Answer:
endothermic
Explanation:
An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.
Explain why it takes much more effort to stop a freight train compared with a car?
Answer:
Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.
A measurement was made of the magnetic field due to a tornado, and the result was 13.00 nT to the north. The measurement was made at a position 8.90 km west of the tornado. What was the magnitude (in A) and direction of the current in the funnel of the tornado? Assume the vortex was a long, straight wire carrying a current.
Answer:
4
Explanation:
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop
Answer:
The correct solution is "122.2211".
Explanation:
Given:
deceleration,
a = 22 ft/sec²
Initial velocity,
[tex]V_i=50 \ m/h[/tex]
Now,
[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]
[tex]=73.333 \ ft/sec[/tex]
Now,
Final velocity,
[tex]V_f=0[/tex]
Initial velocity,
[tex]V_{initial} = 73.333 \ ft/sec[/tex]
hence,
⇒ [tex]V_f^2=V_i^2+2aD[/tex]
By putting the values, we get
[tex]0=(73.333)^2+2\times( -22) D[/tex]
[tex]44D=(73.333)^2[/tex]
[tex]D=\frac{(73.333)^2}{44}[/tex]
[tex]=122.2211[/tex]
The index of refraction for a vacuum is 1.00000. The index of refraction for air is 1.00029. 1) Determine the ratio of time required for light to travel through 1000 m of air to the time required for light to travel through 1000 m of vacuum. (Express your answer to six significant figures.)
Answer:
[tex]\frac{t_{air}}{t_{vaccum}}[/tex] = 1.00029
Explanation:
The refractive index is defined
n = c / v
v = c / n
the speed of light per se wave is constant, so we can use the relations of uniform motion
v = x / t
t = x / v
we substitute
t = x n / c
let's calculate the time
vacuum
t₁ = 1000 1/3 10⁸
t₁ = 3.333333 10⁻⁶ s
air
t₂ = 1000 1.00029 / 3 10⁸
t2 = 3.3343 10⁻⁶ s
the relationship between these times is
t₂ / t₁ = 3.3343 / 3.3333333
t₂ / t₁ = 1.00029
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
C
Explanation:
Similarity
A frictionless spring with a 9-kg mass can be held stretched 1.8 meters beyond its natural length by a force of 80 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after tt seconds. meters
Answer:
the required solution is; x(t) = 0.675sin( 2.222t )
Explanation:
Given the data in the question;
Using both Newton's and Hooke's law;
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5
given that mass m = 9 kg
[tex]x[/tex] = 1.8 m
k is F / x
hence
k = F / x
given that, F = 80 N
we substitute
k = 80 / 1.8
k = 44.44
so
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,
we input
9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,
[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0
so auxiliary equation is,
r² + 4.9377 = 0
r² = -4.9377
r = √-4.9377
r = ±2.222i
hence, the solution will be;
x(t) = A×cos( 2.222t ) + B×sin( 2.222t )
⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )
using initial conditions
x(0) = 0
⇒ 0 = A
[tex]x^t[/tex](t) = 1.5
1.5 = 2.222B
so
B = 1.5 / 2.222 = 0.675
Hence, the required solution is; x(t) = 0.675sin( 2.222t )
A police car travels towards a stationary observer at a speed of 15m/s. the siren on the car emits a sound of frequency 250Hz. Calculate the observer frequency. the speed of sound is 340m/s
Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)
= 250 x (340/( 340-15))
= 261.54 Hz
Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field
Answer:
Option (2) is correct.
The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Explanation:
An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.
So, the correct option is (2).
The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.
The given problem is based on the concept and fundamentals of electromagnetic waves. The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.
In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.
Learn more about the electromagnetic waves here:
https://brainly.com/question/25559554
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?
Answer:
The correct answer is - low pitch
Explanation:
Now for the case it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
if frequency increases then pitch will be increase as well as pitch depends on frequency.
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.
As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
pha của dao động làm hàm
Answer:
pha của dao động là hàm bậc nhất của thời gian.
describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please
Answer:
Force A newton Law first law
F = M.A which Force in 0 N as you Questions Above
Force B
Newton Law 3
Action = -Reaction
Hope you can explain this formula as you want to scribe to explaining
Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel
across a 30 V potential difference to a light bulb.
a. Calculate the current delivered through the light bulb in the two cases.
b. Draw the circuit connection that will achieve the brightest light bulb.
Explanation:
Given that,
Two resistors 4.5 Ω and 2.3 Ω .
Potential difference = 30 V
When they are in series, the current through each resistor remains the same. First find the equivalent resistance.
R' = 4.5 + 2.3
= 6.8 Ω
Current,
[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]
So, the current through both lightbulb is the same i.e. 4.41 A.
When they are in parallel, the current divides.
Current flowing in 4.5 resistor,
[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]
Current flowing in 2.3 ohm resistor,
[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]
In parallel combination, are brighter than bulbs in series.
The reason why a teacher is more important then a farmer
Answer:
A teacher is more important than a famer.
Explanation:
A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.
Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?
a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features
Answer:
c). Easier setup
Explanation:
As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.
How does an airpump work?
An initially motionless test car is accelerated uniformly to 105 km/h in 8.43 s before striking a simulated deer. The car is in contact with the faux fawn for 0.635 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
acceleration before collision:
3.45
m/s2
What is the magnitude of the average acceleration of the car during the collision?
average acceleration during collision:
19.68
m/s2
What is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?
105 km/h ≈ 29.2 m/s
60.0 km/h ≈ 16.7 m/s
Before the collision the test car has an acceleration a of
a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²
During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is
a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²
i.e. with magnitude about 19.7 m/s².
Overall, the car has an average acceleration of
a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²
A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.
1. How many revolutions do the kids make before the constant operational speed is reached ?
2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.
3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?
Answer:
we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer
What is the maximum wavelength, in nm, of light that can eject an electron from a metal with Φ =4.50 x 10–19 J?
[tex]4.4×10^{-7}\:\text{m}[/tex]
Explanation:
The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:
[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]
or
[tex]\dfrac{hc}{\lambda} = \phi[/tex]
Solving for the wavelength [tex]\lambda[/tex],
[tex]\lambda = \dfrac{hc}{\phi}[/tex]
[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]
[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]
Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.
The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.
To find the wavelength, the given values are,
Energy (E) = 4.50×10¯¹⁹ J
What is wavelength?The distance between two consecutive crests and troughs is called the wavelength of a wave.
Here, for the wavelength,
Energy (E) = 4.50×10¯¹⁹ J
Planck's constant (h) = 6.626×10¯³⁴ Js
Speed of light (v) = 3×10⁸ m/s
The wavelength of the light can be obtained as illustrated below:
E = hv / λ
Cross multiply λ,
E × λ = hv
Divide both sides by E,
λ = hv / E
Substituting all the values,
λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹
λ = 0.000000441733 m
λ = 441.73nm
λ - The maximum wavelength of light.
Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm
Learn more about wavelength,
https://brainly.com/question/13047641
#SPJ2
A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.
Answer:
[tex]T=3*10^-3 N/m[/tex]
Explanation:
From the question we are told that:
Radius :
[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]
Work [tex]W=1.5 * 10^{-4}[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=TdA[/tex]
Where
[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]
[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]
[tex]dA=0.050m^2[/tex]
Therefore
[tex]W=TdA[/tex]
[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]
[tex]T=3*10^-3 N/m[/tex]
A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram
Answer:
b
Explanation:
Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil.
a. The first person will gain more velocity as a result of recoll.
b. The second person will gain more velocity as a result of recoll.
c. Both people will gain the same velocity as a result of recoll.
d. The velocity of both people will be zero as a result of recoil
Answer:
The first person will gain more velocity as a result of recoil.
Explanation:
Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''
Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.
Work-Energy Theorem & Power
A 0.5 kg mass sitting on smooth ice is accelerated from rest by a force until is
acquires a speed of 8 m/s. The force acts while the mass moves through a
displacement of 2 m.
A. Calculate the kinetic energy of the mass after the force acts.
B. Calculate the work done by the force.
C. Calculate the magnitude of the force that accelerated the mass.
Answer:
A. 16 J
B. 16 J
C. 8 N
Explanation:
A. Determination of the kinetic energy.
Mass (m) = 0.5 Kg
Velocity (v) =. 8 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 8²
KE = ½ × 0.5 × 64
KE = 0.5 × 32
KE = 16 J
B. Determination of the Workdone by the force.
Kinetic energy (KE) = 16 J
Workdone =.?
Workdone and kinetic energy has the same unit of measurement. Thus,
Workdone = kinetic energy
Workdone = 16 J
C. Determination of the force.
Workdone (Wd) = 16 J
Displacement (s) = 2 m
Force (F) =?
Wd = F × s
16 = F × 2
Divide both side by 2
F = 16 / 2
F = 8 N
Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
electrons in vacuum.
Answer:
thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.
The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.
Answer:
A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J
Explanation:
In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.
In this case let's start by finding the equivalent capacitance.
A) Let's solve the part where C1 and C3 are. These two capacitors are in serious
[tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex] (you has an mistake in the formula)
[tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]
[tex]\frac{1}{C_{eq1}}[/tex] = 0.1 10⁶
[tex]C_{eq1}[/tex] = 10 10⁻⁶ F
capacitors C₂, C₄ and C₅ are in series
[tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]
[tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]
[tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶
[tex]C_{eq2}[/tex] = 5 10⁻⁶ F
the two equivalent capacitors are in parallel therefore
C_{eq} = C_{eq1} + C_{eq2}
C_{eq} = (10 + 5) 10⁻⁶
C_{eq} = 15 10⁻⁶ F
B) the energy stored in C₃
The charge on the parallel voltage is constant
is the sum of the charge on each branch
Q = C_{eq} V
Q = 15 10⁻⁶ 6
Q = 90 10⁻⁶ C
the charge on each branch is
Q₁ = Ceq1 V
Q₁ = 10 10⁻⁶ 6
Q₁ = 60 10⁻⁶ C
Q₂ = C_{eq2} V
Q₂ = 5 10⁻⁶ 6
Q₂ = 30 10⁻⁶ C
now let's analyze the load on each branch
Branch C₁ and C₃
In series combination the charge is constant Q = Q₁ = Q₃
U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]
U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]
U₃ = 3 J
In Branch C₂, C₄, C₅
since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅
U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]
U₄ = 0.5 J
An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m below. Calculate the velocity (in m/s) of the fish just before it hits the water. (Assume that the eagle is flying in the x direction and that the y direction is up.)
Answer:
Explanation:
The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .
v² = u² + 2 g H .
v² = 0 + 2 x 9.8 x 4.7 m
= 92.12
v = 9.6 m /s
The fish's final velocity will have two components
vertical component = 9.6 m/s downwards
Horizontal component = 2.6 m /s .
Resultant velocity = √ ( 9.6² + 2.6² )
= √ ( 92.16 + 6.76 )
= 9.9 m /s
Answer:
The speed of fish at the time of hitting the surface is 9.95 m/s.
Explanation:
Horizontal speed, u = 2.6 m/s
height, h = 4.7 m
Let the vertical velocity at the time of hitting to water is v.
Use third equation of motion
[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]
The net velocity with which the fish strikes to the water is
[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]
An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16
Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed of the CD is equal to 14.73 rad/s.
Explanation:
Given that,
Angular displacement, [tex]\theta=12\ rad[/tex]
Final angular speed, [tex]\omega_f=17\ rad/s[/tex]
The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]
We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,
[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]
Put all the values,
[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]
So, the initial angular speed of the CD is equal to 14.73 rad/s.
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. A
Complete Question
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2
Answer:
[tex]h=1614m[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]
Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]
Density [tex]\rho=1.20kg/m^2[/tex]
Generally the equation for Height climbed is mathematically given by
[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]
[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]
[tex]h=1614m[/tex]
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge
Answer:
I think it is of science is it true na i knew it bro dont take tension
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]
d = 68.5 x 10⁻⁶ m = 68.5 μm
