what is a resultant vector​

Answer: d = st

Explanation:

We know that the distance is equal to the rate (speed) times the time

d = st

Answer:

1. Temperature= 869.35 K

2. Pressure of combustion = 12994.043 kpa

3. Thrust = 127x10⁶N

Explanation:

this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.

1.

The temperature = (273+2400k) - (3800)²/2(4003)

= 2673 - 14440000/8006

= 2673 - 1803.65

= 869.35 K

Approximately 869.4K

2. We first get mach number

= 3800/√1.3(923.8)(869.35)

= 3800/1021.78

= 3.719

Pressure = 100kpa[1+2.07464415]^1.3/0.3

= 12995.043kpa

C. Thrust

Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)

= 12678.621

= 126.781 kN

Thrust is approximately 127kN = 127x10⁶N

Answer:

c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Explanation:

Buoyancy can be defined as a force which is created by the water displaced by an object.

Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.

Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.

The buoyancy of an object is given by the formula;

[tex] Fb = pgV [/tex]

[tex] But, \; V = Ah [/tex]

[tex] Hence, \; Fb = pgAh [/tex]

Where;

Fb = buoyant force of a liquid acting on an object.

g = acceleration due to gravity.

p = density of the liquid.

v = volume of the liquid displaced.

h = height of liquid (water) displaced by an object.

A = surface area of the floating object.

The unit of measurement for buoyancy is Newton (N).

In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.

Answer:

roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge

Explanation:

Answer:a) - 0.4905 m/s²   b) distance = 35.48 m

Explanation:

Given that  

The initial velocity of the skater = 5.90 m/s

 kinetic friction coefficient = 0.0500

final velocity = 0 m/s(since it  comes to rest)

deceleration cause by the kinetic friction = ?

we know that  

F = μN

and N= mg

Therefore;

F = μ m g....................(1)

also  that

F = m a........................(2)

with our common Force, F, equating  (1) and (2), we have that

m a = - μ m g

a = - μ g

a = - 0.05 × 9.81

a = - 0.4905 m/s²

The deceleration cause by the kinetic friction is a = - 0.4905 m/s²

b)

The distance the skater  travels before stopping

is given as

    Vf² = v₀² - 2 a x

  final velocity = 0 m/s(since it  comes to rest)  

Therefore We have that

 0 = v₀² - 2 a x

 x = - v₀² / 2 a

 x = 5.90² / (2 x 0.4905  )

34.81/0.981  

x = 35.48 m

Or

using

v²-u² = 2aS final velocity = 0 m/s(since it  comes to rest)  

0²-5.90² = -2×0.4905×S

34.81=0.981S

S= 34.81/0.981

S=35.48m

Answer:

120 W lightbulb

Explanation:

Let the two lightbulb be A and B respectively.

Given the following data;

Power A = 120W

Power B = 90W

Voltage = 120V

To find the current flowing through each lightbulb;

a. For lightbulb A

Power = current * voltage

120 = current * 120

Current = 120/120

Current = 1 Ampere.

b. For lightbulb B

Current = power/voltage

Current = 90/120

Current = 0.75 Amperes

Therefore, the lightbulb that carries more current is A with 1 Ampere.

The  bulb that  carries more current is :

- A with 1 Ampere.

Let the two lightbulb be A and B respectively.

To find the current flowing through each lightbulb;

Thus, the lightbulb that carries more current is A with 1 Ampere.

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The simplest definition of a black hole is an object that is so dense that not even light can escape its surface. If we squished the Earth's mass into a sphere with a radius of 9 mm, the escape velocity would be the speed of light. Just a wee-bit smaller, and the escape velocity is greater than the speed of light.

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Answer:

.50 M

Explanation:

5*.50=2.5 + 2*.25=.5 = 3n

6*.50= 3N

Final answer is .50M

Air resistance is exerted by an engine

Answer:

By using the most simple velocity equation, velocity = distance / time, meaning the speed would be 247.5 meters per second.

Answer:

The person must be located in the Equator Line. The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

Explanation:

Physically speaking, the centripetal acceleration ([tex]a_{r}[/tex]), measured in meters per square second, experienced by a person is defined by the following expression:

[tex]a_{r} = \omega^{2}\cdot r[/tex] (1)

Where:

[tex]\omega[/tex] - Angular speed of the Earth, measured in radians per second.

[tex]r[/tex] - Distance perpendicular to the rotation axis, measured in meters.

Since rotation axis passes through poles and distance described above is directly proportional to centripetal acceleration. The person must be located in the Equator Line, which is equivalent to the radius of the planet.

In addition, the angular speed of the Earth can be calculated in terms of its period ([tex]T[/tex]), measured in seconds:

[tex]\omega = \frac{2\pi}{T}[/tex] (2)

If we know that [tex]r = 6.371\times 10^{6}\,m[/tex] and [tex]T = 86400\,s[/tex], then the maximum centripetal acceleration experienced by a person is:

[tex]a_{r} = \left(\frac{2\pi}{86400\,s} \right)^{2}\cdot (6.371\times 10^{6}\,m)[/tex]

[tex]a_{r} = 0.0337\,\frac{m}{s^{2}}[/tex]

The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

The person standing still on the surface of the earth must be located in the equator line

Recall: the the centripetal acceleration at the Equator is about 0.03 m/s2.

This then means that the maximum centripetal acceleration of a person standing in the equator line is 0.03 m/s2

The maximum centripetal acceleration as the name implies is the maximum speed of a body or object in a circular path

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Answer:

mgh

Explanation:

Assume the height of the body is 1.8m.

The gravity?of the body is G=mg

the height of the gravity center is about 0.9m

E=mgh

=m*9.8m/s*0.9m

= 8.82mJ

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

          T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]

          T² =  [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g

          g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

           g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2

            g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

             T = t / n

             ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

              ΔT = Δt / n

              ΔT = Δt

now let's look for the uncertainty of g

             Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex]  ΔT

             Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex]   ΔL + 4π²L  (-2  T⁻³) ΔT

a more manageable way is with the relative error

             [tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]

we substitute

              Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation  

let's calculate

               Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])

               Δg = 9.7766 (0.001215 + 0.0184)

               Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

                Δg = 0.2 m / s2

therefore the correct result is

               g ±Δg = (9.8 ± 0.2) m / s²

Answer:

10 mm

Explanation:

We'll begin by calculating the spring constant of the spring. This can be obtained as follow:

Extention (e) = 5 mm

Force (F) = 125 N

Spring constant (K) =?

F = Ke

125 = K × 5

Divide both side by 5

K = 125 / 5

K = 25 N/mm

Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:

Force (F) = 250 N

Spring constant (K) = 25 N/mm

Extention (e) =?

F = Ke

250 = 25 × e

Divide both side by 25

e = 250 / 25

e = 10 mm

Thus, the spring will stretch 10 mm when a 250 N force is applied.

Answer:

POSITIVE CHARGE,  NEGATIVE CHARGE,  ZERO

Explanation:

To solve this completion exercise, we must remember that charges of the same sign repel each other and in a metallic object (frying pan) the charge is mobile.

Let's analyze the situation when we touch the pan, the charges are neutralized, therefore when we bring the pan to the plate that has a positive charge, it attracts the mobile negative charges in the pan, until it is neutralized, therefore on the opposite side of the pan. pan (edge ​​with a glued tape) is left with a positive charge; therefore the edge and the tape, which is very light, have positive charges and repel each other.

We must assume that the frying pan is insulated so that the net charge is zero, since the induction process.

Consequently the words to complete the sentence are

When the pan is on the plate, the edge of the plate has a _POSITIVE CHARGE_____.

This means that the base of the container is loaded NEGATIVE CHARGE_____ because the net charge of the container is ___ZERO_

Answer:

75N

Explanation:

a = v/t = 3/2

F = ma = 50(3/2) = 75

Answer: 0.435 m

Explanation:

Given

mass m=20 kg

initial speed u=2 m/s

coefficient of kinetic friction [tex]\mu_k=0.3[/tex]

deceleration which opposes the motion is given by

[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]

[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]

using [tex]v^2-u^2=2as[/tex]

[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]

Answer:

1. D

Climate is generally defined as the weather condition that prevails in a particular region over a long period of time. Climate is usually measured by examining the pattern of variation in several climatic factors such as rainfall, temperature, relative humidity, wind, pressure, etc. While the weather of a place can change within a space of few hours, it takes years for a change in climatic condition to occur.  

2. d

3. c

4.a.

5. c

6. a.

7. c

Explanation:

The correct answers are (1) d. two of the above (average air temperature and humidity), (2)c. the relative humidity is 100 percent, (3)d. two of the above (water vapor condenses out of the air and air temperature decreases), (4)a. cirrocumulus, (5)c. cumulus, (6)c. steam fog, and (7)c. sleet.

Temperature is a physical quantity that measures the average kinetic energy of the particles in a substance or system. It is a measure of how hot or cold something is, and is typically measured in units such as degrees Celsius or Fahrenheit. Temperature can also be thought of as a measure of the direction in which heat energy flows, with heat energy naturally flowing from areas of higher temperature to areas of lower temperature.

Here in the Question,

1. Weather factors include d. two of the above (average air temperature and annual precipitation are two factors that affect weather, but humidity is also an important factor that can influence the feel of the air).

2. The dew point is the temperature at which b. water vapor starts to condense. When air cools, it can reach a point where it is unable to hold all of its moisture in the form of water vapor. At this point, the water vapor starts to condense into visible droplets, such as dew, and the temperature at which this happens is called the dew point. When the dew point is reached, the relative humidity is at 100 percent.

3. Relative humidity may decrease if d. two of the above (water vapor condenses out of the air and air temperature decreases) occur. If the air cools and reaches the dew point, water vapor will start to condense into droplets, which can reduce the amount of water vapor in the air and lower the relative humidity. Similarly, if the temperature drops without any change in water vapor content, the relative humidity will decrease because colder air can hold less moisture than warmer air.

4. The type of cloud that forms at high altitudes is a. cirrocumulus. These clouds are typically found at altitudes above 18,000 feet and are characterized by small, white, puffy clouds arranged in rows or ripples. They are often a sign of fair weather but can also indicate an approaching storm.

5. The type of cloud that forms when strong air currents carry warm air upward is c. cumulus. Cumulus clouds are large, puffy clouds that can develop vertically, forming a towering cloud with a flat top. They are often associated with thunderstorms and can produce heavy rain, hail, and lightning.

6. The type of fog that forms when cool air moves over a warm lake is c. steam fog. Steam fog, also called evaporation fog or sea smoke, occurs when cold, dry air moves over a warm, moist surface and causes water vapor to rise and condense into fog. This type of fog is often seen over bodies of water during the fall and winter.

7. Rain that passes through a layer of freezing air near the ground becomes c. sleet. Sleet is formed when raindrops fall through a layer of freezing air near the ground and freeze into small ice pellets before hitting the surface. It is different from hail, which forms in strong thunderstorms when updrafts carry raindrops upward into colder air where they freeze and then fall back to the ground, and snow, which forms in clouds when water vapor freezes directly into ice crystals. Glaze is a type of ice that forms when rain falls onto a surface that is below freezing, forming a layer of ice on top of the surface.

Therefore, The correct answers are:1. Weather factors include average air temperature, annual precipitation, and humidity. 2. The dew point is the temperature at which water vapor starts to condense. 3. Relative humidity may decrease if water vapor condenses out of the air or if the air temperature decreases. 4. The type of cloud that forms at high altitudes is cirrocumulus. 5. The type of cloud that forms when strong air currents carry warm air upward is cumulus. 6. The type of fog that forms when cool air moves over a warm lake is steam fog. 7. Rain that passes through a layer of freezing air near the ground becomes sleet, which is different from hail and snow.

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Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

a) The total electric potential is 2282000 V

b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in diagram 1 below is;

[tex]V_P=V_1+V_2[/tex]

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we know that; the Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

[tex]r_1^2=0.015^2+0.0125^2[/tex]

[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]

[tex]r_1 = \sqrt{0.00038125}[/tex]

[tex]r_1 = 0.0195[/tex]

Also

[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]

[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]

[tex]r_2 = \sqrt{0.000549[/tex]

[tex]r_2 = 0.0234[/tex]

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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Answer:

a. Workdone = 44100 Joules

b. Power = 8820 Watts.

Explanation:

Given the following data:

Mass = 225kg

Distance = 20m

Time = 5 seconds

To find the workdone;

Workdone = force * distance

But force = mg

We know that acceleration due to gravity is equal to 9.8m/s²

Force = 225*9.8 = 2205N

Substituting the values into the equation, we have;

Workdone = 2205 * 20

Workdone = 44100 Joules

b. To find the power;

Power = workdone/time

Power = 44100/5

Power = 8820 Watts.

Answer:

 you have to find 4 spring with this elastic constant  k = 316 N / m

Explanation:

In this case for the design of the dispenser the four springs are placed in the four corner at the bottom, therefore we can use the translational equilibrium relationship

                    4 F_e -W = 0

where the elastic force is

                    F_e = k x

we substitute

                   4 kx = mg

                   k = [tex]\frac{mg}{4x}[/tex]

Each tray has a thickness of x = 0.450 cm = 0.450 10⁻² m, this should be the elongation of the spring so that when the tray is in position it will remain fixed.

let's calculate

                  k = [tex]\frac{0.580 \ 9.8}{4 \ 0.450 \ 10^{-2} }[/tex]

                  k = 3.1578 10² N / m

                  k = 316 N / m

therefore you have to find 4 spring with this elastic constant

Answer:

A) 70 x 10^-2

Explanation:

it would be 70 times negative 100 which equals 0.70

Answer:

-the ratio of the speed of light

in air to the speed of light in the substance.

-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.

-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5

Explanation:

Answer:

Numerous studies have shown that the economic costs of divorce fall more heavily on women. After separation, women experience a sharper decline in household income and a greater poverty risk (Smock 1994; Smock and Manning

Answer:

sadness and stress...................

Answer:

Hello your question is incomplete attached below is the complete question

answer :

Total charge enclosed within the sphere : [tex]\frac{q_{r1} }{4\pi e_{0}R^3 } . r[/tex]

Total charge enclosed outside the sphere : [tex]\frac{q}{4\pi e_{0}r^2 } .r[/tex]

Explanation:

Given data:

Total charge of a uniformly charged sphere = Q

radius = R

first step : find the electric field inside and outside the uniformly charged sphere

2nd step : determine the total charge enclosed within and outside the sphere

make a sketch of the uniformly charged sphere

Attached below is a detailed solution

Answer:

[tex]E=30.78\ J[/tex]

Explanation:

The force constant of the spring, k = 76 N/m

The extension in the spring, x = 0.9 m

We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :

[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]

So, 30.78 J of energy is stored in the spring.

Answer:

the impulse must be the same in these two cases    F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

Explanation:

For this exercise we use the relationship between momentum and momentum

         I = Δp

         F t = m v_f - m v₀

To know the speed we use the conservation of energy

starting point. Highest point

       Em₀ = U = m g h

fincla point. Just before the crash

      Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

        m g h = ½ m v²

         v = [tex]\sqrt{2gh}[/tex]

we substitute in the impulse relation

     F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases

Answer:

solar energy

wind power

geothermal energy

hydraulic power

biomass energy

energy storage

(That's all I know).

Answer:

L = 5.08 10⁻¹ m = 50.8 cm

Explanation:

In a double slit experiment for constructive interference is given by

           d sin θ = m λ

let's use trigonometry to find a relationship with the distance

           tan θ = y / L

these experiments are very small angles

           tan θ = [tex]\frac{sin \ \theta}{cos \ \theta}[/tex] = sin θ

when substituting

           sin θ = y / L

substituting in the first equation

           d y / L = m λ

           L = [tex]\frac{d \ y}{m \ \lambda}[/tex]

let's calculate

           L = [tex]\frac{ 0.783 \ 10^{-6} \ 88.2 \ 10^{-2} }{2 \ 680 \ 10^{-9}}[/tex]

           L = 5.08 10⁻¹ m

The distance of viewing screen from the double slits will be L = 5.08 *10⁻¹ m = 50.8 cm

This position, where the resulting wave is larger than either of the two original, is called constructive interference.

In a double slit experiment for constructive interference is given by

d sin θ = m λ

let's use trigonometry to find a relationship with the distance

[tex]tan\theta=\dfrac {y}{L}[/tex]

these experiments are very small angles

tan θ =  = sin θ

when substituting

[tex]sin\theta = \dfrac{y}{l}[/tex]  

substituting in the first equation

[tex]\dfrac{dy}{L}=m\lambda[/tex]

[tex]L=\dfrac{dy}{L\lambda}[/tex]

let's calculate

[tex]L=\frac{0.783\times 10^{-6}\times 88.2\times 10^{-2}}{2.680\times 10^{-9}}[/tex]

L = 5.08 10⁻¹ m

Hence the distance of viewing screen from the double slits will be L = 5.08 10⁻¹ m = 50.8 cm

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