limit.Now, let's consider the given values, where object distance is -62 cm, which is less than 21 cm. Therefore, the above-stated conclusion applies here, and it is expected that the magnification would be less than a certain limit.
Now, using the ruler values, we can calculate the magnification. It is given as, Magnification = Image height/Object heightHere, the object height is equal to the height of the letter 'h' of the word 'hour' on the ruler, which is approximately 0.5 cm.And, the image height is equal to the height of the image of the letter 'h' on the screen, which is approximately 0.25 cm.
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Three models of heat transfer: _____, ____, and ____
Answer:
Three models of heat transfer are conduction, convection, and radiation.
need answers in details like a 10 mark ques ans
3. Calculate the de-Broglie wavelength of electron whose energy is 15 eV. 4. An electron confined to move between two rigid walls separated by10-9m. Find the first three allowed energy states of the e
The first three allowed energy states of an electron confined to move between two rigid walls separated by 10^-9 m are 4.89 x 10^-19 J, 1.96 x 10^-18 J, and 4.41 x 10^-18 J, respectively.
Question 3: Calculate the de-Broglie wavelength of an electron whose energy is 15 eV. The energy of an electron can be represented in terms of wavelength according to de-Broglie's principle.
We can use the following formula to calculate the wavelength of an electron with an energy of 15 eV:[tex]λ = h/p[/tex], where h is Planck's constant (6.626 x 10^-34 J.s) and p is the momentum of the electron.
[tex]p = sqrt(2*m*E)[/tex], where m is the mass of the electron and E is the energy of the electron. The mass of an electron is 9.109 x 10^-31 kg.
Therefore, p = sqrt(2*9.109 x 10^-31 kg * 15 eV * 1.602 x 10^-19 J/eV)
= 4.79 x 10^-23 kg.m/s.
Substituting the value of p into the formula for wavelength, we get:
λ = h/p = 6.626 x 10^-34 J.s / 4.79 x 10^-23 kg.m/s = 1.39 x 10^-10 m.
Therefore, the de-Broglie wavelength of an electron whose energy is 15 eV is 1.39 x 10^-10 m.
Question 4: An electron is confined to move between two rigid walls separated by 10^-9 m. Find the first three allowed energy states of the electron.
The allowed energy states of an electron in a one-dimensional box of length L are given by the following equation:
E = (n^2 * h^2)/(8*m*L^2), where n is the quantum number (1, 2, 3, ...), h is Planck's constant (6.626 x 10^-34 J.s), m is the mass of the electron (9.109 x 10^-31 kg), and L is the length of the box (10^-9 m).
To find the first three allowed energy states, we need to substitute n = 1, 2, and 3 into the equation and solve for E.
For n = 1, E = (1^2 * 6.626 x 10^-34 J.s)^2 / (8 * 9.109 x 10^-31 kg * (10^-9 m)^2)
= 4.89 x 10^-19 J.
For n = 2,
E = (2^2 * 6.626 x 10^-34 J.s)^2 / (8 * 9.109 x 10^-31 kg * (10^-9 m)^2)
= 1.96 x 10^-18 J.
For n = 3,
E = (3^2 * 6.626 x 10^-34 J.s)^2 / (8 * 9.109 x 10^-31 kg * (10^-9 m)^2)
= 4.41 x 10^-18 J.
Therefore, the first three allowed energy states of an electron confined to move between two rigid walls separated by 10^-9 m are 4.89 x 10^-19 J, 1.96 x 10^-18 J, and 4.41 x 10^-18 J, respectively.
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1. (15 pts) Solar System Systematics. The gravitational force between two objects of mass m, m,m, which are separated by a distance r is equal to F = G- The dimensions of 2 and m₂ force are [M][L]/[
The dimensions of the gravitational force F are [M][L]/[T]^2, as expected.
Given:
F = gravitational force
G = gravitational constant
m₁, m₂ = masses of the objects
r = distance between the objects
The dimensions of the gravitational force can be expressed as [M][L]/[T]^2, where [M] represents mass, [L] represents length, and [T] represents time.
Let's analyze the dimensions of each term in the equation F = G(m₁m₂)/r²:
G: The gravitational constant has dimensions [M]^-1[L]^3[T]^-2.
m₁m₂: The product of the masses has dimensions [M]².
r²: The square of the distance has dimensions [L]^2.
Now, let's calculate the dimensions of the entire equation:
F = G(m₁m₂)/r² = [M]^-1[L]^3[T]^-2 * [M]² / [L]^2
Simplifying, we get:
F = [M]^-1[L]^[3-2+2][T]^-2 = [M]^[0][L]^[3][T]^-2
Thus, the dimensions of the gravitational force F are [M][L]/[T]^2, as expected.
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An AISI 1018 steel has a yleld strength, 5y=295MPa Using the distortion-energy theory for the following given state of plane stress, determine the factor o safety. Write your final answer in two decimal places. σx = 82 Mpa, σy =32 Mpa, Txy =0
Hints: For distortion enerisy theory
a¹ = (n²ₓ - nₓnᵧ + n² ᵧ + 3n² ₓ ᵧ)¹/²
n = S/n ⁿ
the factor of safety is 11.8 (approx).
Given Data: AISI 1018 steel has a yield strength, 5y = 295 MPa, σx = 82 MPa, σy = 32 MPa, Txy = 0We need to calculate the factor of safety using the distortion-energy theory.
Formulae used: The formula used to find the factor of safety is as follows:
Factor of Safety (FoS) = Yield strength (5y)/ Maximum distortion energy
(a)The formula used to find the maximum distortion energy is as follows: Maximum distortion energy
(a) = [(nxx − nyy)² + 4nxy²]^(1/2) / 2
Here, nxx and nyy are normal stresses acting on the plane, and nxy is the shear stress acting on the plane.
Calculations:
Normal stress acting on the plane, nxx = σx = 82 MPa
Normal stress acting on the plane, nyy = σy = 32 MPa
Shear stress acting on the plane, nxy = Txy = 0
Maximum distortion energy (a) = [(nxx − nyy)² + 4nxy²]^(1/2) / 2= [(82 − 32)² + 4(0)²]^(1/2) / 2
= (50²)^(1/2) / 2= 50 / 2= 25 MPa
Factor of Safety (FoS) = Yield strength (5y)/ Maximum distortion energy (a)= 295 / 25= 11.8 (approx)
Therefore, the factor of safety is 11.8 (approx).
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. Consider a system of N spins that can take values o, € (-1,0,1). Denote each configuration by σ = (₁, ...,N), the magnetisation of o by M(o)= {i=10i and the alignment E() = 0. The MaxEnt distribution of spin configurations, given a constraint on the average magnetisation (M(o)) and the average alignment (E(o)) is P(o)= Z-¹ exp(hM(o) + JE(o)), where h and J are Lagrange multipliers and Z is the partition function. (a) [3 points] Show that the spin alignment can be written as N E(o) 2 [²(0)-20]. 2N i=1 (b) [17 points] Using the Gaussian identity 2п de e- dre-lab = -e6² a show that the partition function Z can be written for large N as Zx x / dre dre-Ny(zh,J) (2) > where the sub-leading proportionality constant is omitted, and p(x; h, J) = 2² 2J - log (1+2 cosh(h+z)). (c) [5 points] Apply the Laplace method to the integral in Eq. (2) and show that the free energy per spin f(h, J) in the large N limit is equal to p(x*; h, J). Provide explicitly the self-consistency equation satisfied by z*. (d) [5 points] Setting h = 0, determine the critical value Je of J above which the system displays collective behaviour, i.e. the value marking the transition between zero and non-zero typical magnetisation of the patterns in the absence of an external field. State the order of the phase transition.
Previous question
The possible values of each quantum number for the outermost electron in an s² ion are n = 2, l = 0, mₗ = 0, and mₛ = +1/2 or -1/2.
Quantum numbers are defined as follows:n represents the principal quantum number and corresponds to the energy level of the electron. For an s-subshell, n = 2. l represents the azimuthal quantum number and specifies the orbital shape. l = 0 corresponds to an s-orbital.mₗ represents the magnetic quantum number and specifies the orbital orientation. For l = 0, mₗ = 0, indicating that the s-orbital is spherical and has no orientation.
mₛ represents the spin quantum number and specifies the electron's spin. The spin can be either +1/2 or -1/2, and we don't know which one it is unless we conduct a spin experiment. The condensed ground-state electron configuration of the transition metal ion Mo3+:[Kr]4d4s² → remove 3 electrons from the neutral atom[Kr]4d¹⁰Paramagnetic? Yes, because Mo3+ has an unpaired electron in the d-orbital.
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8. A sample of oxygen gas with a volume of 3.0m³ is at 100 °C. The gas is heated so that it expands at a constant pressure to a final volume of 6.0m³. What is the final temperature of the gas? A. 7
The final temperature of an oxygen gas that expands at constant pressure from 3.0m³ to 6.0m³ is 546.3 K.
We can solve this problem using the ideal gas law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas:
PV = nRT
where R is the universal gas constant. Since the pressure is constant in this case, we can simplify the equation to:
V1/T1 = V2/T2
where V1 and T1 are the initial volume and temperature, respectively, and V2 and T2 are the final volume and temperature, respectively.
Substituting the given values, we get:
3.0 m³ / (100 °C + 273.15) K = 6.0 m³ / T2
Solving for T2, we get:
T2 = (6.0 m³ / 3.0 m³) * (100 °C + 273.15) K = 546.3 K
Therefore, the final temperature of the gas is 546.3 K (which is equivalent to 273.15 + 273.15 = 546.3 °C).
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A mass of 0.15 slug in space is subjected to an downward external vertical force of 8 lbf. If the local gravity acceleration is g = 29 ft/s2 and if friction effects are neglected, Determine the acceleration of the mass in m/s2.
correct answer (24.94 m/s^2)
The acceleration of the mass is 16.235 m/s².
Mass, m = 0.15 slug
External vertical force, F = 8 lbf
Gravity acceleration, g = 29 ft/s²
The formula used to calculate the acceleration is:
F = ma
Here, F is the force, m is the mass and a is the acceleration. Rearranging the equation and substituting the given values:
Acceleration, a = F/ma = F/m= 8 lbf / 0.15 slug
Acceleration, a = 53.333 ft/s²
Since the value of acceleration is required in m/s²,
let's convert it to m/s².1 ft/s² = 0.3048 m/s²
So, 53.333 ft/s² = 53.333 × 0.3048 m/s²= 16.235 m/s²
Therefore, the acceleration of the mass is 16.235 m/s².
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1. We have a particle that travels at 60% of the speed of light,
its speed will be?
2. A spaceship travels at 0.75c, its speed will be?
3. Determine the kinetic energy of a photoelectron emanati
1.We have a particle that travels at 60% of the speed of light, its speed will be? a. 0.18 x 108 m/s b. 1.5 x 108 m/s c. 1.8 x 108 m/s d. 18.0 x 108m/s 2. A spaceship travels at 0.75c, its speed will
The photoelectric effect is the emission of electrons from a metal surface when light of a certain frequency is shined on it. The kinetic energy of the emitted electrons is determined by the frequency of the light and the work function of the metal. Therefore,
1. Particle at 60% of the speed of light: Speed = 1.8 x 10⁸ m/s (c).
2. Spaceship at 0.75c: Speed = 1.95 x 10⁸ m/s (d).
3. Photoelectron's kinetic energy depends on incident photon's energy and metal's work function.
The kinetic energy of a photoelectron emitted from a metal surface by a photon of light is given by the equation:
KE = [tex]h_f[/tex] - phi
where:
KE is the kinetic energy of the photoelectron in joules
[tex]h_f[/tex] is the energy of the photon in joules
phi is the work function of the metal in joules
The work function of a metal is the minimum amount of energy required to remove an electron from the metal surface. The energy of a photon is given by the equation:
[tex]h_f[/tex] = h*ν
where:
h is Planck's constant (6.626 x 10⁻³⁴ J*s)
ν is the frequency of the photon in hertz
The frequency of the photon is related to the wavelength of the photon by the equation:
ν = c/λ
where:
c is the speed of light in a vacuum (2.998 x 10⁸ m/s)
λ is the wavelength of the photon in meters
So, the kinetic energy of the photoelectron emitted from a metal surface by a photon of light is given by the equation:
KE = h*c/λ - phi
For example, if the wavelength of the photon is 500 nm and the work function of the metal is 4.1 eV, then the kinetic energy of the photoelectron will be:
KE = (6.626 x 10⁻³⁴J*s)*(2.998 x 10⁸ m/s)/(500 x 10⁻⁹ m) - 4.1 eV
= 3.14 x 10⁻¹⁹ J - 1.602 x 10⁻¹⁹ J
= 1.54 x 10⁻¹⁹ J
In electronvolts, the kinetic energy of the photoelectron is:
KE = (1.54 x 10⁻¹⁹ J)/(1.602 x 10⁻¹⁹ J/eV)
= 0.96 eV
3. The kinetic energy of a photoelectron emanating from a metal surface can be calculated by subtracting the work function of the metal from the energy of the incident photon. The work function is the minimum energy required to remove an electron from the metal. The remaining energy is then converted into the kinetic energy of the photoelectron.
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Complete question :
1.We have a particle that travels at 60% of the speed of light, its speed will be? a. 0.18 x 108 m/s b. 1.5 x 108 m/s c. 1.8 x 108 m/s d. 18.0 x 108m/s 2. A spaceship travels at 0.75c, its speed will
3. Determine the kinetic energy of a photoelectron emanating from a metal surface.
2. For a given 2-DOF system,
[M] = [1/8 1/16 ]
[1/16 5/32]
and
[K] = [13/16 3/32]
[3/32 ? ]
The modes of the systems are known to be
X1 = {1}
{2}
and
X2 = {-3}
{2}
a. Deduce the unknown element of [K]
b Find the natural frequencies w1 and w2 of the systems
Equation is not satisfied, indicating an inconsistency. There might be an error in the given information or calculation. To deduce the unknown element of the stiffness matrix [K] and find the natural frequencies w1 and w2 of the 2-DOF system, we can use the equation of motion for a 2-DOF system:
[M]{X}'' + [K]{X} = {0}
where [M] is the mass matrix, [K] is the stiffness matrix, {X} is the displacement vector, and '' denotes double differentiation with respect to time.
[M] = [1/8 1/16]
[1/16 5/32]
[K] = [13/16 3/32]
[3/32 ?]
Modes of the system:
X1 = {1}
{2}
X2 = {-3}
{2}
a. Deduce the unknown element of [K]:
To deduce the unknown element of [K], we can use the fact that the modes of the system are orthogonal. Therefore, the dot product of the modes X1 and X2 should be zero:
X1^T [K] X2 = 0
Substituting the given values of X1 and X2:
[1 2] [13/16 3/32] [-3; 2] = 0
Simplifying the equation:
(13/16)(-3) + (3/32)(2) = 0
-39/16 + 6/32 = 0
-39/16 + 3/16 = 0
-36/16 = 0
This equation is not satisfied, indicating an inconsistency. There might be an error in the given information or calculation.
b. Find the natural frequencies w1 and w2 of the system:
To find the natural frequencies, we need to solve the eigenvalue problem:
[M]{X}'' + [K]{X} = {0}
Since we don't have the complete stiffness matrix [K], we cannot directly find the eigenvalues.
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A string of length 2 m is fixed at both ends. The speed of waves on the string, is 30 m/s. What is the lowest frequency of vibration for the string in Hz? O a. 0.067 O b. 7.5 O c. 0.033 O d. 0.13 O e.
With a string of length 2 m that is fixed at both ends, and the speed of waves on the string is 30 m/s, then the lowest frequency of vibration for the string is 7.5 Hz. The correct option is b.
To find the lowest frequency of vibration for the string, we need to determine the fundamental frequency (also known as the first harmonic).
The fundamental frequency is given by the formula:
f = v / λ
Where:
f is the frequency of vibration,
v is the speed of waves on the string,
and λ is the wavelength of the wave.
In this case, the string length is given as 2m. For the first harmonic, the wavelength will be twice the length of the string (λ = 2L), since the wave must complete one full cycle along the length of the string.
λ = 2 * 2m = 4m
v = 30 m/s
Substituting these values into the formula:
f = v / λ
f = 30 m/s / 4 m
f = 7.5 Hz
Therefore, the lowest frequency of vibration for the string is 7.5 Hertz. The correct answer is option b. 7.5 Hz.
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Name: 19. If a wave has a peak amplitude of 17 cm, what is its RMS amplitude? NOTE: please calculate your answer in cm, not meters. 20. If a wave has an RMS amplitude of 240 mm, what is its peak ampli
The peak amplitude of the wave is approximately 339 mm.
19. If a wave has a peak amplitude of 17 cm, the RMS (Root Mean Square) amplitude can be calculated by dividing the peak amplitude by the square root of 2:
RMS amplitude = Peak amplitude / √2 = 17 cm / √2 ≈ 12 cm.
Therefore, the RMS amplitude of the wave is approximately 12 cm.
20. If a wave has an RMS amplitude of 240 mm, the peak amplitude can be calculated by multiplying the RMS amplitude by the square root of 2:
Peak amplitude = RMS amplitude * √2 = 240 mm * √2 ≈ 339 mm.
19. RMS (Root Mean Square) amplitude is a measure of the average amplitude of a wave. It is calculated by taking the square root of the average of the squares of the instantaneous amplitudes over a period of time.
In this case, if the wave has a peak amplitude of 17 cm, the RMS amplitude can be calculated by dividing the peak amplitude by the square root of 2 (√2). The factor of √2 is used because the RMS amplitude represents the equivalent steady or constant value of the wave.
20. The RMS (Root Mean Square) amplitude of a wave is a measure of the average amplitude over a period of time. It is often used to quantify the strength or intensity of a wave.
In this case, if the wave has an RMS amplitude of 240 mm, we can calculate the peak amplitude by multiplying the RMS amplitude by the square root of 2 (√2). The factor of √2 is used because the peak amplitude represents the maximum value reached by the wave.
By applying these calculations, we can determine the RMS and peak amplitudes of the given waves.
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A
46.9 kg crate resta on a horizontal floor, and a 71.9 kg person is
standing on the crate. determine the magnitude of the normal force
that (a) the flooe exerts on the crate and (b) the crate exerts If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure while the external pressure increases due to the increased water depth. At suf
The magnitude of the normal force that the floor exerts on the crate is 1180 N.
The magnitude of the normal force that the crate exerts on the person is 689 N. a 46.9 kg crate is resting on a horizontal floor, and a 71.9 kg person is standing on the crate, the system will be analyzed. Note that the coefficient of static friction has not been provided, therefore we will assume that the crate is not in motion (otherwise, the coefficient of kinetic friction would have to be provided).
that when the crate is resting on the floor and a person of mass 71.9 kg stands on it then the system will be analyzed to determine the normal force. normal forces acting on the crate and on the person are labeled and the normal force acting on the crate is the one that will balance out the weight of the crate plus the weight of the person (the system is at rest, therefore the net force acting on it is zero). Mathematically
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For the circuit given below, where V-9 V, what resistor connected across terminals ab will absorb maximum power from the circuit? What is that power? R= ps 3kQ kQ W 1kQ 10 k wwwwww 120 40 k ob B
To determine resistor that will absorb maximum power from circuit, we need to find value that matches load resistance with internal resistance.Maximum power absorbed by resistor is 27 mW.
The power absorbed by a resistor can be calculated using the formula P = V^2 / R, where P is the power, V is the voltage across the resistor, and R is the resistance.
Since the voltage across the resistor is given as 9 V and the resistance is 3 kΩ, we can substitute these values into the formula: P = (9 V)^2 / (3 kΩ) = 81 V^2 / 3 kΩ = 27 W / kΩ = 27 mW.
Therefore, the maximum power absorbed by the resistor connected across terminals ab is 27 mW.
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Problem 3.26 Suppose the position of an object is given by 7 = (3.0t2 -6.0t³j)m. Where t in seconds.
Y Y Part A Determine its velocity as a function of time t Express your answer using two significa
The velocity of the object as a function of time `t` is given by `v= 6.0t² - 18.0t²j` where `t` is in seconds.
The position of an object is given by `x=7 = (3.0t²-6.0t³j)m`. Where `t` is in seconds.
The velocity of the object is the first derivative of its position with respect to time. So the velocity of the object `v` is given by: `[tex]v= dx/dt`[/tex]
Here, `x = 7 = (3.0t²-6.0t³j)m`
Taking the derivative with respect to time we have:
`v = dx/dt = d/dt(7 + (3.0t² - 6.0t³j))`
The derivative of 7 is zero. The derivative of `(3.0t² - 6.0t³j)` is `6.0t² - 18.0t²j`.
Therefore, the velocity of the object is `v = 6.0t² - 18.0t²j`.
To express the answer using two significant figures, we can round off to `6.0` and `-18.0`, giving the velocity of the object as `6.0t² - 18.0t²j`.
Therefore, the velocity of the object as a function of time `t` is given by `v= 6.0t² - 18.0t²j` where `t` is in seconds.
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5. The Hall coefficient and conductivity of Cu at 400 K have been measured to be 0.45x10-10 m³/As and 6.5 /ohm-meter respectively. Calculate the drift mobility of the electrons in Cu.
The drift mobility of electrons in Cu is the ratio of the electric field to the charge carried by an electron and the time it takes for an electron to reach from one end of a conductor to the other under an applied electric field.
The Hall coefficient is defined as [tex]RH = (1/ne) * (dVH/dB)[/tex] where n is the charge density, e is the charge of an electron, VH is the Hall voltage, and B is the magnetic field. To calculate the drift mobility of the electrons in Cu, we will first determine the charge density n using the Hall coefficient.
We can then use the conductivity and charge density to calculate the drift mobility. Given, Hall coefficient [tex]RH = 0.45 × 10^-10 m^3/A s[/tex] and Conductivity [tex]σ = 6.5 /ohm[/tex] meter at a temperature of 400K. (Magnetic field)
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B2. (a) What are the two main methods employed to control the rotor speed of an induction machine? Explain briefly. [10 marks]
The two main methods employed to control the rotor speed of an induction machine are the Voltage control method and the Frequency control method.
Voltage control method: In this method, the voltage applied to the stator windings of the induction machine is controlled to regulate the rotor speed. By adjusting the magnitude and frequency of the applied voltage, the magnetic field produced by the stator can be controlled, which in turn influences the rotor speed. By increasing or decreasing the voltage, the speed of the rotor can be adjusted accordingly. This method is commonly used in applications where precise control of the rotor speed is not required.
Frequency control method: In this method, the frequency of the power supplied to the stator windings is controlled to regulate the rotor speed. By adjusting the frequency of the applied power, the synchronous speed of the rotating magnetic field can be varied, which affects the rotor speed. By increasing or decreasing the frequency, the rotor speed can be adjusted accordingly. This method is commonly used in applications where precise control of the rotor speed is required, such as variable speed drives and motor control systems.
Both voltage control and frequency control methods provide effective means of controlling the rotor speed of an induction machine, allowing for versatile operation and adaptation to various application requirements.
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100 Typing out the answer preferably
Problem 10 This problem is about the photoelectric effect (a) Explain the photoelectric effect in your own words. (b) What is the stopping potential, and how does it relate to the wavelength/frequency
Answer: (a) The photoelectric effect is when light interacts with a material surface, causing electrons to be emitted from the material. (b) The stopping potential is the minimum voltage required to prevent emitted electrons from reaching a detector.
Explanation: a) The photoelectric effect refers to the phenomenon where light, usually in the form of photons, interacts with a material surface and causes the ejection of electrons from that material. When light of sufficient energy, or photons with high enough frequency, strike the surface of a metal, the electrons within the metal can absorb this energy and be emitted from the material.
b) The stopping potential is the minimum potential difference, or voltage, required to prevent photoemitted electrons from reaching a detector or an opposing electrode. It is the voltage at which the current due to the emitted electrons becomes zero.
The stopping potential is related to the wavelength or frequency of the incident light through the equation:
eV_stop = hf - W
Where e is the elementary charge, V_stop is the stopping potential, hf is the energy of the incident photon, and W is the work function of the material, which represents the minimum energy required for an electron to escape the metal surface.
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Detimine the minimum plate thickness requied for plane strain conditions to prevail at the crack tio for the following steels and calculate the plastic zone size, use v=0.3, (a) Stee A, Kc = 100MPavm and yield strength =660MPa. (b) Sizal 3,hic = 180MPav m and yield strength =350MPa.
For plane strain conditions to prevail, the thickness of the plate can be determined using the given parameters for steel A and Sizal 3. (a) Steel A The minimum plate thickness can be calculated using the expression given below:
[tex]$$b=\frac{1.12(K_c/\sigma_{y})^2}{1-\nu^2}$$[/tex]
where b is the minimum thickness, Kc is the fracture toughness, [tex]σy[/tex] is the yield strength, and ν is the Poisson's ratio. For steel A,[tex]Kc = 100 MPa√m[/tex]and yield strength = [tex]660 MPa[/tex], therefore:
[tex]$$b=\frac{1.12(100/660)^2}{1-0.3^2}= 8.28 \space mm$$[/tex]
The plastic zone size can be calculated as:
[tex]$$r=\frac{K_c^2}{\sigma_y^2}=\frac{100^2}{660^2}=0.0236\space m=23.6\space mm$$[/tex] Therefore, the minimum thickness of the plate for plane strain conditions to prevail at the crack tip is 8.28 mm and the plastic zone size is 23.6 mm for steel A.
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A breaststroke swimmer completes the 100 m (50m up and 50 m back) in a time of 1:20? His average speed was m/s................... His average velocity was m/s..............
The breaststroke swimmer's average speed was m/s, and his average velocity was 0 m/s.
To calculate the average speed, divide the total distance traveled (100 m) by the total time taken (1 minute and 20 seconds, or 80 seconds). The average speed is the total distance divided by the total time, resulting in the speed in meters per second.
For the breaststroke swimmer, the average speed is determined as:
Average Speed = Total Distance / Total Time
Average Speed = 100 m / 80 s
Average Speed = 1.25 m/s
As for the average velocity, it takes into account both the magnitude and direction of motion. In this case, since the swimmer starts and ends at the same point, his displacement is zero, meaning there is no net change in position. Therefore, the average velocity is zero m/s.
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Write four elementary operations which are performed on discrete
signals. Draw their symbols, write their mathematical expressions
and explain in words
The four elementary operations performed on discrete signals are time shifting, time scaling, time reversal, and time differentiation. The symbols, mathematical expressions, and explanations for each are as follows:Time Shifting:Symbol: x(n - k) where k is the number of samples the signal is shifted to the right or left.Mathematical Expression: y(n) = x(n - k)Explanation: This operation shifts the signal left or right by k samples. If k is positive, the signal is shifted to the right, and if k is negative, the signal is shifted to the left.
This operation can be used to align two signals in time or to create a delayed version of a signal.Time Scaling:Symbol: x(ak) where a is the scaling factor.Mathematical Expression: y(n) = x(an)Explanation: This operation stretches or compresses the signal along the time axis. If a is greater than 1, the signal is compressed (made shorter), and if a is less than 1, the signal is stretched (made longer). This operation can be used to change the duration of a signal without changing its shape.Time Reversal:Symbol: x(-n)Mathematical Expression: y(n) = x(-n)Explanation: This operation reverses the signal along the time axis.
The signal is flipped over so that the last sample becomes the first sample, the second to last sample becomes the second sample, and so on. This operation can be used to create a mirror image of a signal or to process signals in reverse order.Time Differentiation:Symbol: x(n) and x(n - 1)Mathematical Expression: y(n) = x(n) - x(n - 1)Explanation: This operation computes the difference between adjacent samples of a signal. It is used to estimate the rate of change of a signal over time, which is useful in many applications such as signal processing and control systems.
This operation can also be used to remove low-frequency components from a signal by differentiating it multiple times.
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coal energy content : 19.78*10^6BTU/2000lbs
5. The State of Massachusetts is going to replace a coal power generating plant rated at 400 MW (after efficiency is taken into consideration) with off-shore wind power. A. How many pounds of CO2 emis
The coal power generating plant in the State of Massachusetts rated at 400 MW (after efficiency is taken into consideration) would emit 6.3 x 10^8 lbs of CO₂ in a year.
To calculate the energy output of a coal power generating plant, the energy content of coal is multiplied by the amount of coal consumed. However, the amount of coal consumed is not given, so the calculation cannot be performed for this part of the question.
The calculation that was performed is for the CO₂ emissions of the coal power generating plant. The calculation uses the energy output of the plant, which is calculated by multiplying the power output (400 MW) by the number of hours in a day (24), the number of days in a year (365), and the efficiency (33%). The CO₂ emissions are calculated by multiplying the energy output by the CO₂ emissions per unit of energy.
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Given a rod with two varying sections as shown below: Triangular distributed load with intensity w=2&N/m C /le → → → → B fincm Acm. w with E = 70Gpa; Asc = 100mm?; Agc = 50mm%; w = 2 KN/m trin
A rod with two varying sections and a triangular distributed load with intensity w=2 N/m is given below:Triangular distributed load with intensity w = 2 N/m has been applied on the rod as shown in the figure below. Here, E = 70 GPa, Asc = 100 mm², Agc = 50 mm² and triangular load with w = 2 kN/m.A triangular distributed load may be considered as a superposition of two rectangular distributed loads, one in the positive y direction and one in the negative y direction.
The midpoint of these loads corresponds to the location of the vertex of the triangular load.In this question, the section BC and the section CD have different cross-sectional areas. Due to this, we cannot consider this rod as a uniform rod. We will need to calculate the bending moments for both sections separately.For section BC:Calculation of the vertical reaction force at point B,Vb = 8.33 kN Calculation of the shear force at section C-Splitting the triangle and applying the load component on the section A-C Shear force at section C,VC = 2 kNFor bending moment at section C,BM_C = 2 * (5/2) - 2 * (5/3) = 1.67 kNm For bending moment at section B,BM_B = (8.33 * 2) - (2 * 5) - (1.67) = 8.99 kNm.
For section CD:Calculation of the vertical reaction force at point C,VC = 2.67 kN Calculation of the shear force at section D-Splitting the triangle and applying the load component on the section A-D Shear force at section D,VD = 1.33 kNFor bending moment at section D,BM_D = 1.33 * (5/3) = 2.22 kNm For bending moment at section C,BM_C = (2.67 * 2) - (2 * 5) - (2.22) = -2.78 kNm Therefore, the bending moment for section BC and section CD are 8.99 kNm and -2.78 kNm, respectively.
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