Answer;
A)S(t)=96t-16t² +432
B)it will take 9 seconds for the ball to reach the ground.
C)864feet
Explanation:
We were given an initial height of 432 feet.
And v(t)= 96-32t
A) we are to Find s(t), the function giving the height of the ball at time t
The position, or heigth, is the integrative of the velocity. So
S(t)= ∫(96-32)dt
S(t)=96t-16t² +K
S(t)=96t-16t² +432
In which the constant of integration K is the initial height, so K= 432
b) we need to know how long will the ball take to reach the ground
This is t when S(t)= 0
S(t)=96t-16t² +432
-16t² +96t +432=0
This is quadratic equation, if you solve using factorization method we have
t= -3 or t= 9
Therefore, , t is the instant of time and it must be a positive value.
So it will take 9 seconds for the ball to reach the ground.
C)V=s/t
Velocity= distance/ time
=96=s/9sec
S=96×9
=864feet
By applying the integrations,
(a) [tex]S = 96t-16t^2+432[/tex]
(b) Time will be "t = 9".
(c) Height will be "576"
Given:
Height,
423 feetInitial velocity,
96 feet/secAccording to the question,
(a)
Integrate v:
[tex]S = 96t-16t^2+C[/tex]Initial Condition,
→ [tex]S = 96t-16t^2+432[/tex]
(b)
Hits the ground when,
S = 0→ [tex]0=96t-16t^2+432[/tex]
→ [tex]t =9[/tex]
(c)
Maximum height when,
v = 0→ [tex]0 = 96-32 t[/tex]
→ [tex]t = 3[/tex]
Now,
→ [tex]S = 96\times 3-16\times 3^2+432[/tex]
[tex]= 576[/tex]
Thus the answer above is correct.
Learn more:
https://brainly.com/question/16105731
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .
Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:
[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]
ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:
[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]
[tex]0.8 = e^{-\frac{88}{R} }[/tex]
[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]
[tex]ln(0.8) = -\frac{88}{R}[/tex]
[tex]R = -\frac{88}{ln(0.8)}[/tex]
R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.
Why is it advised not to hold the thermometer by its bulb while reading it?
The fact that Voyager 10 continues to speed out of the solar system, even though its rockets have no fuel, is an example of Group of answer choices Newton's third law of motion. Newton's second law of motion. Newton's first law of motion. the universal law of gravitation. none of these
Answer:
The universal law of gravitation.
PE = m * G M / R^2 potential energy of mass m due to attractive forces
If the kinetic energy of mass m is greater than the energy due to the attractive masses then then mass m can continue indefinitely away from the attracting masses.
g A solenoid 63.5 cm long has 960 turns and a radius of 2.77 cm. If it carries a current of 2.28 A, find the magnetic field along the axis at its center.Find the magnetic field on the solenoidal axis at the end of the solenoid.
Answer:
The value is [tex]B = 0.0043 \ T[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 63.5 = 0.635 \ m[/tex]
The number of turns is [tex]N = 960 \ turns[/tex]
The current is [tex]I = 2.28 \ A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \mu _o * n * I[/tex]
Where n is the number of turn per unit length which is mathematically evaluated as
[tex]n = \frac{N}{l}[/tex]
[tex]n = \frac{960}{0.635}[/tex]
[tex]n = 1512 \ turns /m[/tex]
and [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = 4\pi * 10^{-7} * 1512 * 2.28[/tex]
[tex]B = 0.0043 \ T[/tex]
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses. But he loses them while travelling. Fortunately he has his old pair as a spare. (a) If the lenses of the old pair have a power of 2.25 diopters, what is his near point (measured from the eye) when wearing the old glasses, if they rest 2.0 cm in front of the eye
Answer:
30.93 cm
Explanation:
Given that:
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses
The power of the old pair of lens p = 2.25 diopters
The focal point length = 1/p
The focal point length = 1/2.25
The focal point length = 0.444 m
The focal point length = 44.4 cm
The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens
The near point of the person from the glass = 83 cm
Let consider s' to be the image on the same sides of the lens,
∴ s' = -83 cm
We known that:
the focal length of a mirror image 1/f =1/u +1/v
Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.
Then:
1/f = 1/s + 1/s'
1/s = 1/f - 1/s'
1/s = (s' -f)/fs'
s = fs'/(s'-f)
s =( 44.4× -83)/(-83 - 44.4)
s = - 3685.2 / - 127.4
s = 28.93 cm
Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm
= 30.93 cm
how many electrons do calcium have in their outer shell
Answer:
Calcium has two electrons in its outer shell.
Explanation:
Calcium is defined as a metal due to its physical and chemical traits. The two outer electrons are very reactive. Calcium has a valence of 2.
An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The electron reaches a closest distance of 15 cm before being repelled.
What is the plate's surface charge density?
Answer:
2.29e-9C/m²
Explanation:
Using E = σ/ε₀ means the force on the electron is F = eE = eσ/ε₀.
The work done on the electron is W = Fd = deσ/ε₀. This equals the kinetic energy lost, ½mv².
½mv² = deσ/ε₀
d = 75cm – 15cm = 60cm = 0.6m
σ = mv²ε₀/(2de)
. .= 9.11e-31 * (7.4e6)² * 8.85e-12 / (2 * 0.6 * 1.6e-19)
. .= 2.29e-9 C/m² (i.e. 2.29x10^-9 C/m²)
What is the impedance of an AC series circuit that is constructed of a 10.0-W resistor along with 12.0 W inductive reactance and 7.0 W capacitive reactance
Answer:
11.2 Ω
Explanation:
The impedance of a circuit is given by;
Z= √R^2 +(XL-XC)^2
Since
Resistance R= 10 Ω
Inductive reactance XL= 12 Ω
Capacitive reactance XC= 7 Ω
Z= √10^2 + (12-7)^2
Z= √100 + 25
Z= √125
Z= 11.2 Ω
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
Equal currents of magnitude I travel into the page in wire M and out of the page in wire N. The direction of the magnetic field at point P which is at the same distance from both wires is
Answer:
The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy and pauses (velocity = 0) for a fraction of a second at the very top before heading down the other side.
a) Draw a sankey diagram for a roller coaster's climb.
A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy and pauses for a fraction of a second at the very top before heading down the other side. At the top of the hill total, the kinetic energy of the roller coaster would be zero as the velocity is zero at the top of the hill, therefore the total mechanical energy is only because of potential energy.
What is mechanical energy?Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total stored energy in the system which is represented by total potential energy.
The expression for total mechanical energy is as follows
ME= KE+PE
As total mechanical energy is the sum of all the kinetic as well as potential energy stored in the system.As given in the problem a roller coaster uses 800000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy which means 300000 J of energy is lost in the frictional energy while climbing the hill,
Thus at the top of the hill, the total energy of the roller coasters is only due to the potential energy.
Learn more about mechanical energy from here brainly.com/question/12319302
#SPJ2
A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80.
(a) What is the magnetic field in the core?
(b) What part of the magnetic field is due to atomic currents?
Answer:
A) 0.0267 T
B) 0.0263 T
Explanation:
Given that
The number of turns, N = 400
Radius of the wire, r = 6 cm = 0.06 m
Current in the wire, I = 0.25 A
Relative permeability, K(m) = 80
See the attached picture for the calculation
what effect does decreasing the field current below its nominal value have on the speed versus voltage characteristic of a separately excited dc motor
Answer
The effect is that it Decreases the field current IF and increases slope K1
A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?
Answer:
Distance= 4 miles
Time = 36.3 seconds
Explanation:
80 mph = 178.95 m/s
90 mph = 201.32 m/s
V = u +at
201.32= 0+a(4.5)
201.32/4.5= a
44.738 m/s² = a
Acceleration of the cop car
= 44.738 m/s²
Distance traveled at 4.5seconds
For the cop car
S= ut + ½at²
S= 0(4.5) + ½*44.738*4.5
S= 100.66 meters
Distance traveled at 4.5seconds
For the speeding car
4.5*178.95=805.275
The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal
X/178.95= (704.675+x)/201.32
X-0.89x= 626.37
0.11x= 626.37
X= 5694.3 meters
The time = 5694.3/178.95
Time =31.8 seconds
So the distance they meet
= 5694.3+805.275
= 6499.575 meters
= 4.0 miles
The Time = 4.5+31.8
Time = 36.3 seconds
Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the source, the frequency of the sound will be:________.
a. one-fourth as great.
b. half as great.
c. twice as great.
d. unchanged.
Answer:
d. unchanged.
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
A scientist is carrying out an experiment to determine the index of refraction for a partially reflective material. To do this, he aims a narrow beam of light at a sample of this material, which has a smooth surface. He then varies the angle of incidence. (The incident beam is traveling through air.)
The light that gets reflected by the sample is completely polarized when the angle of incidence is 46.5°.
(a) What index of refraction describes the material?
n =
(b) If some of the incident light (at θi = 46.5°) enters the material and travels below the surface, what is the angle of refraction (in degrees)?
Answer:
a) 1.05
b) 43.6°
Explanation:
a) The index refraction that describes the material can be found using Brewster's law:
[tex] \theta_{1} = arctan(\frac{n_{2}}{n_{1}}) [/tex]
where:
n₁ is the refractive index of the initial medium through which the light propagates (air) = 1
n₂ is the index of the material=?
θ₁ = 46.5°
[tex] n_{2} = n_{1}tan(\theta_{1}) = tan(46.5) = 1.05 [/tex]
Hence, the material's index refraction is 1.05.
b) The angle of refraction can be found as follows:
[tex] n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2}) [/tex]
[tex]sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{sin(46.5)}{1.05} = 0.69[/tex]
[tex] \theta_{2} = arcsin(0.69) = 43.6^{\circ} [/tex]
Therefore, the angle of refraction is 43.6°.
I hope it helps you!
A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Answer:
The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
Explanation:
The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.
When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.
So the centripetal acceleration, a' = 16v²/R.
To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16
a'/a = 16
a' = 16a.
So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
During the data transmission there are chances that the data bits in the frame might get corrupted. This will require the sender to re-transmit the frame and hence it will increase the re-transmission overhead. By considering the scenarios given below, you have to choose whether the packets should be encapsulated in a single frame or multiple frames in order to minimize the re-transmission overhead.
Justify your answer with one valid reason for both the scenarios given below.
Scenario A: Suppose you are using a network which is very prone to errors.
Scenario B: Suppose you are using a network with high reliability and accuracy.
1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.
2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.
Justification:
There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors. The reliable and accurate network environment makes a single frame economically better.
Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.
With encapsulation, each layer:
provides a service to the layer above itcommunicates with a corresponding receiving nodeThus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead. This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.
Learn more about encapsulation of packets here: https://brainly.com/question/22471914
In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit
Answer:
Impedance increases for frequencies below resonance and decreases for the frequencies above resonance
Explanation:
See attached file
Explanation:
You are performing an experiment that requires the highest-possible magnetic energy density in the interior of a very long current-carrying solenoid. Which of the following adjustments increases the energy density?a. Increasing only the length of the solenold while keeping the turns per unit lengh flxed. b. Increasing the number of turns per unit length on the solenold. c. Increasing the cross-sectional area of the solenoid. d. None of these. e. Increasing the current in the solenoid.
Answer:
The correct choice is B & E.
Explanation:
A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.
Cheers!
cyclist always bends when moving the direction opposite to the wind. Give reasons
a. Describe the relationship between the number of batteries and the voltage and explain what you think might be happening
Answer:
Their is a direct relationship between the number of batteries and the increase in power. The voltage is the product of the number of batteries and the voltage which is 9 volts. As the batteries touch ends the voltages of all three combines.
Explanation:
Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m
Answer:
0.03/π m/min
Explanation:
See attached file pls
There is a river in front of you that flows due South at 3.0m/s. You launch a toy boat across the river with the front of the boat pointed due East. When you tested the boat on a still pond, the boat moved at 4.0m/s. Now as it moves to the opposite bank, it travels at some speed relative to you, sitting in your chair. What is this speed
Answer:
5.0 m/sExplanation:
If the river moves towards the south at 3m/s and the both moves towards the east at 4.0m/s, the speed of the boat relative to me will be the resulting displacement of both velocities of the river and that of the boat. This can be gotten using pythagoras theorem.
Let Vr be the relative speed. According to the theorem;
[tex]V_r^2 = V_s^2 + V_e^2\\\\V_r^2 = 3.0^2 + 4.0^2\\\\V_r^2 = 9+16\\\\V_r^2 = \sqrt{25}\\ \\V_r = 5.0m/s[/tex]
Hence this relative speed is 5.0 m/s
An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is vi and second
time he increases the initial sped to 4v. How would you compare the maximum height in the second trial to that in the first trial?
Answer:
The maximum height reached in the second trial is 16times the maximum height reached in the first trial.
Explanation:
The following data were obtained from the question:
First trial
Initial speed (u) = v
Final speed (v) = 0
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Next, we shall obtain the expression for the maximum height reached in each case.
This is illustrated below:
First trial:
Initial speed (u) = v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₁) =.?
v² = u² – 2gh₁ (going against gravity)
0 = (v)² – 2 × 9.8 × h₁
0 = v² – 19.6 × h₁
Rearrange
19.6 × h₁ = v²
Divide both side by 19.6
h₁ = v²/19.6
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₂) =.?
v² = u² – 2gh₂ (going against gravity)
0 = (4v)² – 2 × 9.8 × h₂
0 = 16v² – 19.6 × h₂
Rearrange
19.6 × h₂ =16v²
Divide both side by 19.6
h₂ = 16v²/19.6
Now, we shall determine the ratio of the maximum height reached in the second trial to that of the first trial.
This is illustrated below:
Second trial:
h₂ = 16v²/19.6
First trial:
h₁ = v²/19.6
Second trial : First trial
h₂ : h₁
h₂ / h₁ = 16v²/19.6 ÷ v²/19.6
h₂ / h₁ = 16v²/19.6 × 19.6/v²
h₂ / h₁ = 16
h₂ = 16 × h₁
From the above illustrations, we can see that the maximum height reached in the second trial is 16times the maximum height reached in the first trial.
A single-slit diffraction pattern is formed on a distant screen. Assume the angles involved are small. Part A By what factor will the width of the central bright spot on the screen change if the wavelength is doubled
Answer:
If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
Explanation:
For a single-slit diffraction, diffraction patterns are found at angles θ for which
w sinθ = mλ
where w is the width
λ is wavelength
m is an integer, m = 1,2,3, ....
From the equation, w sinθ = mλ
For the first case, where nothing was changed
w₁ = mλ₁ / sinθ
Now, If the wavelength is doubled, that is, λ₂ = 2λ₁
The equation becomes
w₂ = mλ₂ / sinθ
Then, w₂ = m(2λ₁) / sinθ
w₂ = 2(mλ₁) / sinθ
Recall that, w₁ = mλ₁ / sinθ
Therefore, w₂ = 2w₁
Hence, If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up from chest height to full arm extension, a distance of 0.53 m .
Part A: How much work does the weightlifter do to lift the barbell one time?
Part B: If the weightlifter does 23 repetitions a day, what total energy does she expend on lifting, assuming a typical efficiency for energy use by the body?
Part C: How many 500 Calorie donuts can she eat a day to supply that energy?
Answer:
A) Workdone = 223.57 N-m
B) 22357 J of energy
C) Number of donuts = 10.7 donuts
Explanation:
A) The work done is calculated from the formula;. Work done = Force × Distance
We are given;
Mass; m = 43 kg
Distance = 0.53 m
Force(weight) = mg = 43 × 9.81
Thus;
Work done = 43 × 9.81 × 0.53
Workdone = 223.57 N-m
B) We are told she does 23 repetitions a day.
Thus, we assume 23% efficiency.
So, Work = Energy
Thus;
At 100% efficiency;
Energy = (223.57/100%) × 23 repetitions = 5142.11 J
Now, since she is only 23% efficient, she will expend; 5142.11/0.23 J = 22357 J of energy to do 5390 J of work.
C) from conversions; 4.18 J = 1 calorie
Thus;
22357 J ÷ 4.18 J/cal = 5348.565 calories
We how many 500 calorie donuts she can eat in a day to supply that energy.
Thus;
Number of donuts = 5348.565 cal ÷ 500 cal /donut
Number of donuts = 10.7 donuts
When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.9 nm for the same double slit?
Answer:
The number of interference fringes is [tex]n = 3[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 433 \ nm = 433 *10^{-9} \ m[/tex]
The distance of separation is [tex]d = 6 \mu m = 6 *10^{-6} \ m[/tex]
The order of maxima is m = 5
The condition for constructive interference is
[tex]d sin \theta = n \lambda[/tex]
=> [tex]\theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ][/tex]
=> [tex]\theta = 21.16^o[/tex]
So at
[tex]\lambda_1 = 632.9 nm = 632.9*10^{-9} \ m[/tex]
[tex]6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}[/tex]
=> [tex]n = 3[/tex]