Explanation:
1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.
2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.
Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.
We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.
These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.
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Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh's criterion
Answer:
The distance is [tex]D = 0.000712 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light source is [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]
The distance from a pin hole is [tex]x = 9\ m[/tex]
The diameter of the pin hole is [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]
Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is
mathematically represented as
[tex]D = \frac{1.22 \lambda }{d }[/tex]
substituting values
[tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]
[tex]D = 0.000712 \ m[/tex]
What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N
Answer:
A. 30.38°
B 5.04N
Explanation:
Using
F= ILBsin theta
2 .55N= 8.4Ax 0.5mx 1.2T x sintheta
Theta = 30.38°
B. If theta is 90°
Then
F= 8.4Ax 0.5mx 1.2x sin 90°
F= 5.04N
The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
Answer:
Speed of the electron is 2.46 x 10^8 m/s
Explanation:
momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]
where [tex]M_{0}[/tex] is the rest mass of the electron
V is the velocity of the electron.
under relativistic effect, the mass increases.
under relativistic effect, the new mass M will be
M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]
where
[tex]\beta = V/c[/tex]
c is the speed of light = 3 x 10^8 m/s
V is the speed with which the electron travels.
The new momentum will therefore be
==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have
1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
the equation reduces to
1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]
square both sides of the equation, we have
3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]
3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1
2.0625 = 3.0625[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 0.67
β = 0.819
substitute for [tex]\beta = V/c[/tex]
V/c = 0.819
V = c x 0.819
V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely immersed in water, the scale reads 18. 2 N. What are the volume and density of the block?
Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
Answer:
The volume of the block is 306 cm³
The density of the block is 7.07 g/cm³
Explanation:
Given;
weight of block in air, [tex]W_a[/tex] = 21.2 N
Weight of block in water, [tex]W_w[/tex] = 18.2 N
Mass of the block in air;
[tex]W_a = mg[/tex]
21.2 = m x 9.8
m = 21.2 / 9.8
m = 2.163 kg
mass of the block in water;
[tex]W_w = mg[/tex]
18.2 = m x 9.8
m = 18.2 / 9.8
m = 1.857 kg
Apply Archimedes principle
Mass of object in air - mass of object in water = density of water x volume of object
2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block
0.306 kg = 1000 kg/m³ x Volume of block
Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]
Volume of the block = 3.06 x 10⁻⁴ m³
Volume of the block = 306 cm³
Determine the density of the block
[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]
Four friends push on the same block in different directions. Allie pushes on the block to the north with a force of 18 N. Bill pushes on the block to the east with a force of 14 N. Chris pushes on the block to south with a force of 23 N. Debra pushes on the block to the west with a force of 20 N. Assuming it does not move vertically, in which directions will the block move? north and west south and east south and west north and east
Answer:
South and West
Explanation:
Those people are pushing the hardest. It will move south faster than it moves west.
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave travels at 5 000 m/s. If P-waves are received at a seismic station 1.00 minute before an S-wave arrives, how far away is the earthquake center?
Assuming constant speeds, the P-wave covers a distance d in time t such that
9000 m/s = d/(60 t)
while the S-wave covers the same distance after 1 more minute so that
5000 m/s = d/(60(t + 1))
Now,
d = 540,000 t
d = 300,000(t + 1) = 300,000 t + 300,000
Solve for t in the first equation and substitute it into the second equation, then solve for d :
t = d/540,000
d = 300,000/540,000 d + 300,000
4/9 d = 300,000
d = 675,000
So the earthquake center is 675,000 m away from the seismic station.
Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?
Answer:
Explanation:
This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.
If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).
But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).
A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.Automobile 1: 500kg, 10m/sAutomobile 2: 2000kg, 5m/sAutomobile 3: 500kg, 20m/sAutomobile 4: 1000kg, 20m/sAutomobile 5: 1000kg, 10m/sAutomobile 6: 4000kg, 5m/sRequired:a. Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.b. Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.c. Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.
Answer:
A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.
Give an example of a fad diet that is not healthy and one that is healthy. Explain how you know the difference.
Answer:
Good Diet: ! gallon of water a day, Fruits, Vegetables, White meats(Chicken), Don't eat past 3 PM.
Bad Diet: Pizza, Red meat, Baked goods, Eating at late hours.
Explanation: I know the difference because, When you drink water first thing in the morning it gets your metabolism running. Than means you can digest foods better, you want to feed your body good foods but you should not eat until you feel stuffed. You should eat until you are no longer starving. Than you should drink a cup of water in between meals. I know you should not eat past 3 pm because your body needs time to digest foods because you should never go to sleep with a full stomach. I know the difference between good food and bad food because when you eat healthy food and a balanced diet, your body will have more energy and you wont feel tired afterwards. Eating bad foods and food with artificial sugars will clump up in your kidneys, and your body will have small bursts of energy but you will feel lazy afterwards...Your body is supposed to stay energized from a healthy meal in order to give you the energy your body needs to exercise. If you feel droopy all the time and you don't want to do anything, than you are unhealthy.
Answer:
A vegetarian diet is an example of a good fad diet if you do it correctly. It can help you get lots of veggies and good nutrients from them while still following the non-meat diet you want. This can be effective and good for weight loss becasue you are still eating and getting all the good nutrients and calories from less fatty foods.
Vegan diet (some can be successful but many people fail and do not do good that is why I choose this) The problem with this fad diet is that it can cause nutritional deficiencies and lead to a host of additional health problems, including negatively impacting hormonal health and metabolism. Many people also struggle to find healthy vegan food and end up eating bad and fatty foods instead.
Explanation:
Got a 100
If you stood on a planet having a mass four times higher than Earth's mass, and a radius two times 70) lon longer than Earth's radius, you would weigh:________
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth.
CHECK COMPLETE QUESTION BELOW
you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth
Answer:
OPTION C is correct
The same as you do on Earth
Explanation :
According to law of gravitation :
F=GMm/R^2......(a)
F= mg.....(b)
M= mass of earth
m = mass of the person
R = radius of the earth
From law of motion
Put equation b into equation a
mg=GMm/R^2
g=GMm/R^2
g=GM/R^2
We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have
m= 4M
r=(2R)^2=4R^2
g= G4M/4R^2
Then, 4in the denominator will cancel out the numerator we have
g= GM/R^2
Therefore, g remain the same
A large reflecting telescope has an objective mirror with a 14.0 m radius of curvature. What angular magnification in multiples does it produce when a 3.25 m focal length eyepiece is used? ✕
Answer:
The magnification is [tex]m = -2.15[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 14.0 \ m[/tex]
The focal length eyepiece is [tex]f_e = 3.25 \ m[/tex]
Generally the objective focal length is mathematically represented as
[tex]f_o = \frac{r}{2}[/tex]
=> [tex]f_o = \frac{14}{2}[/tex]
=> [tex]f_o = 7 \ m[/tex]
The magnification is mathematically represented as
[tex]m = - \frac{f_o }{f_e }[/tex]
=> [tex]m = - \frac{7 }{ 3.25 }[/tex]
=> [tex]m = -2.15[/tex]
An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody emitters. For each star she makes a plot of the radiation intensity per unit wavelength as a function of wavelength. She notices that the curve for star A has a maximum that occurs at a shorter wavelength than does the curve for star B. What can she conclude about the surface temperatures of the two stars
Answer:
Star A has a higher surface temperature than star B.
Explanation:
The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:
[tex]T = \frac{2.898x10^{-3} m. K}{\lambda max}[/tex] (1)
Where T is the effective temperature of the star and [tex]\lambda_{max}[/tex] is the maximum peak of emission.
A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.
Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and [tex]\lambda max[/tex] are inversely proportional.
You shine unpolarized light with intensity 52.0 W/m2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 15.0 W/m2. Find the intensity of the light that emerges from the first polarizer.
Answer:
The intensity of light from the first polarizer is [tex]I_1 = 26 W/m^2[/tex]
Explanation:
The intensity of the unpolarized light is [tex]I_o = 52.0 \ W/m^2[/tex]
Generally the intensity of light that emerges from the first polarized light is
[tex]I_1 = \frac{I_o}{2 }[/tex]
substituting values
[tex]I_1 = \frac{52. 0}{2 }[/tex]
[tex]I_1 = 26 W/m^2[/tex]
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W
Complete Question
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.
Answer
a. Approximately [tex]5*10^{10}[/tex] fissions per second.
b. Approximately [tex]6*10^{12 }[/tex]fissions per second.
c. Approximately [tex]4*10^{11}[/tex] fissions per second.
d. Approximately [tex]3*10^{12}[/tex] fissions per second.
e. Approximately[tex]3*10^{14}[/tex] fissions per second.
Answer:
The correct option is d
Explanation:
From the question we are told that
The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]
Thus to generated [tex]100 \ J/s[/tex] i.e (100 W ) the rate of fission is
[tex]k = \frac{100}{3.2 *10^{-11} }[/tex]
[tex]k =3*10^{12} fission\ per \ second[/tex]
B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
A mass m = 0.7 kg is released from rest at the origin 0. The mass falls under the influence of gravity. When the mass reaches point A, it is a distance x below the origin 0; when the mass reaches point B it is a distance of 3 x below the origin 0. What is vB/vA?
Answer:
[tex]v_B/v_A=\sqrt{3}[/tex]
Explanation:
Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:
[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]
Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:
[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]
Let's work in a similar way with the two different positions at those different times, and for which we have some information;
[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]
Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:
[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]
Distinguish between physical and chemical changes. Include examples in your explanations.
Answer:
Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.
Instead the physical changes implies that if you can return to the same substance through a reverse process.
Explanation:
A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.
A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.
A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.
• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.
• In a chemical change always a chemical reaction takes place.
• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.
• In a physical change no new chemical species form.
• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.
• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.
Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.
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which category would a person who has an IQ of 84 belong ?
Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.
Requried:
What is the frequency the horns emit?
Answer: f ≈ 8.5Hz
Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.
For a source moving and a stationary observer, to determine the frequency:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
where:
[tex]f_{0}[/tex] is frequency of observer;
[tex]f_{s}[/tex] is frequency of source;
c is the constant speed of sound c = 340m/s;
[tex]v_{s}[/tex] is velocity of source;
Rearraging for frequency of source:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]
Replacing and calculating:
[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]
[tex]f_{s} = 9.(0.9412)[/tex]
[tex]f_{s} =[/tex] 8.5
Frequency the horns emit is 8.5Hz.
Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s
Answer:
The time difference is 2.97 sec.
Explanation:
Given that,
Tension = 29 N
Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]
Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]
Length of first section = 2.5 m
Length of second section = 2.8 m
We need to total distance of first pulse
Using formula for distance
[tex]d=2.5+2.5[/tex]
[tex]d_{1}=5.0\ m[/tex]
We need to total distance of second pulse
Using formula for distance
[tex]d=2.8+2.8[/tex]
[tex]d_{2}=5.6\ m[/tex]
We need to calculate the speed of pulse in the first string
Using formula of speed
[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]
Put the value into the formula
[tex]v_{1}=\sqrt{\dfrac{29}{19}}[/tex]
[tex]v_{1}=1.24\ m/s[/tex]
We need to calculate the speed of pulse in the second string
Using formula of speed
[tex]v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}[/tex]
Put the value into the formula
[tex]v_{2}=\sqrt{\dfrac{29}{45}}[/tex]
[tex]v_{2}=0.80\ m/s[/tex]
We need to calculate the time for first pulse
Using formula of time
[tex]t_{1}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{1}=\dfrac{5.0}{1.24}[/tex]
[tex]t_{1}=4.03\ sec[/tex]
We need to calculate the time for second pulse
Using formula of time
[tex]t_{2}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{2}=\dfrac{5.6}{0.80}[/tex]
[tex]t_{2}=7\ sec[/tex]
We need to calculate the time difference
Using formula of time difference
[tex]\Delta t=t_{2}-t_{1}[/tex]
Put the value into the formula
[tex]\Delta t=7-4.03[/tex]
[tex]\Delta t=2.97\ sec[/tex]
Hence, The time difference is 2.97 sec.
: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0
Answer:
145060km
Explanation: Given that
speed = dx/dt = 2t^2 +1
integrate
x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)
given that at t=0, x = 1000
so 1000 = 2/3 X (0)^3 + 0 + c
or c = 1000
So x = 2/3t^3 + t + 1000
for t = 1 min = 60s
x = 2/3 X 60^3 + 60 + 1000
x = 2/3×216000+ 1060
x = 144000+1060
= 145060km
At one minute, it will be 145060km far from the earth
The A block, with negligible dimensions and weight P, is supported by the coordinate point (1.1/2) of the parabolic fixed grounded surface, from equation y = x^2/2 If the block is about to slide, what is the coefficient of friction between it and the surface; determine the force F tangent to the surface, which must be applied to the block to start the upward movement.
Answer:
μ = 1
F = P√2
Explanation:
The parabola equation is: y = ½ x².
The slope of the tangent is dy/dx = x.
The angle between the tangent and the x-axis is θ = tan⁻¹(x).
At x = 1, θ = 45°.
Draw a free body diagram of the block. There are three forces:
Weight force P pulling down,
Normal force N pushing perpendicular to the surface,
and friction force Nμ pushing up tangential to the surface.
Sum of forces in the perpendicular direction:
∑F = ma
N − P cos 45° = 0
N = P cos 45°
Sum of forces in the tangential direction:
∑F = ma
Nμ − P sin 45° = 0
Nμ = P sin 45°
μ = P sin 45° / N
μ = tan 45°
μ = 1
Draw a new free body diagram. This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.
Sum of forces in the tangential direction:
∑F = ma
F − Nμ − P sin 45° = 0
F = Nμ + P sin 45°
F = (P cos 45°) μ + P sin 45°
F = P√2
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?
Answer:
a
[tex]\lambda = 1.667 nm[/tex]
b
[tex]\theta = 0.8681^o[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]
The is distance of the screen from the slit is [tex]D = 2.60 \ m[/tex]
The distance between the central bright fringe and either of the adjacent bright [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]
Generally the condition for constructive interference is
[tex]d sin \tha(\theta ) = n \lambda[/tex]
From the question we are told that small-angle approximation is valid here.
So [tex]sin (\theta ) = \theta[/tex]
=> [tex]d \theta = n \lambda[/tex]
=> [tex]\theta = \frac{n * \lambda }{d }[/tex]
Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as
[tex]y = D * sin (\theta )[/tex]
From the question we are told that small-angle approximation is valid here.
So
[tex]y = D * \theta[/tex]
=> [tex]\theta = \frac{ y}{D}[/tex]
So
[tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]
[tex]\lambda =\frac{d * y }{n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]
[tex]\lambda = 1.667 *10^{-6}[/tex]
[tex]\lambda = 1.667 nm[/tex]
In the b part of the question we are considering the next set of bright fringe so n= 2
Hence
[tex]dsin (\theta ) = n \lambda[/tex]
[tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]
[tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]
[tex]\theta = 0.8681^o[/tex]
Determine the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
Answer:
T = 3.14 hours
Explanation:
We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
We know that the radius of Mars is 3,389.5 km.
So, r = 1,787 + 3,389.5 = 5176.5 km
Using Kepler's law,
[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]
M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]
So,
[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]
or
T = 3.14 hours
So, the orbital period is 3.14 hours
Scientists today learn about the world by _____. 1. using untested hypotheses to revise theories 2. observing, measuring, testing, and explaining their ideas 3. formulating conclusions without testing them 4. changing scientific laws
Answer:
Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.
Explanation:
A traditional perception of such a scientist is those of an individual who performs experiments in some kind of a white coat. The reality of the situation is, a researcher can indeed be described as an individual interested in the comprehensive as well as a recorded review of the occurrences occurring in nature but perhaps not severely constrained to physics, chemistry as well as biology alone.The other three choices have no relation to a particular task. So the option given here is just the right one.
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?
Answer:
a) the distance between the interference fringes is reduced by half
b) the distance between stripes is doubled
Explanation:
Interference experiments constructive interference is described by the expression
d sin θ = m λ
let's use trigonometry to find the distance between the interference fringes
tan θ= y / L
dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ / d L
a) it asks us when the screen doubles its distance
L ’= 2 L
subtitute in the equation
y ’= m λ / (d 2L)
y ’=( m λ / d L) /2
y ’= y / 2
the distance between the interference fringes is reduced by half
b) the density of the network doubles
if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half
d ’= d / 2
we substitute
y ’= m λ (L d / 2)
y ’= m λ / (L d) 2
y ’= y 2
the distance between stripes is doubled
an electric device is plugged into a 110v wall socket. if the device consumes 500 w of power, what is the resistance of the device
Answer: R=24.2Ω
Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:
P=V.i
P=R.i²
[tex]P=\frac{V^{2}}{R}[/tex]
The resistance of the system is:
[tex]P=\frac{V^{2}}{R}[/tex]
[tex]R=\frac{V^{2}}{P}[/tex]
[tex]R=\frac{110^{2}}{500}[/tex]
R = 24.2Ω
For the device, resistance is 24.2Ω.
The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is
Complete question is;
The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?
Answer:
13 m
Explanation:
We are given;
Distance between two nearly parallel mirrors; d = 6.5 m
Distance between the face and the nearer mirror; x = 3 m
Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m
Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.
Thus;
Distance of the first reflection of the back of the head in the rear mirror from the object head is;
y' = 2y
y' = 2 × 3.5
y' = 7
The total distance of this image from the front mirror would be calculated as;
z = y' + x
z = 7 + 3
z = 10
Finally, the second reflection of this image will be 10 meters inside in the front mirror.
Thus, the total distance of the image of the back of the head in the front mirror from the person will be:
T.D = x + z
T.D = 3 + 10
T.D = 13m
In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?
Explanation:
Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,
[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]
[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]
[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]
[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]
after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,
[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]
[tex]\qquad \Phi_{B, f}=0[/tex]
(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,
[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]
[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]
[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]
(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb
(b) The induced EMF in the coil is 0.583 mV
Flux and induced EMF:Given that the coil has N = 250 turns
and an area of A = 14cm² = 1.4×10⁻³m².
It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T
(a) The flux passing through the coil is given by:
Ф = NBAcosθ
where θ is the angle between area vector and the magnetic field
The area vector is perpendicular to the plane of the coil.
So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil
So the initial flux is:
Φ = NABcos0° = NAB
Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb
Ф = 1.75 × 10⁻⁵ Wb
Finally, θ = 90°, and since cos90°, the final flux through the coil is 0
(b) The EMF induced is given by:
E = -ΔФ/Δt
E = -(0 - 1.75 × 10⁻⁵)/0.030
E = 0.583 × 10⁻³ V
E = 0.583 mV
Learn more about magnetic flux:
https://brainly.com/question/15359941?referrer=searchResults
What is the reason for the increase and decrease size of the moon and write down in a paragraph.
Answer:
The reason for the increase or decrease of the moon is due to the angular perception of the moon.
Explanation:
Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.
The zenith is the highest part of the sky and the horizon the lowest.
When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.
But when looking up at the sky as there is no reference point there will be a failure in the perception of size.
