Answer:
Campbell-Stokes sunshine recorders or modern sunshine sensors.
Explanation:
Answer:
modern sunshine sensorssunshine recordersCampbell-Stokes
During World War II, mass spectrometers were used to separate the radioactive uranium isotope U-235 from its far more common isotope, U-238. Estimate the radius of the circle traced out by a singly ionized lead atom moving at the same speed.
Answer:
21.55 m
Explanation:
Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?
Answer: Only Tech B is correct.
Explanation:
First, tech A is wrong.
The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.
Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.
which of the following can not happen when a light ray strikes a new medium
Answer:
amplification
Explanation:
reflection can happen
some amount of lighr get absorbed
something gets refracted
but amplification cant
A car is going 8 meters per second on an access road into a highway
and then accelerates at 1.8 meters per second squared for 7.2
seconds. How fast is it then going?
Answer:
20.96 m/s^2 (or 21)
Explanation:
Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.
At first, we know a car is going 8 m/s, that is its initial velocity.
Then, we know the acceleration, which is 1.8 m/s/s
We also know the time, 7.2 second.
Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.
(final velocity - initial velocity) = time * acceleration
final velocity = time*acceleration + initial velocity
After plugging the found values in, we get 20.96 m/s/s, or 21 m/s
krichoffs law of current questions
Answer:
Explanation:
Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.
#I AM ILLITERATE
What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.630 m long and has a mass of 5.69 g.
Answer:
220.698HzExplanation:
The fundamental frequency f₀ is expressed as f₀ =V/2L where;
V is the speed of the string = [tex]\sqrt{\frac{T}{M} }[/tex]
m is the mass of the string
L is the length of the string
T is the tension in the string
f₀ = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
Given datas
m = 5.69g = 0.00569 kg
T = 440N
L = 0.630 m
Required
Fundamental frequency of the steel piano wire f₀
[tex]f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz[/tex]
Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz
What is the power P of the eye when viewing an object 61.0 cm away? Assume the lens-to-retina distance is 2.00 cm , and express the answer in diopters.
Answer:
The power of the eye is 51.64 diopters
Explanation:
The power of the eye is given by;
[tex]P = \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i}[/tex]
where;
P is the power of the eye in diopter
f is the focal length of the eye
[tex]d_o[/tex] is the distance between the eye and the object
[tex]d_i[/tex] is the distance between the eye and the image
Given;
[tex]d_o[/tex] = 61.0 cm = 0.61 m
[tex]d_i[/tex] = 2.0 cm = 0.02 m
[tex]P = \frac{1}{d_o} +\frac{1}{d_i} \\\\P = \frac{1}{0.61} + \frac{1}{0.02} \\\\P = 51.64 \ D[/tex]
Therefore, the power of the eye is 51.64 diopters.
The power P of the eye when viewing an object 61.0 cm away is 51.639D
The power of a lens is a reciprocal of its focal length and it is expressed as:
[tex]P=\frac{1}{f}[/tex]
According to the mirror formula
[tex]\frac{1}{f} =\frac{1}{d_i} +\frac{1}{d_0}[/tex]
where
[tex]d_i[/tex] is the distance from the lens to the image = 61.0cm = 0.61m
[tex]d_0[/tex] is the distance from the lens to the object = 2.00cm = 0.02m
[tex]P=\frac{1}{f} =\frac{1}{0.02} +\frac{1}{0.61}\\P=50+1.639\\P=51.639D[/tex]
Hence the power P of the eye when viewing an object 61.0 cm away is 51.639D
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When light is either reflected or refracted, the quantity that does not change in either process is its
Answer:
Frequency
Explanation:
When waves travel from one medium to another, it is only the frequency of the wave that remains constant . when a wave is refracted at the boundary between two media, the wave will slow down and its wavelength decreases. The wave usually bends at the interface between the two media. The wavelength and speed of a wave may change at the boundary between two media but its frequency remains the same.
Hence the frequency of light is its only property that remains constant.
A generator consists of 1000 turns of wire around a coil of area 10cm2 (0.0010m 2 ). Calculate the peak output voltage of this generator when operating at 100 cycles/second in a 0.50 tesla field.
Answer:
The peak output voltage of this generator is 314.2 V.
Explanation:
Given;
the number of turns of the coil, N = 1000 turns
area of the coil, A = 0.001 m²
angular frequency of the coil, f = 100 cycles/seconds
magnitude of the magnetic field, B = 0.5 T
The peak output voltage of this generator is given by;
E = NBAω
Where;
ω is the angular velocity = 2πf
E = NBA(2πf)
E = 1000 x 0.5 x 0.001(2 x π x 100)
E = 314.2 V
Therefore, the peak output voltage of this generator is 314.2 V.
What happens when two polarizers are placed in a straight line, one behind the other? A. They allow light to pass only if they are polarized in exactly the same direction. B. They block all light if they are polarized in exactly the same direction. C. They allow light to pass only if their directions of polarizations are exactly 90° apart. D. They block all light if their directions of polarizations are exactly 90° apart. E. They block all light if their directions of polarizations are either exactly the same or exactly 90° apart.
Answer:
C
They allow light to pass only if their directions of polarizations are exactly 90° apart.
2. The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?
Answer:
We know that the maximum angle that a light ray can wake with the wall of the core is equipment to the minimum angle with the normal of the core that will give rise in total internal reflection. so using Snell's law the angle is subtracted from 90° to get the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber.
So using
n1sinစ1. = n2sinစ2
1.6sin(x1) = 1.48sin(90),
But sin(90)=1
1.6sin(စ1) = 1.48,
sin(စ1) = 1.48/1.6
စ = 68°
Explanation:
Answer:
i = 67.66⁰Explanation:
Using the Snell's law formula to solve this question which states that the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. This constant is known as the refractive index for the given pair of media. Mathematically,
n = sin(i)/sin(r) where;
i is the angle of incidence
r is the angle of refraction.
n is the refractive index.
Given the refractive index of the optical fibre n₁ = 1.60 and that of cladding n₂ = 1.48
n₂/n₁ = sin(i)/sin(r)
The light ray can make with the wall of the core when its angle of refraction is 90⁰. The angle of incidence at this maximum point is known as the critical angle.
On substitution:
1.48/1.60 = sin(i)/sin90
1.48/1.60 = sin(i)/1
sin(i) = 1.48/1.60
sin(i) = 0.925
i = sin⁻¹0.925
i = 67.66⁰
Hence the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber is 67.66⁰.
Ellen says that whenever the acceleration is directly proportional to the displacement of an object from its equilibrium position, the motion of the object is simple harmonic motion. Mary says this is true only if the acceleration is opposite in direction to the displacement. Which one, if either, is correct
Answer:
Both Ellen and Mary are correct.
Explanation:
Both are correct, it's just different ways of saying the same thing.
When the acceleration is always opposite in direction to the displacement, then, the acceleration is directly proportional to the displacement of an object from its equilibrium position
The distance from the Sun to Earth is approximately 149,600,000 km. The distance from the Sun to Venus is approximately 108,200,000 km. The elongation angle αα is the angle formed between the line of sight from Earth to the Sun and the line of sight from Earth to Venus. Suppose that the elongation angle for Venus is 10∘.10 ∘. Use this information to find the possible distances between Earth and Venus.
Answer:
335206922km
Explanation:
Pls see attached file
a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic field B = 0.4 T at a constant frequency of 55 Hz. What is the maximum induced emf?
a. 625 V
b. 110 V
c. 421 V
d. 332 V
e. 200 V
Answer:
d. 332 V
Explanation:
Given;
number of turns in the wire, N = 40 turns
area of the coil, A = 0.06 m²
magnitude of the magnetic field, B = 0.4 T
frequency of the wave, f = 55 Hz
The maximum emf induced in the coil is given by;
E = NBAω
Where;
ω is angular velocity = 2πf
E = NBA(2πf)
E = 40 x 0.4 x 0.06 x (2 x π x 55)
E = 332 V
Therefore, the maximum induced emf in the coil is 332 V.
The correct option is "D"
d. 332 V
A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.
Answer:
R = 36.885 km
Explanation:
In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other
The diffraction equation for slits is
a sin θ = m λ
the first minimum occurs for m = 1
sin θ = λ a
as the diffraction experiments the angles are very small, we approximate
sin θ = θ
θ = λ / a
This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form
θ = 1.22 λ / a
In this problem they give us the frequency, let's find the wavelength with the relation
c = λ f
λ = c / f
θ = 1.22 c/ f a
since they ask us for the distance between the planes, we can use the definition of radians
θ = s / R
if we assume that the distance is large, we can approximate the arc to the horizontal distance
s = x
we substitute
x / R = 1.22 c / fa
R = x f a / 1.22c
Let's reduce the magnitudes to the SI system
f = 9000 MHz = 9 109 Hz
a = 15 m
x = 100 m
let's calculate
R = 100 10⁹ 15 / (1.22 3 108)
R = 3.6885 10⁴ m
let's reduce to km
R = 3.6885 10¹ km
R = 36.885 km
If 1 kg of each substance in the table absorbs 250 kJ of thermal energy,
which substance will have the smallest change in temperature?
Answer:
Water.
Explanation:
To know which of the substance that will absorbed the greatest amount of thermal energy, we'll simply determine the amount of energy absorbed by each substance.
This is illustrated below:
For Air:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 1.01 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 1.01 x 15
Q = 15.15 KJ
Therefore, the heat absorbed by air is 15.15 KJ.
For Plastic:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 2.60 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 2.60 x 15
Q = 39 KJ
Therefore, the heat absorbed by plastic is 39 KJ.
For Water:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 4.18 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 4.18 x 15
Q = 62.7 KJ
Therefore, the heat absorbed by water is 62.7 KJ.
For Wood:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 1.68 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 1.68 x 15
Q = 25.2 KJ
Therefore, the heat absorbed by Wood is 25.2 KJ.
Summary:
Substance >>>>> Heat Absorbed
Air >>>>>>>>>>>> 15.15 KJ.
Plastic >>>>>>>>> 39 KJ
Water >>>>>>>>>> 62.7 KJ
Wood >>>>>>>>>> 25.2 KJ.
From the above calculations, we can see that water will absorb the greatest amount of thermal energy.
The substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.
Specific heat capacity
The specific heat capacity of each substance can be used to determine the substance with the smallest change in temperature.
Q = mcΔθ
where;
m is massc is specific heat capacityΔθ is change in temperatureΔθ = Q/mc
For airΔθ = (250)/(1.01)
Δθ = 247.5 ⁰C
For plasticΔθ = (250)/(2.6)
Δθ = 96.15 ⁰C
For waterΔθ = (250)/(4.18)
Δθ = 59.81 ⁰C
For woodΔθ = (250)/(1.68)
Δθ = 148.81 ⁰C
Thus, the substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.
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A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish
Answer:
2,66
Explanation:
The refractive index= real depth/ apparent depth
real depth = refractive index * apparent depth
Let's assume index for water is 1.33
real depth = 2*1,33 = 2,66
Alpha particles (charge = +2e, mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of the circular motion of the alpha particles is closest to: A. 0.25 μs B. 0.16 μs C. 0.49 μs D. 0.40 μs E. 0.33 μs
Answer:
Option B: T ≈ 0.16 μs
Explanation:
We are given;
Mass; m = 6.68 × 10^(-27) kg
Magnetic field;B = 0.80 T
Charge;q = 2e
Now, e is the charge on an electron and it has a value of 1.6 × 10^(-19) C
So, q = 2 × 1.6 × 10^(-19)
q = 3.2 × 10^(-19) C
The period of the circular motion of the alpha particles moving along a in the presence of the magnetic field is given by;
T = 2πm/qB
Where ;
m, q and B are as stated earlier.
Plugging in the relevant values, we have;
T = (2π × 6.68 × 10^(-27))/(3.2 × 10^(-19) × 0.8)
T = 0.16395 × 10^(-6) s
This can also be written as;
T ≈ 0.16 μs
Which statement about kinetic and static friction is accurate?
Static friction is greater than kinetic friction, and they both act in conjunction with the applied force.
Kinetic friction is greater than static friction, and they both act in conjunction with the applied force.
Kinetic friction is greater than static friction, but they both act opposite the applied force.
Static friction is greater than kinetic friction, but they both act opposite the applied forcr
Answer:
Static friction is greater than kinetic friction, but they both act opposite the applied force.
Explanation:
Newton's 3rd law states that every action has and equal but opposite reaction.
If an object has static friction, that means it stays in one spot, and it takes a great amount of force to get it moving.
Once the object is moving it has kinetic friction, but it's easier to keep it moving unless you are trying to stop it.
The equal but opposite reaction to something moving it is stopping it, and the equal but opposite reaction to stopping something is moving it.
The same amount of force used to move/stop something is used to stop/move it.
Static friction is greater than kinetic friction, but they both act opposite the applied force.
Friction is the force that opposes motion. Frictional force always acts in opposition to the direction of motion.
There are two kinds of friction;
Static frictionDynamic frictionSince more forces tend to act on a body at rest and prevent it from getting into motion than the forces that tend to stop an already moving body, it follows that static friction is greater than kinetic friction. Both act in opposite direction to the applied force.
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Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequency of 544 Hz. The other cyclist hears the frequency as 563 Hz. If the speed of sound in air is 344 m/s, what is the speed of the motorcycles
Answer:
6ms^-1
Explanation:
Given that the frequency difference is
( 563- 544) = 19
So alsoThe wavelength of each wave is = v/f = 344 /544
and there are 19 of this waves
So it is assumed that each motorcycle has moved 0.5 of this distance
in one second thus the speed of the motorcycles will be
=> 19/2 x 344/544 = 6.0 m/s
An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.
Answer:
Here, m=9×10
−31
kg,
q=1.6×10
−19
C,v=3×10
7
ms
−1
,
b=6×10
−4
T
r=
qB
mv
=
(1.6×10
−19
)(6×10
−4
)
(9×10
−31
)×(3×10
7
)
=0.28m
v=
2πr
v
=
2πm
Bq
=
2×(22/7)×9×10
−31
(6×10
−4
)×(1.6×10
−19
)
=1.7×10
7
Hz
Ek=
2
1
mv
2
=
2
1
×(9×10
−31
)×(3×10
7
)
2
J
=40.5×10
−17
J=
1.6×10
−16
40.5×10
−17
keV
=2.53keV
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
Answer:
A) the moment of inertia of the system decreases and the angular speed increases.
Explanation:
The complete question is
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that
A) the moment of inertia of the system decreases and the angular speed increases.
B) the moment of inertia of the system decreases and the angular speed decreases.
C) the moment of inertia of the system decreases and the angular speed remains the same.
D) the moment of inertia of the system increases and the angular speed increases.
E) the moment of inertia of the system increases and the angular speed decreases
In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as
[tex]I_{1} w_{1} = I_{2} w_{2}[/tex] ....1
where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.
and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.
Also, we know that the moment of inertia of a rotating body is given as
[tex]I = mr^{2}[/tex] ....2
where [tex]m[/tex] is the mass of the rotating body,
and [tex]r[/tex] is the radius of the rotating body from its center.
We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.
From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .
From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.
_____ is a distinct element found in Balinese Gamelan music in which a pair of gongs, one of them de-tuned, is played together, creating a sound similar to a tremolo.
its music btw
Answer:
best song ever will you follow me I will also follow you
Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 500 nm on the slits
Answer:
The values is [tex]m_{max} = 8001 \ bright \ spots[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]
At the first half of the screen from the central maxima
The number of bright spot according to the condition for constructive interference is
[tex]n = \frac{d * sin (\theta )}{\lambda}[/tex]
For maximum number of spot [tex]\theta = 90^o[/tex]
So
[tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]
[tex]n =4000[/tex]
Now for the both sides plus the central maxima we have
[tex]m_{max} = 2 * n + 1[/tex]
substituting values
[tex]m_{max} = 2 * 4000 + 1[/tex]
[tex]m_{max} = 8001 \ bright \ spots[/tex]
What would happen if you changed the position of the screen, but kept the other factors the same?
Answer:
I wish I could help but iam srry
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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What is the emf of this cell under standard conditions? Express your answer using three significant figures.
Complete Question
A voltaic cell utilizes the following reaction and operates at 298 K:
3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).
What is the emf of this cell under standard conditions? Express your answer using three significant figures.
Answer:
The value is [tex]E^o_{cell} = 2.35 V[/tex]
Explanation:
From the question we are told that
The ionic equation is
[tex]3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}[/tex]
Now under standard conditions the reduction half reaction is
[tex]Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V[/tex]
And the oxidation half reaction is
[tex]Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V[/tex]
The emf of this cell under standard conditions is mathematically represented as
[tex]E^o_{cell} = E^o _r - E^o _o[/tex]
substituting values
[tex]E^o_{cell} = 1.61 - (- 0.74)[/tex]
[tex]E^o_{cell} = 2.35 V[/tex]
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish
Answer:
Apparent depth (Da) = 60.15 cm (Approx)
Explanation:
Given:
Distance from fish (D) = 80 cm
Find:
Apparent depth (Da)
Computation:
We know that,
Refractive index of water (n2) = 1.33
So,
Apparent depth (Da) = D(n1/n2)
Apparent depth (Da) = 80 (1/1.33)
Apparent depth (Da) = 60.15 cm (Approx)
The apparent depth of the fish is 60 cm.
To calculate the apparent depth of the fish, we use the formula below.
Formula:
R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1Where:
R.F = Refractive index of waterMake D' The subject of the equation.
D' = D/R.F................... Equation 2From the question,
Given:
D = 80 cmR.F = 1.333Substitute these values into equation 2
D' = 80/1.33D' = 60.01D' = 60 cmHence, the apparent depth of the fish is 60 cm
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