Answer:
The change in the internal energy of the first system is 300 J
The second system will do zero work in order to have the same internal energy.
Explanation:
Given;
heat added to the first system, Q₁ = 500 J
heat added to the second system, Q₂ = 300 J
work done by the first system, W₁ = 200 J
The change in the internal energy of the system is given by the first law of thermodynamics;
ΔU = Q - W
where;
ΔU is the change in internal energy of the system
The change in the internal energy of the first system is calculated as;
ΔU₁ = Q₁ - W₁
ΔU₁ = 500 J - 200 J
ΔU₁ = = 300 J
The work done by the second system to have the same internal energy with the first.
ΔU₁ = Q₂ - W₂
W₂ = Q₂ - ΔU₁
W₂ = 300 J - 300 J
W₂ = 0
The second system will do zero work in order to have the same internal energy.
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
A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 34.0 m/s and the ball is 0.310 m from the elbow joint, what is the angular velocity (in rad/s) of the forearm
Answer:
[tex]\omega=109.67\ rad/s[/tex]
Explanation:
Given that,
The speed of the ball, u = 34 m/s
The ball is 0.310 m from the elbow joint.
We need to find the angular velocity (in rad/s) of the forearm.
We know that,
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]
So, the required angular velocity of the forearm is 109.67 rad/s.
A uniform circular disk has a radius of 34 cm and a mass of 350 g. Its center is at the origin. Then a circular hole of radius 6.8 cm is cut out of it. The center of the hole is a distance 10.2 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.
Answer:
u can ask it to the person who give ot to u i dont no
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.
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.
Answer:
The energy stored in a body due to either it's position or change in shape is called gravitational potential energy.
Question 1 of 10
Which nucleus completes the following equation?
239UHe+?
A. 228 Th
B. 2220
c. 23. Pu
D. 78Th
SUBMIT
Answer:
Option D. ²²²₉₀Th
Explanation:
Let the unknown be ⁿₘZ. Thus, the equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
Next, we shall determine n, m and Z. This can be obtained as follow:
For n:
226 = 4 + n
Collect like terms
226 – 4 = n
222 = n
n = 222
For m:
92 = 2 + m
Collect like terms
92 – 2 = m
90 = m
m = 90
For Z:
ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th
Therefore, the complete equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th
Thus, the unknown is ²²²₉₀Th
The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated in the patellar tendon is 400 N, what is the resulting angular acceleration, in rad/s2), if the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm
Answer:
the resulting angular acceleration is 15.65 rad/s²
Explanation:
Given the data in the question;
force generated in the patellar tendon F = 400 N
patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.
so Torque produced by the knee will be;
T = F × d⊥
T = 400 N × 0.03 m × sin( 20° )
T = 400 N × 0.03 m × 0.342
T = 4.104 N.m
Now, we determine the moment of inertia of the knee
I = mk²
given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )
we substitute
I = 4.2 kg × ( 0.25 m )²
I = 4.2 kg × 0.0626 m²
I = 0.2625 kg.m²
So from the relation of Moment of inertia, Torque and angular acceleration;
T = I∝
we make angular acceleration ∝, subject of the formula
∝ = T / I
we substitute
∝ = 4.104 / 0.2625
∝ = 15.65 rad/s²
Therefore, the resulting angular acceleration is 15.65 rad/s²
A 215 N sign is supported by two ropes. One rope pulls up and to the right 1=29.5∘ above the horizontal with a tension 1 , and the other rope pulls up and to the left 2=44.5∘ above the horizontal with a tension 2 , as shown in the figure. Find the tensions 1 and 2 .
The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:
∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0
(right is positive, left is negative)
∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0
(up is positive, down is negative)
Solve the system of equations. I use elimination here:
• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):
sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0
cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0
T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0
T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)
• Subtract the first equation from the second to eliminate T₁ :
T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
• Solve for T₂ :
T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
T₂ sin(74.0°) = (215 N) cos(29.5°)
… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))
T₂ = (215 N) cos(29.5°) / sin(74.0°)
T₂ ≈ 195 N
• Solve for T₁ :
T₁ cos(29.5°) - T₂ cos(44.5°) = 0
T₁ cos(29.5°) = T₂ cos(44.5°)
T₁ = T₂ cos(44.5°) / cos(29.5°)
T₁ ≈ 160. N
A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________
a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.
Answer:
b. approaches zero.
Explanation:
The phenomenon is known as length contraction.
Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.
let the length of the train = L
Let the length observed when the train is in motion = L₀
Apply Einstein's special theory of relativity;
[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]
from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)
As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L). (L₀ < L)
Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero. (L₀ = 0)
Thus, the length of this object, as measured by a stationary observer approaches zero
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
A boy at a football field practice kicked a 0.500-kg ball with a force of 100N. How fast will the ball move after reaching a distance of 7m?
Answer:
v(7) = 52.915 m/s
Explanation:
First, find the value for acceleration.
F = ma
100 = .5 * a
a = 200 m/s²
Next find the velocity at x = 7 using kinematic equations.
v² = v₀² + 2a(Δx)
v² = (0)² + 2(200)(7)
v = [tex]\sqrt{2800}[/tex]
v = 52.915 m/s
A bullet is fired vertically upward a velocity of 80m/s to what height will the bullet rise above the point of projection
Answer:
The bullet will rise 320 meters above the point of projection.
Explanation:
Assuming that air friction is negligent we can use the kinematic equation:
[tex]v_{2} ^2=v_{1} ^2+2(-a)d\\0\frac{m^2}{s^2} =6400\frac{m^2}{s^2} +2(-10\frac{m}{s^2} )d\\-6400\frac{m^2}{s^2} =(-20\frac{m}{s^2}) d\\320m=d[/tex]
*acceleration is negative (-a) as it is acting in the opposite direction of the motion of the bullet.*
The bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.
Assume the air friction is negligible, the kinematic equation:
[tex]v_f^2 = v_i^2 +2(-a) d[/tex]
Where,
[tex]v_i^2[/tex] - iinitial velocity = 80 m/s
[tex]v_f^2[/tex]- final velocity = 0
[tex]d[/tex]- distance= ?
[tex]a[/tex]- gravitational acceleration = 9.8 m/s² = 10 m/s²
Put the values in the formula,
[tex]\begin {aligne} 0 = (80)^2 + 2 (10)^2 d\\\\d = \dfrac {6400}{ 200}\\\\d &= 3600 \rm \ m\end {aligne}\\[/tex]
Therefore, the bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.
To know more about kinematic equation:
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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Answer:
The ball moves from lowest to highest point:
W = M g h = 3 * 9.8 * 4 = 118 J
This is work done "against" gravity so work done by gravity is -118 J
The tension of the string does no work because the tension does not
move thru any distance W = T * x = 0 because the length of the string is fixed.
When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency. The string's tension and mass per unit length remain unchanged.
If the unfingered length of the string is l=65cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12
x1=
x2=
x3=
x4=
x5=
x6=
Answer:
Explanation:
For frequencies n generated in a string , the expression is as follows
n = 1 /2L√ ( T/m )
n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.
If T and m are constant , then
n x L = constant , hence n is inversely proportional to L or length of string.
Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times
Initial length of string is 65 cm .
x1 = 65 x 1 / 1.75 = 37.14 cm
x2 = 37.14 x 1/ 1.75 = 21.22 cm
x3 = 21.22 x 1 / 1.75 = 12.12 cm
x4= 12.12 x 1 / 1.75 = 6.92 cm
x5 = 6.92 x 1 / 1.75 = 3.95 cm
x6 = 3.95 x 1 / 1.75 = 2.25 cm
A motorcycle of mass 160 kg accelerates from rest to 53 m/s in 9 seconds. Ignore air resistance. Assuming there's no slipping between the wheels and the pavement of the road.
Required:
a. What is the average horizontal component of the force that the road exerts on the wheels (total force on all two wheels, not the force on one wheel)?
b. How far does the motorcycle travel in 9 seconds?
c. In the point-particle analysis of this situation, what is the work done by this force?
d. For the real system, how much work is done by the force of the road on the wheels?
Answer:
a) [tex]F=940.8N[/tex]
b) [tex]S=234.14m[/tex]
c) [tex]W=2.2*10^5J[/tex]
d) [tex]W=0[/tex]
Explanation:
Mass [tex]m=160kg[/tex]
Velocity [tex]v=53m/s[/tex]
Time [tex]t=9seconds[/tex]
a)
Generally the Newton's equation for motion is mathematically given by
[tex]a=\frac{v}{t}[/tex]
[tex]a=\frac{53}{9}[/tex]
[tex]a=5.9m/s^2[/tex]
Therefore
F=ma
[tex]F=160*5.88[/tex]
[tex]F=940.8N[/tex]
b)
Generally the Newton's equation for motion is mathematically given by
[tex]S=0.5at^2[/tex]
[tex]S=0.5*5.9*9^2[/tex]
[tex]S=234.14m[/tex]
c)
Generally the Newton's equation for work done is mathematically given by
[tex]W=Fd[/tex]
[tex]W=940.8*238.14[/tex]
[tex]W=2.2*10^5J[/tex]
d)
Generally the Newton's equation for work done by the force of the road on the wheels is mathematically given by
[tex]W=Fdcos\theta[/tex]
[tex]W=0[/tex]
when we jump on a concrete surface,the feet get injured.Why
Answer:
Explanation:
Bhjb
Explanation:
its because a concrete surface is a hard surface which doesn't absorb the energy of gravitation when we fall down so we get hurt more badly..
hope this helps
A certain heating element is made out of Nichrome wire and used with the standard voltage source of V=120 V. Immediately after the voltage is turned on, the current running through the element is measured at I1=1.28 A and its temperature at T1=25°C. As the heating element warms up and reaches its steady-state (operating) temperature, the current becomes I2=1.229 A.
Required:
Find this steady-state temperature T2.
Answer:
T₁ = 232.5 ºC
Explanation:
For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law
V = i R
R = V / i
i₀ = 1.28 A
R₀ = 120 / 1.28
R₀ = 93.75 ohm
i₁ = 1.229 A
R₁ = 120 / 1.229
R₁ = 97.64 or
Resistance in a metal is linear with temperature
ΔR = α R₀ ΔT
where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹
ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]
ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]
ΔT = 2,075 10² C
ΔT = T₁-T₀ = 2,075 10²
T₁ = T₀ + 207.5
T₁ = 25+ 207.5
T₁ = 232.5 ºC
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
* 1a Average speed
Carl Lewis runs the 100 m sprint in about 10 s.
His average speed in units of m/s would be:
of
Answer:
Explanation:
[tex] \implies v_{av} = \dfrac{total \: displacement}{total \: time} [/tex]
[tex] \implies v_{av} = \dfrac{100}{10} [/tex]
[tex]\implies v_{av} =10 \: {ms}^{ - 1} [/tex]
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.
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.
Some copper wire has a resistance of 200 ohms at 20 degrees C . A current is then passed through the same wire and the temperature rises to 90 degrees C. Determine the resistance of the wire at 90 degrees correct to the nearest ohm assuming the coefficient of resistance is 0.004/degree C at 0 degrees
Answer:
256 ohms
Explanation:
Applying,
R = R'[1+α(T-T')]............. Equation 1
Where R = Final resistance of the wire, R' = Initial resistance of the wire, T = Final temperature, T' = Initial temperature, α = Temperature coefficient of resistance
From the question,
Given: R' = 200 ohms, T = 90 degrees, T' = 20 degrees, α = 0.004/degree
Substitute these values into equation 1
R = 200[1+0.004(90-20)]
R = 200[1+0.28]
R = 200(1.28)
R = 256 ohms
The resistance of the wire at 90 °C correct to the nearest ohm assuming the coefficient of resistance is 0.004 °C¯¹ is 256 ohm
Data obtained from the question Original resistance (R₁) = 200 ohmOriginal temperature (T₁) = 20 °C Coefficient of resistivity (α) = 0.004 °C¯¹New temperature (T₂) = 90 °C New resistance (R₂) =? How to determine the new resistanceα = R₂ – R₁ / R₁(T₂ – T₁)
0.004 = R₂ – 200 / 200(90 – 20)
0.004 = R₂ – 200 / 200(70)
0.004 = R₂ – 200 / 14000
Cross multiply
R₂ – 200 = 0.004 × 14000
R₂ – 200 = 56
Collect like terms
R₂ = 56 + 200
R₂ = 256 ohm
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The average 8-18 year old spends how many hours per day average in front of a screen doing little physical activity
Nearly four hours every day, doing little to no physical activity.
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
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:
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If four students separately measure the density of a rock, and they all have very low percent
differences between their measurements, what can you say for certain about the accuracy of their
results?
Answer:
Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.
Explanation:
Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.
Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.
Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.
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
