The star and its planet could be a maximum of 0.806 light-years away from the telescope if the wavelength used was 1400 nm.
Part G) One important goal of astronomers is to have a telescope in space that can resolve planets like the earth orbiting other stars. If a planet orbits its star at a distance of 1.5×1011m (the radius of the earth's orbit around the sun) and the telescope has a mirror of diameter 8.0m, the maximum distance from the telescope the star and its planet could be if the wavelength used was 690nm can be calculated using Rayleigh criterion.Rayleigh criterion states that when two sources of light are separated by a certain angle, the smallest resolvable detail is when the maximum of one coincides with the first minimum of the other.
Mathematically, this means the following:θ = 1.22λ/Dwhere:θ is the angular resolutionλ is the wavelength of the lightD is the diameter of the mirrorFor the star and planet to be resolvable, the angular resolution must be less than or equal to the angle between the star and planet from the telescope.θ ≤ angle between star and planet from the telescopeThe angle between the star and planet from the telescope can be calculated using basic trigonometry as follows:tan θ = (1.5×1011 m) / (distance from telescope)θ = tan⁻¹[(1.5×1011 m) / (distance from telescope)]Therefore,θ ≤ tan⁻¹[(1.5×1011 m) / (distance from telescope)]Substituting for θ in the Rayleigh criterion equation, we get:tan⁻¹[(1.5×1011 m) / (distance from telescope)] ≤ 1.22(690×10⁻⁹ m) / (8.0 m)tan[tan⁻¹[(1.5×1011 m) / (distance from telescope)]] ≤ 1.22(690×10⁻⁹ m) / (8.0 m)tan[tan⁻¹[(1.5×1011 m) / (distance from telescope)]] ≤ 1.050x10⁻⁹distance from telescope = 1.5×10¹¹m / tan [1.050x10⁻⁹]≈ 1.429x10¹⁷mConverting to light-years,1.429x10¹⁷ m = 1.51 ly (1 ly = 9.46×10¹⁵m).
Therefore, the star and its planet could be a maximum of 1.51 light-years away from the telescope if the wavelength used was 690nm.Part H) Similarly, for the wavelength used as 1400 nm,θ ≤ tan⁻¹[(1.5×1011 m) / (distance from telescope)]tan⁻¹[(1.5×1011 m) / (distance from telescope)] ≤ 1.22(1400×10⁻⁹ m) / (8.0 m)tan[tan⁻¹[(1.5×1011 m) / (distance from telescope)]] ≤ 1.94x10⁻⁹distance from telescope = 1.5×10¹¹m / tan [1.94x10⁻⁹]≈ 7.62x10¹⁶mConverting to light-years,7.62x10¹⁶ m = 0.806 ly (1 ly = 9.46×10¹⁵m). Therefore, the star and its planet could be a maximum of 0.806 light-years away from the telescope if the wavelength used was 1400 nm.
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IDEAL AND REAL GASES 5. (a)Determine the specific volume of superheated water vapor at 10 MPa and 400°C, using the ideal-gas equation, (b) the generalized compressibility chart, and (c) the steam tables. Also determine the error involved in the first two cases (b) Explain the terms (i) Compressibility factor(ii) reduced pressure (iii) reduced temperature (iv) Principle of corresponding states (v) Critical pressure (vi) Critical temperature and (vii) Compressibility chart. (c) explain the terms involved in (i) vanderwaal's equation of state (ii) Beatie- bridgeman equation of state.
In order to determine the specific volume of superheated water vapor at 10 MPa and 400°C, three methods are used: the ideal-gas equation, the generalized compressibility chart, and the steam tables.
Each method provides different approaches for calculating the specific volume, and the errors associated with the first two methods are also determined.
(i) Compressibility factor: The compressibility factor, denoted as Z, is a measure of how much a real gas deviates from ideal gas behavior. It is the ratio of the actual volume of a gas to the volume predicted by the ideal-gas equation at the same pressure and temperature. Z is influenced by intermolecular forces and the size of the gas molecules.
(ii) Reduced pressure: The reduced pressure, denoted as P_r, is a dimensionless quantity defined as the actual pressure of a gas divided by its critical pressure. It is used in the principle of corresponding states to compare the behavior of different gases at similar reduced temperatures.
(iii) Reduced temperature: The reduced temperature, denoted as T_r, is a dimensionless quantity defined as the actual temperature of a gas divided by its critical temperature. It is used in the principle of corresponding states to compare the behavior of different gases at similar reduced pressures.
(iv) Principle of corresponding states: The principle of corresponding states states that the thermodynamic properties of different gases can be compared when their reduced pressures and reduced temperatures are the same. This principle allows for the use of generalized charts and equations that are valid across a wide range of gases.
(v) Critical pressure: The critical pressure is the highest pressure at which a substance can exist as a distinct liquid and vapor phase. Above the critical pressure, the substance exists as a supercritical fluid, where the distinction between liquid and vapor phases disappears.
(vi) Critical temperature: The critical temperature is the highest temperature at which a substance can exist as a distinct liquid and vapor phase. Above the critical temperature, the substance exists as a supercritical fluid, regardless of pressure.
(vii) Compressibility chart: A compressibility chart is a graphical representation of the compressibility factor Z as a function of reduced pressure and reduced temperature. It is used to estimate the compressibility factor for real gases based on experimental data or correlations.
(i) Van der Waals equation of state: The Van der Waals equation of state is a modification of the ideal gas equation that accounts for intermolecular forces and finite molecular size. It includes correction terms for the attractive forces between gas molecules and the excluded volume occupied by the molecules.
(ii) Beattie-Bridgeman equation of state: The Beattie-Bridgeman equation of state is an empirical equation used to describe the behavior of gases at high pressures. It incorporates additional parameters to account for the effects of pressure on the compressibility factor and specific volume.
In summary, the specific volume of superheated water vapor at 10 MPa and 400°C can be determined using the ideal-gas equation, the generalized compressibility chart, or the steam tables.
The terms mentioned in the question, such as compressibility factor, reduced pressure, reduced temperature, principle of corresponding states, critical pressure, critical temperature, compressibility chart, Van der Waals equation of state, and Beattie-Bridgeman equation of state, are all important concepts in thermodynamics and gas behavior.
They help us understand the deviations of real gases from ideal gas behavior and provide methods for analyzing and predicting their properties.
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A beam of 40keV electrons strikes a copper target. What is the expected cut-off wavelength? Given that h = 6.626 x 10-34 J.s e = 1.6 x 10-19C c = 3 x 108ms-¹ Select one: O a. 3.22 x1013m O b. 3.22 x
The expected cut-off wavelength for a beam of 40 keV electrons striking a copper target is approximately 3.11 x [tex]10^{-11}[/tex] m. The correct answer is option c.
The cut-off wavelength (λc) can be determined using the equation:
λc = (hc) ÷ E
Where:
λc = Cut-off wavelength
h = Planck's constant (6.626 x [tex]10^{34}[/tex] J·s)
c = Speed of light (3 x [tex]10^{8}[/tex] m/s)
E = Energy of the electrons
To calculate the cut-off wavelength, we need to convert the given energy from kiloelectronvolts (keV) to joules (J):
E = 40 keV × (1.6 x [tex]10^{-9}[/tex] C/keV) = 6.4 x [tex]10^{-18}[/tex] J
Substituting the values into the equation:
λc = [(6.626 x [tex]10^{-34}[/tex] J·s) * (3 x [tex]10^{8}[/tex] m/s)] / (6.4 x [tex]10^{-18}[/tex] J)
Simplifying the equation:
λc ≈ 3.11 x [tex]10^{-11}[/tex] m
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The complete question is:
A beam of 40keV electrons strikes a copper target. What is the expected cut-off wavelength? Given that h = 6.626 x [tex]10^{-34}[/tex] J.s e = 1.6 x [tex]10^{-19}[/tex] C c = 3 x [tex]10^{8}[/tex]ms-¹
Select one:
a. 3.22 x[tex]10^{13}[/tex] m
b. 3.22 x [tex]10^{10}[/tex] m
c. 3.11 x [tex]10^{-11}[/tex] m
d. 3.11 x [tex]10^{-14[/tex] m
please solve
The expression of MHC class II molecules by B cells Select one: a. is down-regulated by IL-4 b. is up-regulated by IFN-gamma c. None of the answers is true d. All answers are true e. is down-regulated
Answer:
the correct answer is: b. The expression of MHC class II molecules by B cells is up-regulated by IFN-gamma, while it is not down-regulated by IL-4.
Explanation:
The expression of MHC class II molecules by B cells is up-regulated by IFN-gamma. MHC class II molecules are involved in antigen presentation, where they display peptide antigens to helper T cells. This interaction is crucial for the activation of adaptive immune responses. IFN-gamma, which is primarily produced by activated T cells and natural killer cells, plays a significant role in the immune response by enhancing the expression of MHC class II molecules on various cell types, including B cells. This up-regulation of MHC class II molecules allows B cells to present antigens effectively and activate helper T cells, leading to an efficient immune response against pathogens.
On the other hand, IL-4 does not down-regulate the expression of MHC class II molecules by B cells. IL-4 is a cytokine that primarily promotes the differentiation of B cells into antibody-secreting plasma cells and enhances antibody production. It is involved in the humoral immune response. While IL-4 plays a role in B cell activation and antibody production, it does not directly affect the expression of MHC class II molecules. Instead, IL-4 primarily influences the expression of other molecules involved in B cell activation, proliferation, and antibody production.
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a spherical surface surrounds a point charge 2.06nc. find the electric flux (unit in nm2/c) through the surface when the charge at the position r/2 distance away from the center of the spherical surface.
The electric flux through the spherical surface surrounding a point charge of 2.06 nC is 5.15 x 10^5 nm^2/C when the charge is at a position r/2 distance away from the center of the spherical surface.
The electric flux through a closed surface is given by Gauss's Law, which states that the electric flux (Φ) is equal to the total charge enclosed (Q) divided by the permittivity of free space (ε₀).
Φ = Q / ε₀
In this case, the charge enclosed is the point charge of 2.06 nC. The permittivity of free space (ε₀) is a constant value.
To determine the electric flux when the charge is at a position r/2 distance away from the center of the spherical surface, we need to consider the symmetry of the situation. Since the charge is spherically symmetric, the electric flux through the spherical surface will be the same regardless of the charge's position.
Therefore, the electric flux through the spherical surface is 5.15 x 10^5 nm^2/C, regardless of the charge's position at a distance of r/2 from the center of the spherical surface.
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What happens when you add heat to a mass of ice which is at \( 0^{\circ} \mathrm{C} \) ? The ice will increase in temperature. Nothing will happen. Some of the ice will melt. The heat will disappear.
When you add heat to a mass of ice that is at 0∘C, some of the ice will melt. At this temperature, ice is in a solid state, and by adding heat energy to it, you increase the temperature of the ice.
The added heat energy will be used to break the intermolecular bonds holding the ice molecules together, causing the ice to melt and transform into liquid water. Therefore, the correct answer is: Some of the ice will melt.
It's important to note that during this phase change from solid to liquid, the temperature of the system remains constant at 0 C until all the ice has melted. This phenomenon is known as the latent heat of fusion, where heat energy is absorbed by the substance without causing a change in temperature.
Therefore, the correct answer is: Some of the ice will melt.
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In a counting experiment, the radiation γ emitted by a source of (137,55)Cs is measured. The counting rate in the presence of the fund is 10 pulses/s. Remove the source and measure the counting rate for the fund. 8 pulses/s are obtained. The same source is used in a βγ coincidence measurement experiment. The counting rate in coincidence in the presence of the fund obtained in this case is 3 pulses/s.It is required to calculate the maximum fund counting rate in the coincidence measurement method.
Indication: It should be borne in mind that for the measurement of the fund and for the measurement of sources and in the presence of the fund, in both types of experiments equal measurement times were used for the two types of counting rates. Make any other assumptions necessary, if any, for performing the calculation.
The maximum fund counting rate in the coincidence measurement method is 10.64 pulses/s
In a counting experiment, it is given that the radiation γ emitted by a source of (137,55)Cs is measured. The counting rate in the presence of the fund is 10 pulses/s.
Remove the source and measure the counting rate for the fund.
8 pulses/s are obtained.
The same source is used in a βγ coincidence measurement experiment.
The counting rate in coincidence in the presence of the fund obtained in this case is 3 pulses/s.
To find: The maximum fund counting rate in the coincidence measurement method.
Assumptions made for performing the calculation:- Equal measurement times were used for the two types of counting rates. Also, we can assume that there is no dead time in the detection system, and the detectors have 100% efficiency.
Formula Used: Maximum counting rate = Counting rate in the presence of the fund / (1 - ε)
Where, ε = Coincidence efficiency
For a βγ coincidence measurement experiment,
ε can be defined as: ε = R(coincidence) / [R(β) × R(γ)]
Now, Let's find ε:
We have R(coincidence) = 3 pulses/s
We know, for γ counting experiment, the counting rate in the presence of the fund is 10 pulses/s
Hence, R(γ) = 10 pulses/s
For βγ coincidence measurement experiment, let's assume R(β) = 5 pulses/s (It is an assumption.)
Now, we can find ε as:
ε = 3 / (5 × 10)ε = 0.06
Now, Maximum counting rate = Counting rate in the presence of the fund / (1 - ε)Maximum counting rate = 10 / (1 - 0.06)
Maximum counting rate = 10.64 pulses/s
Therefore, the maximum fund counting rate in the coincidence measurement method is 10.64 pulses/s.
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Figure 101 > Determine the moment of force F about pont O Express the result as a Cartesian vedor Equ11 Assume P 150 lb. Enter the z, y and components of the moment of force separated by commas VAZO v
In this question, we are to determine the moment of force F about point O in the given figure, 101. We are to express the result as a Cartesian vector equation. Let's solve this problem step by step.
Step 1: Observe the given figure carefully and identify the values of the given quantities. Here, we have a force F, a distance d from point O, and a perpendicular distance h from point O to the line of action of force F. Given, the force F, P = 150 lb, the perpendicular distance h = 6 in, and the distance d = 8 in. Observe the given figure:Here, we can see that the force F creates a moment about point O. The moment is a product of the force F and the perpendicular distance h from point O to the line of action of force F.
Step 2: We can find the position vector r from point O to the line of action of force F using the distance d and the perpendicular distance h. Since the perpendicular distance h is in the y-direction, we can write the position vector as:r = -i * d + j * hwhere i and j are unit vectors in the x and y directions, respectively. Here, we have used a negative sign for the x-component of the position vector since the vector is in the opposite direction to the unit vector i.
Step 3: We can now use the vector equation M = r x F to find the moment about point O. We can first find the cross product r x F as:r x F = (-i * d + j * h) x F = -d * k * Fwhere k is the unit vector in the z-direction. Here, we have used the fact that i x j = k, j x k = i, and k x i = j. Note that the z-component of the cross product is non-zero, while the x and y components are zero due to the choice of position vector r.
Step 4: Finally, we can express the result as a Cartesian vector equation. The moment of force F about point O is given by:M = -d * k * F where d = 8 in, h = 6 in, and F = P = 150 lb. Therefore,M = -8 k * 150 = -1200 k lb.in
This is the required moment of force F about point O expressed as a Cartesian vector equation. The z-component of the moment is -1200 lb.in.
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the transmissibility ratio for a system with a given c, k, and m, is equal to 1.2, explain what that means in terms of the motion of the car relative to the ground. would the transmissibility increase or decrease if we added an additional passenger to the vehicle? what could you do to reduce the amplitude of vibrations in a vehicle, while driving? justify your answers.
The transmissibility ratio of a system with given c, k, and m is the ratio of the amplitude of vibration of the vehicle to the amplitude of vibration of the ground. The transmissibility ratio will increase if we add an additional passenger to the vehicle, and the amplitude of vibrations in a vehicle while driving can be reduced by applying any of the techniques mentioned above.
The transmissibility ratio for a system with given c, k, and m equal to 1.2 means that the system will amplify a sinusoidal ground excitation by a factor of 1.2.
In terms of the motion of the car relative to the ground, it implies that the vertical acceleration of the vehicle will be 1.2 times greater than the ground's acceleration.
So, the car will experience more vibrations than the ground.
Therefore, it is desirable to have a low transmissibility ratio, indicating that the vehicle is less affected by the vibrations and the ride is smoother.
If we add an additional passenger to the vehicle, the transmissibility ratio will increase.
This is because an additional passenger would increase the mass of the system, causing the system's natural frequency to reduce.
When the natural frequency of the system decreases, the transmissibility ratio increases, implying that the system becomes more vulnerable to vibrations.
Therefore, to reduce the amplitude of vibrations in a vehicle while driving, one could adopt any of the following methods: Increase the damping of the vehicle
Suspension Tuning Use of Dynamic Vibration Absorbers Isolating the source of vibration Reducing the weight of the vehicle
Using a more flexible material for the tires.
In conclusion, the transmissibility ratio of a system with given c, k, and m is the ratio of the amplitude of vibration of the vehicle to the amplitude of vibration of the ground.
The transmissibility ratio will increase if we add an additional passenger to the vehicle, and the amplitude of vibrations in a vehicle while driving can be reduced by applying any of the techniques mentioned above.
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QUESTION 5 Find Resultant F direction Fi-50 N 30° G Fj-50 N Click Save and Submit to save and submit. Click Save All Answers to save all answers. a 80 F3 F1 F2 F4 Fy-75 N # FS Mac
The magnitude of the resultant force F is approximately 134.73 N, and its direction is approximately 68.2° (measured counterclockwise from the positive x-axis).
To find the resultant force vector F and its direction, add the given force vectors together.
Let's break down the given force vectors into their respective components:
Fi = -50 N (along the x-axis)
Fj = -50 N (along the y-axis)
Fy = -75 N (along the y-axis)
Now, let's calculate the resultant force vector F by adding the corresponding components:
Fx = Fi
Fy = Fj + Fy
F = sqrt(Fx^2 + Fy^2)
Substituting the values:
Fx = -50 N
Fy = (-50 N) + (-75 N) = -125 N
F = sqrt((-50 N)^2 + (-125 N)^2)
Calculating the magnitude of the resultant force F:
F = sqrt(2500 N^2 + 15625 N^2) = sqrt(18125 N^2) ≈ 134.73 N
To find the direction of the resultant force, use trigonometry:
tan(theta) = Fy / Fx
Substituting the values:
tan(theta) = (-125 N) / (-50 N) = 2.5
Taking the inverse tangent of both sides:
theta = arctan(2.5) ≈ 68.2°
Therefore, the magnitude of the resultant force F is approximately 134.73 N, and its direction is approximately 68.2° (measured counterclockwise from the positive x-axis).
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details will be appreciated
You are an astronaut, living for a long time interval in the International Space Station (ISS). During your off-duty hours, you have run out of books to read and video games to play. So, your mind wan
The length of the string needed to produce a fundamental frequency of 262 Hz and play a perfect middle C is approximately 0.8 cm.
To determine the length of the string needed to produce a fundamental frequency of 262 Hz, we can use the formula for the fundamental frequency of a vibrating string:
f = (0.5L) × (T/μ)[tex]^{\frac{1}{2} }[/tex]
Where:
f = Frequency (262 Hz)
L = Length of the string
T = Tension in the string
μ = Linear mass density
We need to soL = (1/2f) * (T/μ)
First, we need to calculate the tension in the string. The tension is equal to the electrostatic force between the charged spheres. According to Coulomb's law, the electrostatic force (F) between two point charges Q1 and Q2 separated by a distance r is given by:
F = (k × Q1 × Q2) ÷ [tex]r^{2}[/tex]
Where:
k = Coulomb's constant (k = 9 x [tex]10^{9}[/tex] N·[tex]\frac{m^{2} }{C^{2} }[/tex])
Q1 = Charge on sphere 1 (65.0 µC = 65.0 x [tex]10^{-6}[/tex] C)
Q2 = Charge on sphere 2 (65.0 µC = 65.0 x [tex]10^{-6}[/tex] C)
r = Distance between the spheres (2r = L)
Since the charges are uniformly spread over the surface of the spheres, we can assume the charges act as point charges concentrated at the center of each sphere. Thus, the distance between the spheres is equal to twice the radius of a sphere.
r = 2 * 1.80 cm = 3.60 cm = 0.036 m
Substituting the values into Coulomb's law:
F = (9 x [tex]10^{9}[/tex] N·[tex]\frac{m^{2} }{C^{2} }[/tex]) * (65.0 x [tex]10^{-6}[/tex] [tex]C^{2}[/tex] ÷ (0.036 [tex]m^{2}[/tex])
Simplifying the equation:
F ≈ 8.158 N
Now we can substitute the values of f, T, and μ into the formula for L:
L = (1 ÷ (2 × 262 Hz)) × (8.158 N ÷(8.00 x [tex]10^{-3}[/tex]kg/m))[tex]^{\frac{1}{2} }[/tex]
Simplifying the equation:
L ≈ 0.008 m = 0.8 cm
Therefore, the length of the string needed to produce a fundamental frequency of 262 Hz and play a perfect middle C is approximately 0.8 cm.
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The complete question is:
You are an astronaut, living for a long time interval in the International Space Station (ISS). During your off-duty hours, you have run out of books to read and video games to play. So, your mind wanders to your hobby of music. The last book you read discussed Gauss's law, and you get an inspiration. You plan to attach two nonconducting spheres of radius r = 1.80 cm together using a light insulating string of length L and linear mass density μ = 8.00 ✕ 10^−3 kg/m, with the string attached at the surface of each sphere. Then, using the electrical system on the ISS, you will be able to electrify each sphere to a charge of Q = 65.0 µC, uniformly spread over the surface of the sphere. The combination will then be allowed to float freely in the ISS. The spheres will repel, creating a tension in the string. When you pluck the string, you wish it to play a perfect middle C, at 262 Hz. Determine the length of the string (in cm) that you need. (Assume the frequency of 262 Hz is the fundamental frequency. Round your answer to at least one decimal place.)
A ball falls from a height hand strikes the ground. Immediately after bouncing, the ball leaves the ground with velocity v, = 0.58 m/s and vy = 4.25 m/s. (a) How high will the ball bounce? (b) How far will the ball travel horizontally before hitting the ground again? (c) If the coefficient of restitution between the ball and ground is 0.75, what height was the ball originally dropped from?
(a) The ball will bounce to a height of approximately 1.04 meters. (b) The ball will travel a horizontal distance of approximately 1.19 meters before hitting the ground again. (c) The ball was originally dropped from a height of approximately 3.11 meters.
(a) To determine how high the ball will bounce, we can use the conservation of mechanical energy. The initial potential energy of the ball when it was dropped is equal to the sum of its final potential energy and kinetic energy after the bounce.
Since the ball leaves the ground with zero vertical velocity, all of its initial potential energy is converted to its final potential energy. Therefore, we can equate the initial potential energy to the final potential energy to find the maximum height. Using the equation PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the maximum height, we can solve for h.
(b) The horizontal distance traveled by the ball before hitting the ground again can be calculated using the horizontal component of its velocity. Since there are no horizontal forces acting on the ball, its horizontal velocity remains constant. We can use the equation d = vt, where d is the horizontal distance, v is the horizontal velocity, and t is the time it takes for the ball to hit the ground again.
To find the time, we can use the vertical component of the initial velocity and the acceleration due to gravity. By dividing the initial vertical velocity by the acceleration due to gravity, we obtain the time it takes for the ball to reach the ground. Substituting this value into the equation for distance, we can find the horizontal distance.
(c) The coefficient of restitution (COR) is a measure of the elasticity of a collision. It is defined as the ratio of the final relative velocity to the initial relative velocity along the line of impact. In this case, the COR is given as 0.75. To determine the height from which the ball was originally dropped, we can use the concept of energy conservation.
The kinetic energy of the ball just before impact is equal to the sum of the gravitational potential energy at the original height and the kinetic energy just after the bounce. By applying the equation for kinetic energy and considering the COR, we can solve for the original height.
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es 1.9 and 1.10 (Section sese problems VP 1.10.3 Part F Vector A has components A, -5.00. A, 3.00, and A,-0. Vector B has components B, -2.50, B, = 4.00, and B, -1.50. Find the angle between the two v
Given vector A has components A,-5.00, A,3.00, and A,-0, and vector B has components B,-2.50, B,=4.00, and B,-1.50.
Here's the corrected solution:
Let vector A = (A1, A2, A3) = (A,-5.00, A,3.00, A,-0)
Let vector B = (B1, B2, B3) = (B,-2.50, B,4.00, B,-1.50)
The magnitude of vector A is:
|A| = √(A1² + A2² + A3²) = √((A)² + (-5.00)² + (3.00)² + (A)² + (-0)²) = √(2A² + 34)
The magnitude of vector B is:
|B| = √(B1² + B2² + B3²) = √((B)² + (-2.50)² + (4.00)² + (-1.50)²) = √(B² + 25.25)
The dot product of vectors A and B is:
A · B = A1B1 + A2B2 + A3B3
= (A)(B) + (-5.00)(-2.50) + (3.00)(4.00) + (A)(-1.50) + (-5.00)(-2.50) + (A)(-1.50)
= 7.50A + 22.50
The angle between two vectors can be calculated using the formula:
cos θ = A · B / |A||B|
Putting the respective values, we have:
cos θ = (7.50A + 22.50) / (√(2A² + 34) * √(B² + 25.25))
Therefore, the angle between the two vectors is given by:
θ = cos⁻¹((7.50A + 22.50) / (√(2A² + 34) * √(B² + 25.25)))
In this way, we have found the angle between vectors A and B.
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motivation based on the pleasure one will experience from mastering a task is called:
The motivation based on the pleasure one will experience from mastering a task is called intrinsic motivation.
Intrinsic motivation refers to the internal drive and enjoyment that comes from engaging in an activity or task. It is based on the inherent satisfaction, interest, or pleasure derived from the activity itself, rather than external rewards or incentives.
When someone is intrinsically motivated, they are motivated by the process of learning, improving skills, or experiencing personal growth and satisfaction. The pleasure and sense of accomplishment that come from mastering a task or achieving a goal become the primary driving force.
This type of motivation is often associated with activities that are personally meaningful, enjoyable, or align with an individual's values and interests. It can lead to increased engagement, persistence, and a sense of fulfillment.
Intrinsic motivation can be contrasted with extrinsic motivation, which involves engaging in an activity for external rewards such as money, praise, or recognition. While external rewards can be effective in some cases, intrinsic motivation tends to be more sustainable and can foster a deeper sense of satisfaction and fulfillment.
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Can you please answer it within 30 minutes?
A 5 kg toy car is travelling at 3 m/s goes down a slope with a height of 1.5 m. a toy block weighing 8 kg sits at the bottom of block of the slope. find their respective velocities
The mass of the toy car m₁=5 kg, velocity of the car u₁=3m/s, height of the slope h=1.5m, mass of the toy block m₂=8 kg. The potential energy of the toy car is converted to kinetic energy when it slides down the slope.
Therefore, potential energy of the toy car = m₁ghwhere g= 9.8 m/s² (acceleration due to gravity) , the potential energy of the toy car = 5 × 9.8 × 1.5 = 73.5 Joules. This potential energy of the car is converted into kinetic energy and the formula for kinetic energy is, K.E = 1/2 m v²where m is the mass of the body, v is its velocity. We can calculate the velocity of the toy car using the above formula. K.E of toy car = P.E of toy car => 1/2 m₁ v₁² = m₁ g h On substituting the given values in the above equation,1/2×5×v₁² = 5×9.8×1.5=>v₁ = 7.67 m/s.
The velocity of the toy car is 7.67 m/s when it reaches the bottom of the slope. Similarly, let's find the velocity of the toy block at the bottom of the slope. Initial velocity of the toy block is 0 m/s.
The formula for finding the final velocity can be given asv₂ = √((2gh) / 7),where g= 9.8 m/s² (acceleration due to gravity).
On substituting the given values, we getv₂ = √((2×9.8×1.5) / 7) => 2.4 m/s.
Hence, the respective velocities of the toy car and the toy block are 7.67 m/s and 2.4 m/s respectively.
Note: The calculation above is just a sample answer to the given question. You can recheck the answer using the same method.
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a 142 n force is the net force acting on a 26.0 kg object that starts from rest. at the instant the object has gone 5.00 m the rate at which the force is doing work is
The rate at which the force is doing work on the object is 710 watts.
To calculate the rate at which the force is doing work on the object, we need to use the formula for work:
Work = Force × Distance × cos(θ)
where θ is the angle between the force and the direction of motion. In this case, the object starts from rest, so the angle between the force and the direction of motion is 0 degrees (cos(0) = 1).
Net force (F) = 142 N
Mass of the object (m) = 26.0 kg
Distance (d) = 5.00 m
First, we calculate the magnitude of the force:
F = ma
where a is the acceleration of the object. Since the object starts from rest, the initial velocity (u) is 0, and we can use the equation:
v^2 = u^2 + 2ad
where v is the final velocity.
Since the object starts from rest, the equation simplifies to:
v^2 = 2ad
Solving for v:
v = √(2ad)
v = √(2 × 142/26.0 × 5.00)
v ≈ 5.18 m/s
Now, we can calculate the rate at which work is done:
Work = F × d × cos(0)
Work = 142 N × 5.00 m × 1
Work ≈ 710 joules
Since work is the energy transferred per unit time, the rate at which work is done, or power (P), is given by:
Power = Work / time
However, the time is not given in the question. Therefore, we cannot calculate the exact power value.
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A long cylindrical wire of radius 4 cm has a current of .8 amps flowing through it. Calculate the the magnetic field at r= 4cm and r=6cm at t = 0 seconds, if the current is changing as a function of time and equal to I = .8sin(200t). Assume the wire is made of copper and current density as a function of r is uniform.
To calculate the magnetic field at radial distances of 4 cm and 6 cm from a long cylindrical wire with a changing current, we need to use Ampere's law.
Given that the wire is made of copper and has a current of 0.8 A flowing through it, we can determine the magnetic field at these distances at t = 0 seconds using the formula B = (μ₀I)/(2πr), where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r is the radial distance from the wire.
Ampere's law states that the magnetic field around a closed loop is directly proportional to the current passing through the loop. In this case, we are interested in calculating the magnetic field at radial distances of 4 cm and 6 cm from the wire.
First, let's calculate the magnetic field at r = 4 cm:
B₁ = (μ₀I)/(2πr₁)
= (4π × 10^(-7) T·m/A) × (0.8 A)/(2π × 0.04 m)
= 10^(-6) T
Next, let's calculate the magnetic field at r = 6 cm:
B₂ = (μ₀I)/(2πr₂)
= (4π × 10^(-7) T·m/A) × (0.8 A)/(2π × 0.06 m)
= 6.67 × 10^(-7) T
Therefore, at t = 0 seconds, the magnetic field at a radial distance of 4 cm from the wire is 10^(-6) T, and at a radial distance of 6 cm, the magnetic field is 6.67 × 10^(-7) T.
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special relativity tells us that options: a) light moves faster if emitted by a source moving in the direction of emission and light moves slower if emitted by a source moving in the direction opposite the emission. b) light moves slower if emitted by a source moving in the direction opposite the emission. c) light moves at the same speed independent of the motion of the source. d) light moves faster if emitted by a source moving in the direction of emission. e) the speed of light is variable.
According to special relativity, the correct option is c) light moves at the same speed independent of the motion of the source. This principle is one of the fundamental tenets of special relativity and is known as the constancy of the speed of light.
In special relativity, the theory developed by Albert Einstein, the speed of light in a vacuum, denoted by 'c', is considered a fundamental constant of nature. It is the same for all observers, regardless of their motion relative to the source of light. This principle is known as the constancy of the speed of light.
One of the key consequences of this principle is that the speed of light is independent of the motion of the source. Whether the source of light is moving towards or away from an observer, the speed of light remains unchanged. This has been experimentally verified and has significant implications for our understanding of space, time, and the nature of reality.
Calculations are not required to explain this principle. The constancy of the speed of light is a fundamental postulate of special relativity, which has been validated through numerous experimental observations and is supported by a wide range of evidence and theoretical consistency.
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The drag force F acting on a globe depends on the speed of
fluid flow, globe diameter, fluid density and
dynamic viscosity of that fluid. Determine the number of independent parameters Π
that can be formed and obtain a set as a function of the first Π. NOTE:
use the FLT system and compare with the MLT
The number of independent dimensionless parameters (Π) that can be formed is 2, and the set of dimensionless parameters as a function of the first Π (Π1) is the same in both the FLT and MLT systems.
To determine the number of independent parameters, we can use the Buckingham Pi theorem. According to this theorem, the number of independent dimensionless parameters (Π) that can be formed is equal to the number of variables (n) minus the number of fundamental dimensions (k).
In this case, we have the following variables:
n = 4 variables (speed of fluid flow, globe diameter, fluid density, dynamic viscosity)
The fundamental dimensions (k) in the FLT (Fundamental, Length, Time) system are:
Length (L)Time (T)So, k = 2.
Therefore, the number of independent dimensionless parameters (Π) that can be formed is given by:
Π = n - k = 4 - 2 = 2.
We can form two independent dimensionless parameters. Let's express them in terms of the given variables:
Using the FLT system:
Π1 = (Fluid density * (Fluid flow speed)² * Globe diameter) / (Dynamic viscosity²)
Π2 = (Fluid density * Fluid flow speed * Globe diameter) / Dynamic viscosity
These two dimensionless parameters (Π1 and Π2) can be used to express the drag force (F) as a function of the first Π (Π1) in terms of the FLT system.
Now, let's compare this with the MLT (Mass, Length, Time) system.
In the MLT system, the fundamental dimensions are:
Mass (M)Length (L)Time (T)Since we are given variables related to fluid dynamics, the fundamental dimension of mass (M) is not involved.
Therefore, in the MLT system, k = 2.
Hence, using the MLT system, we can still form two independent dimensionless parameters (Π1 and Π2), which are the same as in the FLT system. The expressions for Π1 and Π2 remain unchanged.
In summary, the number of independent dimensionless parameters (Π) that can be formed is 2, and the set of dimensionless parameters as a function of the first Π (Π1) is the same in both the FLT and MLT systems.
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b) () can be recovered from () using a single modulator followed by an ideal lpf with frequency response ((). answer the following questions:
Given a signal () and a frequency response () of an ideal low-pass filter (LPF), we can recover the original signal () by using a single modulator followed by the LPF.
The modulator is responsible for modulating the signal and converting it to a form suitable for transmission or processing. The LPF then filters out high-frequency components, leaving behind the original signal. By applying this process, we can retrieve () from ().
To recover the original signal (), we first use a modulator to modulate the signal. Modulation involves manipulating the original signal in some way to encode information or prepare it for transmission. The modulated signal is then passed through an ideal LPF, which has a frequency response () that allows only low-frequency components to pass through while attenuating high-frequency components.
The LPF acts as a filter, removing any high-frequency noise or unwanted signals that may have been introduced during the modulation process. By removing these high-frequency components, we are left with the original signal () that can be recovered.
This process of using a modulator followed by an ideal LPF is commonly employed in communication systems, where signals are transmitted and received. It allows for the extraction of the original signal from the modulated version by selectively filtering out unwanted frequency components.
In short , by applying a single modulator followed by an ideal LPF with a frequency response (), we can recover the original signal () from (). This process involves modulating the signal and then filtering out high-frequency components, resulting in the retrieval of the original signal.
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4. |-/1.66 Points DETAILS MY NOTES ASK YOUR TEACHER A centrifuge in a medical laboratory rotates at an angular speed of 3450/min. When switched or rotates 54.0 times before coming to rest. Find the co
The coefficient of kinetic friction is calculated to be μk = 0.060, and the friction force acting on the centrifuge is determined as Ff = 0.288 N. The total centrifugal force acting on the contents inside the centrifuge is found to be Fc = 205.98 N.
The net force acting on the centrifuge is then calculated as Fnet = Fc - Ff, resulting in Fnet = 205.692 N. Using Newton's second law (F = ma), the acceleration of the centrifuge is determined as a = Fnet / m, where m is the mass of the centrifuge plus its contents. The acceleration is calculated to be a = 34.282 m/s^2.
Finally, the time taken for the centrifuge to come to rest is found using the formula t = ω / a, where ω is the initial angular speed of the centrifuge. Substituting the given values, the time is calculated as t = 3.025 s.
Therefore, the time taken for the centrifuge to come to rest is approximately 3.025 seconds.
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(Problem 11.15 of textbook) Consider the Hückel molecular orbitals for butadiene given in equation 11.75. Each ; is an atomic p₂ orbital on carbon atom i, so there is a nodal plan in the xy plane for each molecular orbital. There are other nodes in these orbitals as we move from atom 1 to atom 4 since the orbitals change sign. How many nodes are there are there for ₁. 2. 3 and 4? Where are they? Compare to Fig. 11.21.
Hückel molecular orbitals for butadiene, and their nodal planes Each "; " in the molecular orbital is an atomic p₂ orbital on carbon atom i. As a result, each molecular orbital has a nodal plane in the xy-plane. As we go from atom 1 to atom 4, the orbitals change sign, resulting in other nodes.
The number of nodes in 1, 2, 3, and 4 are as follows: Node in 1: 1 node located on the plane of the carbon atoms 1 and 2.A node in 2: 0 nodes
A node in 3: 1 node on the plane of the carbon atoms 2 and 3, and another on the plane of the carbon atoms 3 and 4. A node in 4: 2 nodes, one on the plane of carbon atoms 1 and 2 and the other on the plane of carbon atoms 3 and 4.
The nodal planes of the molecular orbitals, as well as their relative energies. The figure is in good agreement with the findings above. It depicts two degenerate molecular orbitals and two higher-energy molecular orbitals that are also degenerate. The nodal planes of the former lie on the planes of carbon atoms 1 and 2 and carbon atoms 3 and 4, while those of the latter lie on the planes of carbon atoms 2 and 3.
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what is the strength of the electric field 4.2 cm from a small plastic bead that has been charged to -7.2 nc ? express your answer to two significant figures and include the appropriate units.
The strength of the electric field 4.2 cm from the charged plastic bead is 171 N/C.
To calculate the strength of the electric field, we can use Coulomb's law. Coulomb's law states that the electric field created by a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.
Charge of the plastic bead (q) = -7.2 nC (negative charge indicates an excess of electrons)
Distance from the bead (r) = 4.2 cm = 0.042 m
Coulomb's law equation is given as:
Electric field (E) = (k * q) / r^2
Where:
k is the electrostatic constant, approximately equal to 9 × 10^9 Nm²/C².
Substituting the given values into the equation:
E = (9 × 10^9 Nm²/C² * -7.2 × 10^(-9) C) / (0.042 m)²
Calculating the expression inside the parentheses:
E = (9 × 7.2 × 10^9 × 10^(-9)) / (0.042 × 0.042)
Simplifying the calculation:
E = (9 × 7.2) / (0.042 × 0.042) × 10^(9-9)
E = 171 N/C
Therefore, the strength of the electric field 4.2 cm from the charged plastic bead is 171 N/C.
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1. [-/3 Points] DETAILS OSCOLPHYS2016 5.1.WA.004. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Mary applies a force of 75 N to push a box with an acceleration of 0.41 m/s². When she increases the pushi
Mary applies a force of 75 N to push a box with an acceleration of 0.41 m/s². When she increases the pushing force, the box’s acceleration increases to 1.18 m/s².
Force applied, F = 75 N, Acceleration, a1 = 0.41 m/s²a2 = 1.18 m/s².
As we know,F = ma, Where, F is the force applied, m is the mass of the body, a is the acceleration of the body.
Therefore,The mass of the body can be given as,m = F/a.
On substituting the given values,
we get,m1 = F/a1 = 75 N/0.41 m/s²= 182.93 kgm2 = F/a2 = 75 N/1.18 m/s²= 63.56 kg.
Thus, the mass of the box is 63.56 kg.
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Boyle's Law: Pressure-Volume Relationship in Gases- PhET The primary objective of this experiment is to determine the relationship between the pressure and volume of a confined gas. Using a online simulation (screen shown in Figure 1) from the University of Colorado: PHET Simulation - Gas Properties The temperature of the gas sample and the number of molecules it contains will be kept constant. Pressure and volume data pairs will be collected during this experiment and then analyzed. From the data and graph, you should be able to determine what kind of mathematical relationship exists between the pressure and volume of the confined gas. Historically, this relationship was first established by Robert Boyle in 1662 and has since been known as Boyle's law. OBJECTIVES In this experiment, you will . Determine the relationship between pressure and volume of the gas. • Describe the relationship between gas pressure and volume in a mathematical equation. Use the results to predict the pressure at other volumes. Ges Properties (315) Fie Help Heat Centret Remor 100 Oas in Pung Heavy Species Light Spaces PHET Constant Parameter Volume Temperature None Gas in Chamber Hean Species Light Species Gravity Tools & Options Pressure **Hide Tools 2000 09 Layer lool Rule Speces information Advanced Opon Energy stograms Center of mass makar Rest PROCEDURE 1. Start up your browser and go to PHET Simulation - Gas Properties. Click on the "Run Now" button and after the simulation starts, select Temperature for the Constant Parameter. Click on "Show Tools" and select "Ruler". Drag the ruler to the position shown in figure one 2. We will generate data for a heavy species of gas molecules. 3. Pump approximately 1 atm of pressure into the container. (Note this can be done by manually by typing in "200" for the number of heavy species gas molecules). Note the Temperature of the gas. 4. Drag the handle of the box as far left as it goes. Wait for the Temperature to return to its initial value. Move the ruler to measure the size of the box and record this Length in nm and Pressure in Atmospheres (Atm) in the data table. 5. Drag the handle of the box to the right 1 nm, and record the new box length in nm and pressure in Atm in the data table 6. Repeat the process until you have a total of 8 data values.
Boyle's law states that the pressure exerted by a given quantity of gas varies inversely with its volume at a fixed temperature and is expressed mathematically as PV = k, where P is pressure, V is volume, and k is a constant value when temperature is constant.
Boyle's law is a law of physics that describes the behavior of an ideal gas. It claims that the volume of a fixed mass of gas at a fixed temperature is inversely proportional to the pressure exerted on it. This relationship is known as the Boyle's law relationship, and it is one of the most fundamental laws of physics when it comes to gas behavior.
Boyle's law is significant for a variety of reasons. The most significant of these is that it explains the behavior of a large number of gases under a variety of circumstances.
Furthermore, Boyle's law is used to explain the behavior of numerous different phenomena in the world of physics, including air pressure and weather systems.
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A 1cm thick stainless steel plate has different temperature on both sides (T1 = 100 °C and T2 = 90 °C). Assuming steady-state conditions, what is the rate of heat transfer per unit area through the plate, if the thermal conductivity is 10 W/(m °C)?
0.1 J/s
1.0 J/s
10 J/s
100 J/s
Option 3 is correct. The rate of heat transfer per unit area through the 1cm thick stainless steel plate with temperature difference [tex]T1 = 100 ^0C[/tex] and [tex]T2 = 90 ^0C[/tex], and thermal conductivity of 10 W/(m[tex]^0C[/tex]) is 10 J/s.
For calculating the rate of heat transfer per unit area, use Fourier's law of heat conduction, which states that the heat transfer rate (Q) is directly proportional to the temperature difference (ΔT) and the thermal conductivity (k), and inversely proportional to the thickness of the material (L). Mathematically, the equation is given as:
Q/A = k * (T1 - T2) / L
where Q/A is the rate of heat transfer per unit area, k is the thermal conductivity, T1 and T2 are the temperatures on both sides of the plate, and L is the thickness of the plate.
Plugging in the given values into the equation,
[tex]Q/A = 10 W/(m ^0C) * (100 ^0C - 90 ^0C) / 0.01 m[/tex]
Simplifying the equation,
Q/A = 10 J/s
Therefore, the rate of heat transfer per unit area through the stainless steel plate is 10 J/s.
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The boiler develops 450 hp and uses 4 lb of coal per hp-hr. The coal contains 13800 Btu/lb, of steam pressure 150 psig. Feedwater temperature is 120°F. A feedwater heater is added raising the temperature of water to 195°F. Heater cost P7')00. Pant operates 10 hours a day, 300 days a year. The cost of coal is P2(10 per ton of 2000 lbs. Allowing 10% depreciation and repairs, 15% interest end 4% insurance, a. how much would the owner save by installing the heater b. % profit
The amount of coal burned in one hour, 1 hp-hour = 4 lb of coal
450 hp x 1 hour = 450 hp-hours450 hp-hours x 4 lb/hp-hour
= 1,800 lb of coal in one hour
With a feed water temperature of 120°F and the steam condensing at 212°F (atmospheric pressure), the heat required to raise the feed water temperature is:
b. % profit : The percentage of profit is calculated as the difference between the amount of profit made by the heater installation, P617,010.00, and the cost of the heater, P7,500.00, divided by the cost of the heater multiplied by 100.% profit
= (P617,010.00 - P7,500.00) / P7,500.00 x 100%Profit percentage
= 8053.47%
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Water vapor enters the adiabatic diffuser with a
velocity of 540 m/s at a temperature of 160 kPa and 125 C.
Calculate the lowest velocity giving the pressure value of 330 kPa
at the outlet.
The lowest velocity that gives a pressure value of 330 kPa at the outlet is 540 m/s.
In an adiabatic diffuser, the principle of conservation of energy can be applied to relate the initial and final states of the fluid. We'll assume ideal gas behavior for water vapor and use the ideal gas law equation: PV = nRT.
Initial conditions:
Velocity (V1) = 540 m/s
Temperature (T1) = 125°C = 398 K
Pressure (P1) = 160 kPa = 160,000 Pa
Final conditions:
Pressure (P2) = 330 kPa = 330,000 Pa
Using the ideal gas law, we can write the equation for the initial and final states as follows:
P1 * V1 = n * R * T1 (1)
P2 * V2 = n * R * T2 (2)
Since the process is adiabatic, the mass flow rate (n) remains constant. Dividing equation (2) by equation (1) and substituting the given values, we can solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
= (160,000 * 540 * T2) / (330,000 * 398)
Assuming the outlet temperature remains the same as the initial temperature (T2 = T1 = 398 K), we can calculate the lowest velocity (V2) that gives a pressure value of 330 kPa at the outlet.
V2 = (160,000 * 540 * 398) / (330,000 * 398) = 540 m/s
Therefore, assuming the outlet temperature remains the same as the initial temperature, the lowest velocity that gives a pressure value of 330 kPa at the outlet is 540 m/s.
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Part A
A distant point source of light (like a star) emits light of wavelength 575 nm. When this light enters a camera whose lens has a focal length of 135 mm, the diffraction pattern formed on the camera's detector has an Airy disk of radius 0.0112 mm. Find the diameter of the lens aperture.
Part B
Find the f-number.
(a) The diameter of the lens aperture is approximately 0.0224 mm.
(b) The f-number is approximately 12.05.
(a) The radius of the Airy disk, which represents the diffraction pattern formed by the lens aperture, is given as 0.0112 mm. The diameter of the aperture can be calculated by multiplying the radius by 2:
Diameter = 2 * Radius = 2 * 0.0112 mm = 0.0224 mm.
(b) The f-number is the ratio of the focal length of the lens to the diameter of the lens aperture. Given the focal length of 135 mm and the aperture diameter of 0.0224 mm, we can calculate the f-number:
f-number = Focal Length / Aperture Diameter = 135 mm / 0.0224 mm ≈ 12.05.
Therefore, the diameter of the lens aperture is approximately 0.0224 mm, and the f-number is approximately 12.05. The f-number is commonly used to represent the speed of a lens, with smaller values indicating larger aperture openings and faster lenses.
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The energy that tidally heats a satellite ultimately derives
from the satellite’s orbit.
True
or
False
The energy that tidally heats a satellite ultimately derives from the satellite’s orbit. This statement is false.
The energy that tidally heats a satellite does not ultimately derive from the satellite's orbit. Tidal heating occurs due to gravitational interactions between the satellite and the gravitational field of another celestial body, such as a planet or a moon. These gravitational interactions cause tidal forces, which deform the satellite and generate internal friction within its structure. This internal friction converts gravitational potential energy into heat energy, resulting in tidal heating.
The energy source for tidal heating is the gravitational potential energy of the satellite-moon system, not the satellite's orbit itself. Therefore, the energy that tidally heats a satellite does not derive directly from its orbit but from the gravitational interaction with the celestial body it orbits around.
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From the list of aircraft components below, select all those which correctly describe those typically, or commonly, powered by the hydraulic system on most large modern airliners. Only select answers you are sure are correct. Partial credit is available for each correct answer but negative marking is applied within this questions (but it is not possible to score a negative mark for the question overall). 4 Fuel pumps Wing and Engine anti-icing Engine starter motors Primary flight controls (Elevators, Ailerons and Rudder). Secondary flight controls (e.g. flaps and slats) Undercarriage retraction/extension Air Cycle Machines/Packs Brakes and nose wheel steering
Based on the information available up until September 2021, the aircraft components that are typically powered by the hydraulic system on most large modern airliners are:
1) Primary flight controls (Elevators, Ailerons, and Rudder).
2) Secondary flight controls (e.g., flaps and slats).
3) Undercarriage retraction/extension.
4) Brakes and nose wheel steering.
- Therefore, the correct options are:
Primary flight controls (Elevators, Ailerons, and Rudder).Secondary flight controls (e.g., flaps and slats).Undercarriage retraction/extension.Brakes and nose wheel steering.Learn more about aircraft components here : brainly.com/question/29382256
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