Decibels are used to measure sound (dB). A motorcycle engine operating is roughly 95 dB louder than regular conversation, which is around 60 dB louder than a whisper.
What do you mean by decibel level?A relative unit of measurement, the decibel (symbol: dB), is equal to one-tenth of a bel (B). It uses a logarithmic scale to express the ratio of two values of a power or root-power quantity. The power ratio between two signals with a one-decibel difference in level is 101/10 (or roughly 1.26), and the root-power ratio is 10120. (approximately 1.12). The unit can convey an absolute value or a change in the relative value. When used in this method, the unit symbol is frequently followed by letter codes that denote the reference value. In the latter situation, the numeric value reflects the ratio of a value to a fixed reference value. A typical suffix is "V" for the reference value of 1 volt, for instance (e.g., "20 dB").
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A toy doll and a toy robot are standing on a frictionless surface facing each other. The doll has a mass of 0. 20 kg, and the robot has a mass of 0. 30 kg. The robot pushes on the doll with a force of 0. 30 n. The magnitude of the acceleration of the robot is.
The acceleration of the robot is 1.43 m/s2 in magnitude.
What is acceleration?The acceleration of an object is defined as the variation of the velocity with respect to time. The acceleration is calculated by finding the ratio of the change in velocity to the change in time.
The acceleration is also calculated by using the force formula. The force is actually defined as the acceleration produced in the body by applying the force on an object of mass m.
On a non-stick surface, a toy robot and doll are standing side by side. The robot weighs 0. 30 kg, while the doll weighs 0. 20 kg. With a force of 0. 30 n, the robot pushes against the doll.
According to Newton's second law.
F = mass x acceleration
Given that
Force applied = 1N
Acceleration = Force/mass
Substitute the values and get acceleration.
Acceleration = 1/0.7
Acceleration = 1.43m/s²
Hence, the acceleration of the robot is 1.43 m/s2 in magnitude.
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Why is the value of 'g' taken as negative when a body is thrown vertically upwards?
Answer:
See below
Explanation:
[tex]\displaystyle {\textsf {Acceleration is the rate of change of velocity with time:}}[/tex]
[tex]\boxed {g = \frac{dv}{dt}}[/tex]
[tex]\sf {where \;dv = \;change \;in \;velocity\; and \;dt \;the \;change \;in \;time}[/tex]
here g is the gravitational force
When a body is thrown vertically upward, it has an initial speed. As it keeps going up, it speed keeps decreasing until it becomes zero, then the ball starts dropping down
On the upward movement, the change in velocity is negative while the change in time is always positive
So dv/dt = a negative number divided by a positive number which is a negative value for g in the equation for g
three resistors having values of 4 ω, 6 ω, and 10 ω are connected in parallel. this circuit is connected to a 12 v battery. what is the current in the 10 ω resistor?
While current is the same in series and parallel, voltage is not. Therefore, because these three resistors are connected in parallel, their voltage is the same, or 12 volts. Using the equation V=IR, 12V=I*10Ω gives you a current of 1.2A.
What is resistor?A resistor is a passive two-terminal electrical component used in circuits to implement electrical resistance. Resistors have a variety of purposes in electronic circuits, reducing current flow, modifying signal levels, dividing voltages, biassing active components, and terminating transmission lines are a few examples. High-power resistors that can generate many watts of heat instead of electrical energy can be utilised as test loads for generators, power distribution systems, and motor controls. Fixed resistors' resistances only sporadically vary as a result of changes in operating voltage, time, or temperature. Variable resistors can be utilised as force sensors, heat sensors, light sensors, volume controls, lamp dimmers, humidity sensors, and chemical activity sensors.
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6
Which of the following is NOT something that would lead astronomers to believe a black hole might be close-by in space?
A.
an object that has gas and dust appear to be funneling around an object
B.
an object that has spinning particles around an emptiness in space that emit x-rays
OC.
an object that has height and depth but no width
OD. an object in space with highly elevated temperatures
Reset
Next
The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width.
What is a black hole that can be close-by in space?There is said to be a wandering black hole that can be seen in an approximately 5,000 light-years away from earth.
It is one that can be seen in the Carina-Sagittarius spiral aspect of our galaxy.
Note that, its discovery gives room for astronomers to state that the nearest form of isolated stellar-mass black hole to Earth can be as close as about 80 light-years away.
Therefore, The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width because it has no width, no depth and no height.
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3. According to Newton's First Law, unless acted upon by an unbalanced force, an object at rest will
Answer:
An object at rest remains at rest
Explanation:
An object at rest stays at rest and an object in motion stays in motion with the same speed
a test charge of 1µc is placed halfway between a charge of 4.7µc and another of 7.7 µc separated by 10 cm. what is the magnitude of the force (in newtons) on the test charge? your answer should be a number with two decimal places, do not include the unit.
The magnitude of the force on the test charge is 1.
What is magnitude?Greatness is the quantitative worth of seismic energy. It is a particular worth having no connection with distance and heading of the focal point. The idea of extents traces all the way back to the Ancient Greeks, when stars overhead was sorted into six sizes. Extent alludes to the evaluating of the splendor of stars, the first being the most splendid. It has been moved to different issues since basically the seventeenth 100 years. In Physics, greatness is characterized as the most extreme degree of size and the course of an item. Greatness is utilized as a typical consider vector and scalar amounts. By definition, we realize that scalar amounts are those amounts that have size as it were.
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A boat travels 12. 0 m while it reduces its velocity from 9. 5 m/s to 5. 5 m/s. What is the magnitude of the boat’s acceleration while it travels the 12. 0 m?.
A boat travels 12. 0 m while reduces its velocity from 9. 5 m/s to 5. 5 m/s, while it travels the 12. 0 m, the acceleration of the boat is -2.5 m/s²
What is acceleration?The rate at which an object's velocity with respect to the time changes is referred to as acceleration. It is a vector quantity to accelerate (in that they have magnitude and direction). The direction of the net force acting on an object determines the direction of its acceleration. According to Newton's Second Law, the amount of an object's acceleration is the combined result of two causes:
The size of the net balance of all external forces acting on that object is directly proportional to the magnitude of this net resulting force; The magnitude of the mass of that object, depending on the materials out of which it is built, is inversely proportional to the magnitude of the mass.The given parameters;
distance traveled by boat, d = 12 m
initial velocity of boat, u = 9.5 m/s
final velocity of boat, v = 5.5 m/s
The acceleration of boat is calculated as;
[tex]v^{2}=u^{2}+2as\\ a=\frac{v^{2}-u^{2} }{2s}\\ a=\frac{5.5^{2}-9.5^{2} }{2*12}[/tex]
a=-2.5m/[tex]s^{2}[/tex]
Thus, the deceleration of boat is 2.5 m/s².
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a scientist measures the speed of sound in a monatomic gas to be 449 m/s at 20° c. what element does this gas consist of?
Given:
[tex]speed,\;v=449\;m/s[/tex]
We know that,
[tex]adiabatic index, \gamma = 1.667[/tex]
[tex]gas constant,\;R=8.314\;J[/tex]
The form speed of the sound in a monoatomic gas is,
[tex]v=\sqrt{\frac{\gamma\;R\;T}{M} }[/tex]
Substitute the known values in the above equation,
[tex]449=\sqrt{x} \frac{1.667 \times 8.314 \times 293}{M}[/tex]
[tex]M=\sqrt{x} \frac{1.667 \times 8.314 \times 293}{449}[/tex]
[tex]M=0.02014 \times (\frac{1000}{1})[/tex]
[tex]M=20.14\;g/mol[/tex]
Therefore, monoatomic gas is Neon gas (Ne)
Explain monoatomic gas?
Monatomic gas, which differs from diatomic, triatomic, or generally polyatomic gases in that it contains particles (molecules) made up of just one atom, includes gases like helium or sodium vapor. Because a monatomic gas lacks the rotational and vibrational energy components that characterize polyatomic gases, its thermodynamic behavior in the normal temperature range is incredibly straightforward. As a result, its heat capacity is independent of temperature and molecular weight (in this case, atomic weight), and its entropy (a measure of disorder) only depends on temperature and molecular weight.
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Three solid plastic cylinders all have radius 2.50cm and length 6.00cm. Find the charge of each cylinder given the following additional information about each one. Cylinder (c) carries charge with uniform density 500nC/m³ throughout the plastic.
By the charge density, The charge throughout te plastic is 0.059 nC.
We need to know about charge density to solve this problem. The charge density can be determined as
λ = Q / V
where λ is charge density, Q is charge and V is volume.
The parameter given is the charge density and the solid cylinder shape which are :
λ = 500nC/m³
r = 2.5 cm = 0.025 m
L = 6.0 cm = 0.06 m
Find the volume of solid cylinder
V = π . r² . L
V = π . 0.025² . 0.06
V = 1.18 x 10¯⁴ m³
Find the charges
Q = λ x V
Q = 500 x 1.18 x 10¯⁴
Q = 0.059 nC
Hence, the charge throughout the plastic is 0.059 nC.
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The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT.(c) A t what distance is it one-tenth as large?
The distance should be 1.27 m from the wire in order to get magnetic field one-tenth as large .
The magnitude and direction of the magnetic field due to a straight wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by [tex]B=\frac{u_0I}{2\pi r}[/tex] ..(1)Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.Let [tex]B_1[/tex] be inside the plane and [tex]B_2[/tex] be outside the plane . It is required to calculate magnetic field at point A .
Direction of magnetic field at A is calculated using Right hand rule.
[tex]B_{net}=B_2-B_1\\\\B_{net}=\frac{u_0I}{2\pi (0.4-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(0.4+\frac{3\times10^{-3}}{2})} \\\B_{net}=7.5\times 10^{-9}T[/tex]....(1)
Let a distance "R" from the wire so that the magnetic field is [tex]\frac{B_{net}}{10}[/tex]
Using the magnetic field equation , we get
[tex]\frac{B_{net}}{10} =B_2-B_1\\\\\\frac{7.5\times 10^{-9}}{10} =\frac{u_0I}{2\pi (R-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(R+\frac{3\times10^{-3}}{2})} \\\\\R^2-2.25\TIMES10^{-6}=1.6\\\\R=1.27m[/tex]
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Charges of 3.00nC,-2.00 nC,-7.00 nC, and 1.00nC are contained inside a rectangular box with length 1.00m, width 2.00m, and height 2.50m. Outside the box are charges of 1.00nC and 4.00nC . What is the electric flux through the surface of the box? (a) 0(b) -5.64× 10² N . m²/C(c) -1.47 ×10³ N . m²/C(d) 1.47× 10³ N . m²/C(e) 5.64× 10² N . m²/C
The electric flux through the surface is 5.65×〖10〗^2 Nm^2/C ≈5.6×〖10〗^2 Nm^2/C
The electric flux through a surface describes the number of electric field lines that cross that surface. It is expressed as the surface integral of the electric field through that surface.
Gauss's law states that for a closed surface, the electric flux through that surface depends only on the total charge present inside that surface. Hence, for a closed surface, the electric flux can be computed without evaluating any integrals.
The charges are: q1=3.0nC, q2=−2.0nC, q3=−7.0nC, and q4=1.0nC
Gauss's Law provides that the net electric flux through a closed surface depends only on the net charge enclosed by the surface, by the equation:
ϕ=Qnet÷ϵ0
where,
Φ is the net electric flux through the closed surface.
Qnet is the net charge enclosed by the surface.
∈_0=8.85×〖10〗^(-12) C^2/Nm^2 is the permittivity of free space.
The net charge inside the box is:
Qnet = q1+q2+q3+q4
=3.0nC−2.0nC−7.0nC+1.0nC
= −5.0nC=−5.0×10−9C
The net electric flux through the box is:
Φ=Q_net/∈_0
Φ=(-5.0×〖10〗^(-9) C)/(8.85×〖10〗^(-12) C^2/Nm^2 )
Φ=5.65×〖10〗^2 Nm^2/C ≈5.6×〖10〗^2 Nm^2/C
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The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin b through an angle of 2°. Determine the average normal strain in each wire. The wires are unstretched when the lever is in the horizontal position.
The average normal strain in wires A, C and D is found to be 0.035, 0.035 and 0.059 respectively when there is a change in the angle by 2° clockwise.
A physical measurement called strain illustrates how much an object's dimensions vary while it is under stress. The ratio between the object's changed length and its initial length is known as the linear strain.
All wires are stretched when the stiff lever rotates 2° in a clockwise direction around pin B.
Elongation would be R x angle in radian
Elongation in the wire A = 200 mm x 3.142 x 2/180 = 6.982 mm
Elongation in the wire C = 300 x 3.142 x 2/180 = 10.473 mm
Elongation in the wire D = 500 x 3.142 x 2/180 = 17.455 mm
Therefore,
Average normal strain in the wire A = 6.982/200 = 0.03491
Average normal strain in the wire C = 10.473/300 = 0.03491
Average normal strain in the wire D = 17.455/300 = 0.05818
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1: The ball is on the 50 yard line. The ball travels west 5 yards before it's
handed off and ran forward (EAST) 15 yards where they're tackled.
What is the distance the ball traveled?
Your answer
This is a required question
* 1 point
The ball is on the 50-yard line. The ball travels west 5 yards. The distance ball traveled was 60 yards.
WHAT IS DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT ?Distance can be defined as the length of any path between any two locations. Displacement is the direct distance between any two points when calculated via the shortest path between them. When computing distance, the direction is disregarded. The displacement computation takes the direction into consideration.
CALCULATIONThe 5 yard gain would bring the ball to the 45 yard line (ball was tackled at the 50 yard line, so 50 - 5 = 45), but the 15 yards lost due to the tackle and push back would bring ball to the 60 yard line (45 + 15 = 60).
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A wire carries a steady current of 2.40A . A straight section of the wire is 0.750 m long and lies along the x axis within a uniform magnetic field, →B = 1.60kT . If the current is in the positive x direction, what is the magnetic force on the section of wire?
The force on the wire will be
F = 2.88 Newton
We have a current carrying wire in a uniform magnetic field.
We have to determine the magnitude of the magnetic force on this section of wire.
What is the magnitude of force acting on a current (I) carrying wire of length (L) placed in a Magnetic field (B)?The force on the current carrying wire will be -
F = IBL sinθ
According to the question, we have -
Length (L) = 0.75 m (along +x - axis)
Current (I) = 2.4 A (along + x - axis)
Magnetic Field (B) = 0.39 k T (along +z axis)
Therefore, the angle between Magnetic field and Length = θ = 90°.
Therefore -
F = IBL sin(90)
F = 2.4 x 1.6 x 0.75 x sin(90)
F = 2.88 x sin(90)
F = 2.88 Newton
Hence, the force on the wire will be
F = 2.88 Newton
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I need help with this ASAP... Please and thank you
Answer: 1) D & 2) H
Explanation:
D. Tropical rainy &
H. Long-term average temperature and precipitation for the area
4. A student has 500 identical, rectangular sheets of paper. The mass of 1.0m² of the paper is 0.080 kg. (2) Using a metre rule, she measures the length of one sheet of paper and its width. The length is 0.300 m and the width is 0.210m. (1) Calculate the mass of one sheet of paper.
mass of one sheet of paper is 0.00504 kg or 5.04g.
First we will find the surface area of a single sheet of paper
surface area = length ×width
surface area =0.3×0.21
surface area =0.063m²
now 0.063m² is the surface area of a single sheet of paper .
so now we will find number of papers required to form 1m² of area
which will be given by dividing 1m² by surface area of a single sheet
i.e.
number of sheets required = 1/0.063m²
number of sheets required =15.87
mass of one sheet = mass of 1 m^2 of paper / number of times narrower
mass of one sheet = 0.080 kg / 15.873
mass of one sheet = 0.00504 kg
mass of one sheet of paper is 0.00504 kg or 5.04g.
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A flat surface of area 3.20m² is rotated in a uniform electric field of magnitude E=6.20 × 10⁵N . m²/C . Determine the electric flux through this area (a) when the electric field is perpendicular to the surface
The electric flux is 0 N.m²/C.
We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as
Φ = E . S = E . Scosθ
where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.
From the question above, we know that
E = 6.20 × 10⁵ N/C
θ = 90⁰
S = 3.20 m²
By substituting the following parameter, we get
Φ = E . Scosθ
Φ = 6.20 × 10⁵ . 3.20cos(90⁰)
Φ = 0 N.m²/C
Hence, the electric flux is 0 N.m²/C
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Liam notices that his phone battery is depleting faster than he remembers it should. He suspects that running certain apps on his phone is using up his phone battery. He thinks it might be app 1, app 2, or app 3.
Find the independent, dependent, and controlled variables. What is your hypothesis.
Answer:
he would tap on the "close tab "button and see old the apps he has used in that period
Explanation:
__ which is a acidic mass of pratially decomped organic matter
Answer: partially decomposed
Explanation:
The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT. (a) At what distance is it 0.100μ
The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .
The magnitude and direction of the magnetic field due to a straight wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by [tex]B=\frac{u_0I}{2\pi r}[/tex] ...(1)Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.It is given that magnetic field 40.0 cm away from a straight wire is 1.00μT having current 2.00 A .
From equation (1) magnetic field 40.0 cm = 0.4m away from a straight wire is 1.00μT which is given by [tex]1.00=\frac{u_0I}{2\pi \times0.4}[/tex] .....(2)
From equation (1) magnetic field 'r' m away from a straight wire is 0.100μT which is given by [tex]0.100=\frac{u_0I}{2\pi \times r}[/tex] ...(3)
On dividing equation (2) by (3) , we get
[tex]\frac{1}{0.1} =\frac{r}{0.4} \\\\r=4m[/tex]
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A car starts from rest and accelerates at 4 m/s2 until it reaches 36 m/s. How far did it travel at this time?
The distance of a car that accelerates at 4 m/s² until it reaches 36 m/s is 162m.
How to calculate distance?The distance moved by a body can be calculated using the following formula:
S = ut + ½at²
Where;
S = distancet = time a = accelerationu = velocityHowever, the time taken must be calculated first as follows:
t = 36m/s ÷ 4m/s²
t = 9s
S = 0 × t + ½ × 4 × 9²
S = 0 + ½ × 81 × 2
S = 162m
Therefore, the distance of a car that accelerates at 4 m/s² until it reaches 36 m/s is 162m.
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The wavelength of UV light absorbed strongly by DNA is 263 nm. What is the energy in Joules of a quantum of this light?
Explanation:
E = hc / wavelength
Given that:
h = 6.626 × 10-34 Js
c = 2.998 x 10^8 m/s
wavelength = 263 x 10^-9 m
Which conclusion can be made from Gay-Lussac’s law?
For a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin is the conclusion which can be made from Gay-Lussac’s law and is denoted as option D.
What is Pressure?This is referred to as the perpendicular force applied on a body per unit area and the unit is Pascal.
Gay-Lussac was a scientist who discovered through numerous experiments and observations that at a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin.
This is denoted as the following equation below:
P ∝ T
P = kT where p is pressure, k is constant and t is temperature.
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The options are:
At a constant temperature, the pressure of a gas is directly proportional to the volume.At a constant temperature, the pressure of a gas is indirectly proportional to the volumeFor a constant volume, the pressure of a gas is indirectly proportional to the temperature in KelvinFor a constant volume, the pressure of a gas is directly proportional to the temperature in KelvinAssume that the variable name has the value 33. What is the value of name after the assignment name = name * 2 executes?
When the variable name has been assigned with the value 33, then the value of the name after the assignment name will be 66.
In the programming language the sign ' = ', is referred to as the assignment operator which is used to assign values to a particular variable, where the variable referred to the name of the memory location. As it is given the assignment name = name * 2, it means that name on the right-hand side is the assignment name, so whatever value it gets is replaced by twice that previous value according to the operation.
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If you double the wavelength of light, what happens to the energy of the photons?.
Answer:
the energy of photon will be half.
Explanation:
If you need a deeper explanation please let me know. Hope I helped!
When thermal energy is removed from particles, what action occurs?
A The particles' energy is destroyed.
B The particles' temperature increases.
C The particles move more quickly.
D The particles' kinetic energy decreases.
When thermal energy is removed from particles, then the particles' kinetic energy decreases (Option D).
What is particles' kinetic energy?The expression particles' kinetic energy makes reference to the amount of motion energy (i.e. energy in movement) that contain the particles of an object, which depends on the temperature of the material.
In conclusion, when thermal energy is removed from particles, then the particles' kinetic energy decreases (Option D).
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an object is dropped from a bridge and allowed to freefall to the ground. the height of the object over time can be modeled using h(t)
Dropped from a bridge and allowed to fall freely to the ground is an object. Modeling the object's height over time using h(t) = 144 - 16t2 is possible
What do you mean by Freefall?An item that is falling only due to gravity is said to be in free fall. This means that any moving object that is simply being affected by gravity is said to be "in a condition of free fall." An item of this kind will accelerate downhill at a rate of 9.8 m/s/s. If an item is just moving in one direction—downward or upward toward its peak—and is only affected by gravity, then its acceleration value is 9.8 m/s/s.
The initial velocity of an object being dropped from a height is zero meters per second (m/s) as opposed to being hurled.
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Perform the calculation and report your answer using sig figs. 3.42 + 4 + 5.2 + 12
Answer:
3.42 + 4 + 5.2 + 12= 24.62
So since we have 24.62 it will be 4 sig figs
Explanation:
The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n = 1.33 is the index of refraction of water.(d) Evaluate u in the limit as the speed of the water approaches c .
The speed of light u is equal to c.
Given,
The motion of a transparent medium influences the speed of light.
The water moves with speed v in a horizontal pipe.
Assume the light travels in the same direction as the water moves.
The speed of light with respect to the water is c /n.
The index of refraction of water, n = 1.33
Let us assume u' be the speed of light in water.
u' is related to the refractive index of water,as u' = c/n
where, c is the speed of light.
Let, u be the speed of light in water in the lab frame.
Now, u and u' are related as : u = ( u'+v) /(1+u'v/c^2)
Here, v is the speed of water in the horizontal pipe.
We know the value of u'. so by substituting the value, we will get ,
u = (c/n+v) /(1+cv/nc^2)
u = c/n (1+nv/c) /(1+v/nc)
u in the limit as the speed of the water approaches c.
Thus,
u = lim v-->c [c/n (1+nv/c) /(1+v/nc) ]
u= c/n (1+n) /(1+1/n)
u= c .
Hence, the speed of light u is equal to c.
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Disclaimer: incomplete question. here is the complete question.
Question:The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n = 1.33 is the index of refraction of water.
(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc).
(d) Evaluate u in the limit as the speed of the water approaches c .
At one location on the Earth, the rms value of the magnetic field caused by solar radiation is 1.80 μT. From this value, calculate (b) the average energy density of the solar component of electromagnetic radiation at this location.
The average energy density of the solar component of electromagnetic radiation at this location is determined as 2.58 x 10⁻⁶ J.
Average energy densityThe average energy density of the solar component of electromagnetic radiation at this location is calculated as follows;
U(avg) = (Brms)²/μ₀
U(avg) = (1.8 x 10⁻⁶)² / (4π x 10⁻⁷)
U(avg) = 2.58 x 10⁻⁶ J
Thus, the average energy density of the solar component of electromagnetic radiation at this location is determined as 2.58 x 10⁻⁶ J.
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