The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
a An object is tarown up with a velocity v = 6.02 +7.0j. Calculate the (1) time taken reach the maximum height (ii) the horizontal range (s = 10m/s2).
Answer:
(i) 0.6s (ii) 8.42m
Explanation:
U² = 6.02² + 7²
U = 9.23
angle of projection
tanø = 6.02/7
ø = 40.7
Time of fligt
t = Usinø/g
t = 9.23 sin 40.7/10
t = 0.6
H range = U²sin2ø/g
H = 9.23²sin 81.4/10
H = 8.42m
The time taken reach the maximum height is 0.6s and the horizontal range is 8.42m.
What is Velocity?Velocity is defined as the directional motion of an object which is indicated by the rate of change of position as observed from a particular frame of reference. It is measured by a particular standard of time. It is a vector quantity as it has both magnitude and direction.
It can be expressed as:
v= d/t
Where. v is the velocity in m/s
d is the displacement measured in meter 'm'
t is the time measured in seconds 's'
For above given information,
v = 6.02 +7.0j, so the initial velocity will be u
u² = 6.02² + 7²
u = 9.23m/s
Angle of projection, tanø = 6.02/7
ø = 40.7
Time taken= t = u sinø/g
t = 9.23 sin 40.7/10
t = 0.6s
Horizontal range = u²sin2ø/g
H = 9.23²sin 81.4/10
H = 8.42m
Thus, the time taken reach the maximum height is 0.6s and the horizontal range is 8.42m.
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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the universe be in that case?
Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]
Age of the universe; [tex]t = \ ?[/tex]
We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:
[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]
Now,
Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]
We know that;
[tex]1\ light\ years = 9.46*10^{15}m[/tex]
so
[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]
Therefore;
[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]
Now, we input this Hubble's constant value into our equation;
[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]
Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
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Which is NOT a function of the
cell wall?
A. Protects cell from bursting
B. Provides support for plant cells
C. Protects cell from harsh internal
environments
D. Absorbs sunlight to give energy to the cell
Answer:
The answer is D
Explanation:
The chloroplast absorbs sunlight for energy not the cell wall
A 500 kg cart is rolling to the right at 1.3 m/s. a 60 kg man is standing on the right end of the cart. what is the speed of the cart if tha man suddenly starts running to the left with a speed 10.0 m/s relative to the cart
Answer:
P1 = 1.3 (500 + 60) = 728 kg-m total momentum to right at start
P2 = (v2 - 10) 60 + 500 v2
total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart
728 = 560 v2 - 600
v2 = 1328 / 560 = 2.37 m/s new speed of cart
Check:
After: p2 for cart = 500 * 2.37 = 1186
p1 for man = (2.37 - 10) * 60 = -458
P2 = p1 + p2 = 728 total momentum unchanged
a mass of 7.5kg has a weight of 30n on a certain planet calculate the acceleration due to gravitt on this planet
Answer:
Acceleration due to gravity on a certain planet = 4 m/s²
Explanation:
According to the question,
Weight = 30 N
Mass = 7.5 kg
Let acceleration due to gravity be 'a'
Formula:
Weight = Mass × Acceleration due to gravity
30 = 7.5 × a
a = 30/7.5
a = 4 m/s²
2.(01.01 LC)
Which of the following is true for gravitational force? (3 points)
Decreases with increase in mass
Increases with increase in mass
Increases with increase in distance
Decreases with decrease in distance
Answer:
Increases with increase in mass
Explanation:
gravity is proportional to mass and inversely proportional to the square of the distance between them
F = GMm/d²
A projectile is launched with a horizontal velocity of 20 m/s and an initial vertical velocity of 20 m/s. What is the projectile's acceleration in the Horizontal direction? Verticle direction?
Answer:
Vertical acceleration 9.8 m/s² downward
Horizontal acceleration 0.0 m/s²
assuming no air resistance.
How are the two types of power plants similar how are they different
Answer:
iIn a nuclear plant, the heat source is from the nuclear reaction whereas in a thermal power plant it is from the combustion of coal. The difference is in the inlet steam parameters to the turbine in a nuclear plant. Thermal power plants use steam at superheated conditions. ... The nuclear plant uses a 'wet steam turbine'.
Explanation:
The heat source in a nuclear power plant is the nuclear reaction, whereas the heat source in a thermal power plant is coal combustion. The difference is in the turbine's input steam characteristics.
What is a power plant?Power plant is an industrial structure that generates electricity. The majority of power plants are linked to the electrical grid.
Nuclear power bare a form of thermal power plant. You have a reactor where fission takes place and heat is generated, a heat exchanger that transports this heat to where it is needed.
Thermal power plant equipment converts this heat into electric energy, usually via a steam turbine.
The reactor, heat exchanger, and thermal conversion technology all have different designs and technologies, but the overall architecture is quite similar to other types of thermal power plants.
The heat source in a nuclear power plant is the nuclear reaction, whereas the heat source in a thermal power plant is coal combustion. The difference is in the turbine's input steam characteristics.
Hence, the two types of power plants differ in difference is in the turbine's input steam characteristics.
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Which is a force that wears away landforms? Select three options.
A. weathering
B. erosion
C. humans
D. clouds
E. light
Question below...........................
[tex]\boxed{\sf PE=mgh}[/tex]
[tex]\boxed{\sf KE=\dfrac{1}{2}mv^2}[/tex]
[tex]\boxed{\sf ME=KE+PE}[/tex]
#1
[tex]\\ \sf\longmapsto PE=60(0)(10)=0J[/tex]
[tex]\\ \sf\longmapsto KE=\dfrac{1}{2}(60)(8)^2=30(64)=1920J[/tex]
[tex]\\ \sf\longmapsto ME=1920+0=1920J[/tex]
#2
[tex]\\ \sf\longmapsto PE=60(10)(1)=600J[/tex]
[tex]\\ \sf\longmapsto KE=600J[/tex]
[tex]\\ \sf\longmapsto ME=1200J[/tex]
Now
[tex]\\ \sf\longmapsto \dfrac{1}{2}mv^2=600\implies 30v^2=600\implies v^2=20\implies v=4.2m/s[/tex]
20 cm long 10 cm wide and 5 cm thick as a mass of 500 g determine the greatest pressure that can be exerted by block on the flat surface?
100000 Pascal
Explanation:
pressure= force/area
Max pressure= force/min area
so f=5
min area= 5×10^-5
5÷5*10^-5 = 100000pascal
in vacuum , the shorter the wavelength of an electromagnetic wave is , the:
A. lower its frequency
B. higher its energy
C. longer its period
D. slower its speed
Answer:
Higher its Energy
Explanation:
A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Nm-1
. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple
harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy.
The conservation of energy allows to find the result for the point where the kinetic and potential energy are equal is;
x = 0.026 m
Given parameters
The mass of the body m = 220 g = 0.220 kg The force constant k = 7.0 N / m The initial displacement or amplitude xo = 5.2 cm = 0.052 mTo find
The point where scientific and potential energy are equal.
The law of the conservation of mechanical energy is one of the most important in physics, stable that if there is no friction, the mechanical energy of the system is conserved. The mechanical energy is formed by the sum of the kinetic energy and the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. With maximum compression.
Em₀ = U = ½ k x²
Final point. Where the kinetic and potential energy are equal.
[tex]Em_f = K +U[/tex]
Since the mechanical energy is constant at this point K = U, therefore we can write the energy.
[tex]Em_f = 2U = 2 ( \frac{1}{2} \ k \ x_f^2 )[/tex]
Energy is conserved.
[tex]Em_o = Em_f \\\frac{1}{2} \ k x_o^2 = 2 ( \frac{1}{2} \ k x_f^2)[/tex]Emo = Emf
½ k x² = 2 (½ k xf²)
[tex]x_f = \frac{x_o}{2}[/tex]
let's calculate.
[tex]x_f = \frac{0.052}{2} \\x_f = 0.026 m[/tex]
In conclusion using the conservation of energy we can find the point where the kinetic and potential energy are equal is;
x = 0.026 m
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An archer's bow is drawn at its midpoint until the tension in the string is 0.842 times the force exerted by the archer. What is the angle between the two halves of the string
Consult the attached free body diagram.
If we take the direction of F to be the positive horizontal axis, and upward to be the positive vertical axis, then using Newton's second law we have net forces
• ∑ F [horizontal] = F [archer] + T cos(180° - θ) + T cos(180° + θ) = 0
• ∑ F [vertical] = T sin(180° - θ) + T sin(180° + θ) = 0
since the bow is held in place while it's drawn. T is the magnitude of the tension in the string, and it can be shown to be equal in both strings since they both make the same angle with the negative horizontal axis (the dashed line).
We only really need the first equation. Simplifying it, we get
F [archer] - T cos(θ) - T cos(θ) = 0
F [archer] - 2T cos(θ) = 0
F [archer] = 2T cos(θ)
cos(θ) = F [archer] / (2T)
We're given that the tension T in the string is 0.842 times the force exerted by the archer, which is to say
T = 0.842 F [archer]
and from this we have
cos(θ) = F [archer] / (2 • 0.842 F [archer])
cos(θ) = 1/1.684
cos(θ) ≈ 0.593
Solving for θ gives an angle of θ ≈ arccos(0.593) ≈ 53.6°. Then the angle between the two tension forces is twice this, or about 2θ ≈ 107°.
Example 4.16
An object of mass 3 kg rests on a plane. The coefficient of static friction and that of kinein
friction are given by Hs = 0.3 and pk = 0.2.
The plane is inclined at angle o to the horizontal.
(i) Find the maximum value of 0 for which the object remains at rest on the plane.
(ii) Find the acceleration of the object if it started sliding from rest down the plane at
angle Omax to the horizontal.
(ii) How long does it take the object to move, from rest, a distance of Imetre under the
conditions of (ii).
Answer:
Explanation:
(i) μs = F/N = mgsinθ/mgcosθ = tanθ
tanθ = 0.3
θ = 16.7°
(ii) a = F/m
a = (mgsinθ - (μk)mgcosθ) / m
a = g(sinθ - (μk)cosθ)
a = 9.8(sin16.7 - (0.2)cos16.7)
a = 0.94 m/s²
(iii) s = ½at²
t = √(2s/a)
t = √(2(1)/0.94)
t = 1.5 s
QUESTION 5 When an instrument is sounded together with a turning fork of frequency 260Hz , 2 beats are heard. When the same instrument is sounded with a fork of frequency 256H2, , 6 beats are heard. Find the frequency of the instrument .
Answer:
Explanation:
4126h2.
PLEASE HELP! I USED 100 POINTS!
Complete the sentence.
_____ are cracks in rock layers, and the pressures within the crust can push one of these layers past another.
A. Faults
B. Tectonic plates
C. Stalagmites
Subject: Science
Answer:
A i think its right im postivite
Answer: A. Faults
Explanation: ;-; its easy because faults are cracks and they can push layers past each other making tectonic plates rub against each other then creating a earthquake
What was different about the molecules you needed to make protein 3 compared to the molecules you used to make protein 2?
Answer:
the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.
An object weighs 573.0 N on planet Xyleneer. If the object's mass is 92.1 kg, what is the acceleration due to gravity on planet Xyleneer?
Answer:
a = 6.22 m//s²
Explanation:
F = ma
a = F/m
a = 573.0 / 92.1
a = 6.221498...
Answer:
[tex]\boxed {\boxed {\sf 6.22 \ m/s^2}}[/tex]
Explanation:
We are asked to find the acceleration due to gravity on another planet.
Weight is the measure of the force of gravity. Therefore, we can use the following version of the force formula:
[tex]F_g=mg[/tex]
In this formula, [tex]F_g[/tex] is the weight, m is the mass, and g is the acceleration due to gravity.
The object weights 573.0 Newtons (or 573.0 kg*m/s²) on the planet. The object has a mass of 92.1 kilograms.
[tex]F_g[/tex]= 573.0 kg* m/s²m= 92.1 kgSubstitute these values into the formula.
[tex]573.0 \ kg*m/s^2 = 92.1 \ kg * g[/tex]
We are solving for g, so we must isolate the variable. It is being multiplied by 92.1 kilograms. The inverse of multiplication is division, so divide both sides of the equation by 92.1 kg.
[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}= \frac{92.1 \ kg*a}{92.1 \ kg}[/tex]
[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}=a[/tex]
The units of kilograms cancel.
[tex]6.22149837 \ m/s^2=a[/tex]
The original measurements of weight and mass have 4 and 3 significant figures. Our answer must have the least number of sig figs, or 3. For the number we found, that is the hundredth place. The 1 in the thousandths place tells us to leave the 2 in the hundredth place.
[tex]6.22 \ m/s^2=a[/tex]
The acceleration due ot gravity on planet Xyleneer is approximately 6.22 meters per second squared.
Describe what you think energy is in physics and what does it do.
In Physics, energy can be defined as the ability and capacity to do work by an object or physical body.
What is energy?In Physics, energy can be defined as the ability to do work. Thus, energy must be possessed or transferred to a physical object (body) before it can be used in doing a work or heating a system.
The types of energy.Generally, there are two (2) main types of energy and these are;
Potential energy (P.E): it is an energy that is possessed by an object or body due to its height (position) above the Earth surface.Kinetic energy (K.E): it is an energy possessed by an object or body due to its motion.For example, you require a sufficient amount of energy to move a crate of egg across a given distance.
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Answer:. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another. ... All forms of energy are associated with motion.
Explanation:. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another. ... All forms of energy are associated with motion.
A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.what are the Component of vectors B and it's direction
Answer:
I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
What is magnitude of the resultant?IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird. A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.
Therefore, My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
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A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the mass of the beaker? The specific heat of glass is 0.18 cal/g ∙ °C, and that of water is 1.0 cal/g ∙ C°.
From the information provided in the question, the mass of the beaker is 144.4 g.
From the information provided in the complete question;
volume of water = 74 mL
Mass of water = 74 g
specific heat of glass = 0.18 cal/g ∙ °C
specific heat of water = 1.0 cal/g ∙ C°
Mass of glass = x g
Total heat gained by the system = 2000.0cal
Temperature rise = 20.0°C
Heat gained by system = Heat gained by glass + Heat gained by water
Heat gained by glass = x × 0.18 × 20
Heat gained by water = 74 × 1.0 × 20
Hence;
2000 = (x × 0.18 × 20) + ( 74 × 1.0 × 20)
2000 - 1480 = (x × 0.18 × 20)
x = 520/3.6
x = 144.4 g
Missing parts;
A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18
cal/g °C, and that of water is 1.0 cal/g °C.
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A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1 point)
A. The forecaster may have done the calculations a second time because he made a mistake in the calculation.
B. Someone may have reported the weather incorrectly before the first computation.
C. The forecaster may have had the computer model do the calculations a second time, and he found that the prediction changed.
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Answer:
(d) is your answer for your question
What is the neutron number for 75 32Ge?
Answer:
If the mass of geranium is 75 and the atomic number is 32 then it must have
N = 75 -32 = 43 neutrons
5 What is the maximum speed at which a car round a curve of 25m radius on a level road if the coefficient of static friction between the tires and the road is 0.80?
Hi there!
On a level road:
∑F = Ff (Force due to friction)
The net force is the centripetal force, so:
mv²/r = Ff
Rewrite the force due to friction:
mv²/r = μmg
Cancel out the mass:
v²/r = μg
Solve for v:
v = √rμg
v = √(25)(9.81)(0.8) = 14.01 m/s
Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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Which light is most sensitive to the eyes?
Answer:
Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.
what are the greenhouse gasses in the earth that are primarily responsible for the greenhouse effect on Earth
Answer:
Water Vapour, Carbon dioxide, methane,nitrous oxide,ozone.
Calculate the amount of work done (in joules) to raise the 0.2kg mass 0.5 m
Answer:
1
Explanation:
A=mgh=0.2*10*0.5=1 J
What diameter telescope is needed to resolve the separation between an Earth-like planet and its star at 550 nm if the linear separation between them is 1 AU and the star system is 3 pc from Earth
The Rayleigh criterion allows finding the result for the diameter of the telescope that allows solving the separation of the star and the planet is:
The diameter of the telescope is D = 0.415 m
The Rayleigh criterion is used to find the separation of two points, it is based on the fact that the diffraction maximuum pattern of the first object coincides with the first minimum of the second object.
By entering in the diffraction ratio for slits you will find.
sin θ = [tex]\frac{\lambda}{a}[/tex]
In general in diffraction experiments the angles are very small,
[tex]tan \theta = \frac{y}{x} = \frac{sin \theta}{cos \theta} \\sin \theta = \frac{y}{x}[/tex]
For the case of circular apertures, when solving in polar coordinates, a constant appears.
[tex]\frac{y}{x} = 1.22 \frac{\lambda}{D}[/tex]
[tex]D = 1.22 \frac{\lambda \ x}{y}[/tex]
Where λ is the wavelength of light and D is the diameter of the aperture.
They indicate that the separation between the star and the planet is 1 AU and the distance from the system to the Earth is 3 parce.
Let's reduce the parce to astronomical units
x = 3 pc ( [tex]\frac{206264 AU}{1 pc}[/tex] )
x = 6.18 10⁵ AU
Let's calculate
D = [tex]D = 1.22 \ \frac{550 \ 10^{-9 } \ 6.18 \ 10^5 }{1}[/tex]
D = 0.415 m
In conclusion, using the Rayleigh criterion we can find the result for the diameter of the telescope that allows solving the separation of the star and the planet is:
The diameter of the telescope is D = 0.415 m
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