Answer:
The minimum thickness of the material is 114.5 nm
Explanation:
The remaining part of the question is as follows
What should the minimum thickness of the material be for it to be nonreflecting for light of wavelength 600 nm (in a vacuum)? Please give details of your reasoning.
Both reflections occur against a surface of higher index of refraction, so there is a phase shift at each reflection. The phase shifts cancel each other out, so if we want the emerging rays to be out of phase, we must have
Given
Index of refraction = 1.31
Index of refraction = 1.43
As we know
[tex]2 t = \frac{\Lambda}{n} (m+\frac{1}{2})\\[/tex]
Here t is the thickness
[tex]\frac{1}{2}[/tex] represents the phase shift
n is the refraction index
Substituting the given values we get
[tex]t = \frac{\lambda}{4n} \\t = \frac{600}{4*1.31}\\t = 114.5[/tex] nm
Astronauts on a distant planet set up a simple pendulum of length 1.2 m. The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet
Answer:
If you use P = 2 * pi * (L / g)^1/2 for the period of the simple pendulum
g = 4 * pi^2 * 1.2 / 2.8^2 = 6.04 m/s2
Note: omega = 2 pi * f = 2 pi / P and omega = (g / L)^1/2
Answer:
6.0426 m/s^2
Explanation:
The period of a simple pendulum is equal to 2pi sqroot L/g
T = 280s/100 rev = 2.8s
Plug in 1.2 m for L and 2.8s for T, then solve for g to get 6.04 m/s^2 (dependent on how you round)
An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose the electron is released from rest a distance 0.050 m from the ring center. It then oscillates through the ring center. Calculate its period. (The electron is always much closer to the ring center than a radius.)
Answer:
T = 1.12 10⁻⁷ s
Explanation:
This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.
All the charge dq is at a distance r
dE = k dq / r²
Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry
cos θ =[tex]\frac{dE_x}{dE}[/tex]
dEₓ = dE cos θ
cos θ = x / r
substituting
dEₓ = [tex]k \frac{dq}{r^2 } \ \frac{x}{r}[/tex]
DEₓ = k dq x / r³
let's use the Pythagorean theorem to find the distance r
r² = x² + a²
where a is the radius of the ring
we substitute
dEₓ = [tex]k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq[/tex]
we integrate
∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} } ∫ dq
Eₓ = [tex]k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}[/tex]
In the exercise indicate that the electron is very central to the center of the ring
x << a
Eₓ = [tex]k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}[/tex]
if we expand in a series
[tex](\ 1+ (x/a)^2 \ )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2[/tex]
we keep the first term if x<<a
Eₓ = [tex]\frac{ k Q}{a^3} \ x[/tex]
the force is
F = q E
F = [tex]- \frac{kQ }{a^3} \ x[/tex]
this is a restoring force proportional to the displacement so the movement is simple harmonic,
F = m a
[tex]- \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }[/tex]
[tex]\frac{d^2 x}{dt^2} = \frac{keQ}{ma^3} \ x[/tex]
the solution is of type
x = A cos (wt + Ф)
with angular velocity
w² = [tex]\frac{keQ}{m a^3}[/tex]
angular velocity and period are related
w = 2π/ T
we substitute
4π² / T² = \frac{keQ}{m a^3}
T = 2π [tex]\sqrt{\frac{m a^3 }{keQ} }[/tex]
let's calculate
T = 2π [tex]\sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19} \ 0.021 \ 10^{-3} } }[/tex]
T = 2π pi [tex]\sqrt{320.426 \ 10^{-18} }[/tex]
T = 2π 17.9 10⁻⁹ s
T = 1.12 10⁻⁷ s
3 نقطة (نقاط)
السؤال 2
A block slides on a rough horizontal surface from paint A to point B. A force (magnitude P -
2.0 N) acts on the block between A and B, as shown. If the kinetic energies of the block at A
and B are 5.0 J and 4.0 J, respectively and the work done on the block by the force of friction
as the block moves from A to B is -4.5 J what is the distance between point A and B?
40°
B
The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is
[tex]W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J[/tex]
Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.
The only other force acting on the block as it moves is the force P. Let [tex]W_P[/tex] be the work done by the force P. Then the total work done on the block is
[tex]W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}[/tex]
1 point
What is the speed of a wave that has a frequency of 200 Hz and a
wavelength of 2 meters? Express your answer to the nearest whole
number. (wave speed = wavelength * frequency) *
400 m/s
100 m/s
0.01 m/s
Answer:
just multiply the frequency and wavelength
200× 2
400
The distance a freely falling object falls from rest is one-half second is?
Answer:
[tex]s = ut + \frac{1}{2}a {t}^{2} \\ = \frac{1}{2} \times 9.81 \times ({ \frac{1}{2} })^{2} \\ s = 1.22m[/tex]
wha are the role of competition policy authorities
Answer: The competition policy helps in development of the competition in the authority.
Explanation:
In consumer market competition exists between the producers. The main role of the competition policy is to promote competition in the market. It helps multiple businesses or individuals to utilize their resources efficiently and produce products at best quality and at cheapest prices. It allows the consumers to compare and purchase the better deal. It helps in starting of new businesses and also helps in increase in total production output. It is the result of the natural economic growth.
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanescent wave attenuator to attenuate all the modes except the TEM mode along the guide. Find the minimum length of this attenuator needed to attenuate each mode by at least 100 dB. Assume perfect conductor plates.
Answer:
the required minimum length of the attenuator is 3.71 cm
Explanation:
Given the data in the question;
we know that;
[tex]f_{c_1[/tex] = c / 2a
where f is frequency, c is the speed of light in air and a is the plate separation distance.
we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s
plate separation distance a = 7.11 mm = 0.0711 cm
so we substitute
[tex]f_{c_1[/tex] = 3 × 10¹⁰ / 2( 0.0711 )
[tex]f_{c_1[/tex] = 3 × 10¹⁰ cm/s / 0.1422 cm
[tex]f_{c_1[/tex] = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }
Now, we determine the minimum wavelength required
Each non propagating mode is attenuated by at least 100 dB at 15 GHz
so
Attenuation constant TE₁ and TM₁ expression is;
∝₁ = 2πf/c × √( ([tex]f_{c_1[/tex] / f)² - 1 )
so we substitute
∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )
∝₁ = 3.1079 × 10⁻⁷
∝₁ = 310.79 np/m
Now, To find the minimum wavelength, lets consider the design constraint;
20log₁₀[tex]e^{-\alpha _1l_{min[/tex] = -100dB
we substitute
20log₁₀[tex]e^{-(310.7np/m)l_{min[/tex] = -100dB
[tex]l_{min[/tex] = 3.71 cm
Therefore, the required minimum length of the attenuator is 3.71 cm
When playing tennis, if you hit the ball off of the top end of the racket, an uncomfortable standing wave vibration is produced in the racket. This wave is
Answer:
Transverse.
Explanation:
Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.
An electromagnetic spectrum refers to a range of frequency and wavelength that an electromagnetic wave is distributed or extends. The electromagnetic spectrum comprises of gamma rays, visible light, ultraviolet radiation, x-rays, radio waves, and infrared radiation.
A transverse wave can be defined as a type of wave in which particles of the medium of propagation oscillates or vibrates in a direction that is perpendicular to the direction that the wave moves i.e at right angle to the direction of propagation of the wave.
Basically, sound is a transverse and all transverse wave are the direct opposite of a longitudinal wave that usually travel in the same direction of its oscillation.
Hence, this wave is transverse.
A 1460 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 18.0 m/s. They stick together. In what direction and with what speed do they move after the collision?
Answer in degrees north of east
AND
the speed after the collision in m/s
Answer:
Solution given:
North car
mass[m1]=1460kg
velocity[u1]=27 m/s
mass[m2]=2165kg
velocity [u2]=18m/s
let v be velocity after collision
we have
From the principle of conservation of linear momentum
m1u1+m2u2=(m1+m2)v
1460*27+2165*18=(1460+2165)v
v=[tex] \frac{78390}{3625} [/tex]
v=21.6m/s
the speed after the collision in 21.6 m/s.
For angle.
Tan angle =[tex] \frac{m1u1}{m2u2} [/tex]
Tan angle =[tex] \frac{1460*27}{2165*18} [/tex]
Tan angle=327.74
angle=Tan-¹(327.74)=89.82=90°
in degrees north of east is 90°
17. This is used to produce a rough sketch with thick and dark ink to make the figure of an image more distinct. A. shading B. Painting C. Outlining D. Sketching
How are soil and air similar?
Answer:
The air in the soil is similar in composition to that in the atmosphere with the exception of oxygen, carbon dioxide, and water vapor. In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.036%.
hope this helps
have a good day :)
Explanation:
What is the total current in a parallel circuit containing a 12-V battery, a 2 Ω resistor, and a 4 Ω resistor?
Answer:
V=IR
since the circuit is parallel both resistor have same voltage and different current value according to OHMs law.
total resistance in parallel=
1/R=1/R1+1/R2+...+1/Rn
since we have two resistor in parallel
Rt=R1R2/R1+R2
2*4/2+4=4/3 ohms
I=V/R
12/4/3=36/4=9Amp
OR
I=12/2=6amp
I=12/4=3amp
total current
I=6+3
9amp
Can someone help me
BRAINLIST A wave travels at a constant speed. How does the wavelength change if the
frequency is reduced by a factor of 3? Assume the speed of the wave remains
unchanged.
A. The wavelength does not change.
B. The wavelength increases by a factor of 3.
C. The wavelength decreases by a factor of 3.
D. The wavelength increases by a factor of 9.
how dose an exam question outed from text book
Answer:
In which school you are???
Explanation:
The ratio of the focal length of a lens to it's diameter is the
Answer:
The f-ratio describes the relationship between the lens diameter and the focal length and is calculated by dividing the focal length by the diameter of the lens. For example, if a lens were to have a focal length of 50mm and a diameter of 10mm, then the f-ratio would be 50mm/10mm=5 or otherwise referred to as f5.
Explanation:
Answer:
The f-ratio describes the relationship between the lens diameter and the focal length and is calculated by dividing the focal length by the diameter of the lens. For example, if a lens were to have a focal length of 50mm and a diameter of 10mm, then the f-ratio would be 50mm/10mm=5 or otherwise referred to as f5.
Explanation:
i just want a thank you :)
What effect does the Duck Velocity have on the waves seen by the observer?
Towards the boat:
Away from the boat:
Same as the boat:
3. Suppose you take a pendulum with length L and mass m having a period T to a
planet where the value of g is 176 of the value on Earth. What would be the period
of the pendulum on the planet? *
T/6
OT
O 1.6T
2.4 T
4.6 T
help fast
(C)
Explanation:
[tex]t = 2\pi \sqrt{ \frac{l}{g} } [/tex]
If g is only 1/6 on another planet, then
[tex]t = 2\pi \sqrt{ \frac{l}{ \frac{g}{6} } } = 2\pi \sqrt{ \frac{6l}{g} } [/tex]
[tex] = \sqrt{6} \: (2\pi \sqrt{ \frac{l}{g} } ) = 2.4 \times t(on \: earth)[/tex]
On a rectangle with length 100 m, width 50m and 2 vehicles stationed together. Find the time that they meet given that car A travels 5 m/s and leaves 3 seconds before car B, and car B is traveling at 3 m/s in the opposite direction. Can you create a generic equation from the previous scenario
Answer:
[tex]T=35.625sec[/tex]
Explanation:
From the question we are told that:
Length [tex]L=100 m[/tex]
Width [tex]W=50m[/tex]
Velocity of Car A [tex]V_A=5m/s[/tex]
Velocity of Car B [tex]V_B=3m/s[/tex]
Distance traveled by car A before car B moves
[tex]d_l=5*3[/tex]
[tex]d_l=15[/tex]
Therefore total distance traveled at same time interval
[tex]D=total\ distance-d_l[/tex]
Where
Total distance=Perimeter of rectangle
[tex]P=2(L+B)[/tex]
[tex]P=2(100+50)[/tex]
[tex]P=300[/tex]
Therefore
[tex]D=total\ distance-d_l[/tex]
[tex]D=300-15\\D=285m[/tex]
Generally the equation for time taken to meet is mathematically given by
[tex]T=\frac{Distance D}{Relative\ speed V_r}[/tex]
Where
Relative speed = Speed of car A +Speed of car B
[tex]V_r=V_A+V_B[/tex]
[tex]V_r=5+3[/tex]
[tex]V_r=8m/s[/tex]
Therefore the time taken to meet
[tex]T=\frac{ D}{ V_r}[/tex]
[tex]T=\frac{ 285}{ 8}[/tex]
[tex]T=35.625sec[/tex]
The average velocity of blood flowing in a certain 4-mm-diameter artery in the human body is 0.28 m/s. The viscosity and density of blood are approximately 4 cP and 1.06 Mg/m3, respectively. Determine the volumetric flow rate of blood in the artery. (m3/s)
Answer:
V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s
Explanation:
The volume flow rate of the blood in the artery can be given by the following formula:
[tex]V = Av[/tex]
where,
V = Volume flow rate = ?
A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²
v = velcoity = 0.28 m/s
Therefore,
[tex]V = (1.26\ x\ 10^{-5}\ m^2)(0.28\ m/s)[/tex]
V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s
the removal of rock in solution by acidic rain water________
Answer:
Solution
Explanation:
Solution is a process in chemical weathering and it is the process of dissolving or removing rock in a solution through the activities of acid rain or solution. Chemical weathering refer to the process where rocks interact with chemical solutions or acid rain which can result in His integration.
when applied behavior analysis is used properly what happens???
Answer:
Applied Behavior Analysis therapy (ABA) is a type of intensive therapy that focuses on the principles and techniques of learning theory to help improve social behavior. ABA therapy helps to (1) develop new skills, (2) shape and refine previously learned skills, and (3) decrease socially significant problem behaviors.
Explanation:
The atmospheric features of Neptune are easier to see than those of Uranus because A. Neptune has greater warmth and less haze. B. Neptune has more methane. C. The atmosphere of Uranus rotates differentially. D. Uranus has no significant atmosphere.
Answer:
Option B is the correct answer (Neptune has more methane)Explanation:
From the options given,
The atmospheric features of Neptune are easier to see than those of Uranus because Neptune has more methane
Neptune has small amount of methane and water which gives it blue colour and white patches which distinguish it from uranus
For more information, visit
http://abyss.uoregon.edu/~js/ast121/lectures/lec20.html
A 70 kg stunt pilot begins pulling out of a dive into a vertical circle. If the plane's speed at the lowest point of the circle is 80 m/s, what is the apparent
weight of the pilot at the lowest point of the pullout? The pilot experiences a force of 5g from the centripetal acceleration at the bottom of the dive.
O
3880 N
о
3430 N
0
4116 N
3986 N
Answer:
Apparent weight of pilot due to centripetal acceleration:
m v^2 / R = 5 g m = 5 * 70 * 9.8 = 3430 N
Weight of pilot = 70 * 9.8 = 686 N
Total = 3430 + 686 = 4116 N
A retired bank president can easily read the fine print of the financial page when the newspaper is held no closer than arm's length, 59.1 cm from the eye. What should be the focal length of an eyeglass lens that will allow her to read at the more comfortable distance of
Answer:
Explanation:
comfortable distance is 25 cm .
He must be using convex lens . In that case rays coming from object placed at 25 cm appears to be coming from 59.1 cm due to converging nature of convex lens.
object distance u = -25 cm
image distance v = -59.1 cm
Lens formula
1 / v - 1 /u = 1 /f
-1 / 59.1 + 1 / 25 = 1/f
- .0169 + .04 = 1 / f
.0231 = 1 / f
f = 43.3 m
Please help!!!!!!!!!
Answer:
try S or Q
Explanation:
100 POINTS AND BRAINLIEST
Answer:
b
Explanation:
The materials are most likely used to change the surface area of the ramp. For example, the sand paper can change the ramp's texture. The wax paper can also change the surface area of the ramp by texture, it can be made smoother.
Answer:
B.
Explanation:
Hope it helped!
A ballistic pendulum is a device for measuring bullet speeds. One of the simplest versions consists of a block of wood hanging from two long cords. (Two cords are used so that the bottom face of the block remains parallel to the floor as the block swings upward.) A 0.013-kg bullet traveling at 320 m/s hits a 3.0-kg ballistic pendulum. However, the block is not thick enough for this bullet, and the bullet passes through the block, exiting with one-third of its original speed.
Required:
How high above its original position does the block rise?
Answer:
0.043 m upwards
Explanation:
The mass of the bullet, [tex]$m_1$[/tex] = 0.013 kg
Mass of the ballistic pendulum, [tex]$m_2$[/tex] = 3 kg
Velocity of the bullet, [tex]$v_1$[/tex] = 320 m/s
Therefore, from the law of conservation of momentum, we get
[tex]$m_1v_1+m_2v_2=m_1v_1^1+m_2v_2^1$[/tex]
[tex]$(0.013 \times 320)+(3 \times 0) = \left(0.013 \times \frac{320}{0}\right) + (3 \times v_2^1)$[/tex]
[tex]$3 \times v_2^1=2.774$[/tex]
[tex]$v_2^1=0.92 \ m/s$[/tex]
Therefore the required height to rise the block is given by :
[tex]$(v_2^1)^2-v_2^2=2gh$[/tex]
[tex]$h=\frac{(v_2^1)^2-v_2^2}{2g}$[/tex]
[tex]$h=\frac{(0.92)^2-0}{2(-9.81)}$[/tex]
[tex]$h=-0.043 \ m$[/tex]
Therefore, the block moves upwards for 0.043 meters.
Look at the Position vs. Time and Velocity vs. Time plots. What is the person's velocity when his position is at its maximum value (around 6 m )
Answer:
The person's velocity is zero.
Explanation:
An athlete is performing squats in the weight room. The knee is going from anatomical position to 92 degrees and then back to anatomical position each squat. The athlete performs a total of 10 squats. This is done over a time period of 30 seconds. What is the angular acceleration (rad/sec 2) of the knee
Answer:
α = 0.357 ras / s²
Explanation:
This is a rotational kinematics exercise, it tells us that it performs 10 squats in 30 s, for which it performs one squat at t = 3 s, also indicates that the angle of the squat is θ = 92º
θ = θ₀ + w₀ t + ½ α t²
the athlete starts from rest, whereby w₀ = 0 and the initial angle in the vertical position is zero (θ₀=0)
θ = ½ α t²
α = 2 θ /t²
let's reduce the magnitudes to the SI system
θ = 92º (π rad /180º) = 0.511π rad
let's calculate
α = 2 0.5111π /3²
α = 0.1136π rad / s²
α = 0.357 ras / s²