Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.83 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (Hint: The angle to the fourth dark band and the angle to the first bright band are small enough that tan θ ≈ sin θ.)

Answers

Answer 1

Answer:

a)1984.5nm

b)523mm

Explanation:

A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference

θ=(m+0.5)λ where m= 1,2.3....

Where given from the question the 4th dark Fringe which will take place at m= 3

θ=7/2y

Where y= 567nm

= 7/2(567)=1984.5nm

But

B)tan θ ≈ y/d

And sinθ = mλ/d

y=mλd when m= 1 which is the first bright we have

Then y=(1× 567.D)/d

But the distance from Central to the 4th dark Fringe is 1.83cm then

y= 7λD/2d= 1.83cm

D/d=(2)×(1.83×10^-2)/(7×567×10^-9)

=92221.5

y= (567×10^-9)× (92221.5)

=0.00523m

Therefore, the distance between the first and center is y1-y0= 523mm


Related Questions

Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is

Answers

Answer:

the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle

An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .

Answers

Answer:

364days

Explanation:

Pls see attached file

Explanation:

The moon will take 112.7 days to make one revolution around the planet.

What is Kepler's third law?

The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.

Given, the distance from the moon's center to the planet's surface,

h = 2.165 × 10⁵ km,

The radius of the planet, r = 4175 km  

The mass of the planet = 6.70 × 10²² kg

The total distance between the moon's center to the planet's center:

a = r +h = 2.165 × 10⁵ + 4175

a = 216500 + 4175

a = 220675

a = 2.26750 × 10⁸ m

The period of the planet can be calculated as:

[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]

[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]

T = 9738253.26 s

T = 112.7 days

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Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?
A. Increasing the number of lines per length.
B. Decreasing the number of lines per length.
C. Increasing the distance to the screen.
D. Increasing the wavelength of the laser.

Answers

Answer:

Answer:

A. Increasing the number of lines per length.

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?

Answers

Answer:

a

    [tex]z= 2.5 \ m[/tex]

b

   [tex]z = (1 \ m , 4 \ m )[/tex]

Explanation:

From the question we are told that

     Their distance apart is  [tex]d = 5.00 \ m[/tex]

      The  wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              [tex]z - (d-z) = m \lambda[/tex]

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        [tex]z - (5-z) = 0[/tex]

=>     [tex]2 z - 5 = 0[/tex]

=>     [tex]z= 2.5 \ m[/tex]

Generally the path difference for destructive  interference is

           [tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]

=>         [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]

=>        [tex]|2z - d| =\frac{\lambda}{2}[/tex]

substituting values

          [tex]|2z - 5| =\frac{6}{2}[/tex]

=>      [tex]z = \frac{5 \pm 3}{2}[/tex]

So  

      [tex]z = \frac{5 + 3}{2}[/tex]

      [tex]z = 4\ m[/tex]

and

      [tex]z = \frac{ 5 -3 }{2}[/tex]

=>   [tex]z = 1 \ m[/tex]

=>    [tex]z = (1 \ m , 4 \ m )[/tex]

For a beam of light in air (n = 1) reflecting off glass (n = 1.5), what is Brewster's angle to the nearest degree?

Answers

Answer: 56°

Explanation:

Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.

The formula is;

[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]

[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]

= 56.30993247

= 56°

A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscillating electric field

Answers

Answer:

The amplitude of the oscillating electric field is 1316.96 N/C

Explanation:

Given;

frequency of the wave, f = 2.4 Hz

intensity of the wave, I = 2300 W/m²

Amplitude of oscillating magnetic field is given by;

[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

I is intensity of wave

c is speed of light = 3 x 10⁸ m/s

[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]

The amplitude of the oscillating electric field is given by;

E₀ = cB₀

E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶

E₀ = 1316.96 N/C

Therefore, the amplitude of the oscillating electric field is 1316.96 N/C

Suppose a 58-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.150 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.10 T? magnitude V direction ---Select--- †\

Answers

Answer:

95.7v

Explanation

Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

E= Ndစ/dt

E= N BA-0/ deta t

Given that

N = 58turns

B = 1.10T

A = 0.150m^²

Deta t= 0.1s

now we have

E = 58(1.10x0.150)/0.1

= 95.7v

Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE

Explanation:

please help !!!!!!!!!!

Answers

Answer:

Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.

Explanation:

As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.

What did the results of photoelectric-effect experiments establish?

Answers

Answer:

Option A

Electrons are emitted if low intensity, high-frequency light hits a metal surface.

Explanation:

From the experiments conducted to study the photoelectric effect, conclusions were made that the key factor that contributes to the emission of electrons from the surface of the metal is the frequency of the beam of light. This frequency has to be beyond a minimum threshold, if not, there will be no emission of electrons from the metal surface no matter the intensity of the beam of light or the length of time it is incident upon the metal surface.

This makes option A correct because it highlights the contributions made by the threshold frequency to the photoelectric effect.

within which type of system is the total mass conserved but not the total energy

Answers

In a closed system the mass is conserved, but the energy is not conserved.

To find the answer, we have to study about different systems in thermodynamics.

What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.

Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.

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Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine

Answers

It depends on what stage of the mission you're talking about.

==>  While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.

==>  When the engines ignite, their thrust (d) is greater than the force of gravity.  So the net force on the rocket is upward, and the spacecraft accelerates upward.

==>  After the engines shut down, the net force acting on the rocket is due to Gravity (c).

. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.  

. . . If it has enough horizontal speed, it enters Earth orbit.  

. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.    

A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.

What unbalanced force bills for the course of the internet pressure of the rocket?

A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.

What's the net pressure of unbalanced?

If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.

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Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is1000 kg/m-3. What is the water pressure on the upper pipe (II).

Answers

Answer:

The water pressure on the upper pipe is 92.5 kPa.

Explanation:

Given that,

Pressure in lower pipe= 120 kPa

Speed of water in lower pipe= 1 m/s

Acceleration due to gravity = 10 m/s²

Density of water = 1000 kg/m³

Radius of lower pipe = 12 m

Radius of uppes pipe = 6 m

Height of upper pipe = 2 m

We need to calculate the velocity in upper pipe

Using continuity equation

[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]

[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]

[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]

Put the value into the formula

[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]

[tex]v_{2}=4\ m/s[/tex]

We need to calculate the water pressure on the upper pipe

Using bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]

Put the value into the formula

[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]

[tex]120500=P_{2}+28000[/tex]

[tex]P_{2}=120500-28000[/tex]

[tex]P_{2}=92500\ Pa[/tex]

[tex]P_{2}=92.5\ kPa[/tex]

Hence, The water pressure on the upper pipe is 92.5 kPa.

A 1300-turn coil of wire 2.40 cm in diameter is in a magnetic field that increases from 0 T to 0.120 T in 9.00 ms . The axis of the coil is parallel to the field. What is the emf of the coil?

Answers

Answer:

The induced emf in the coil is 7.843 V

Explanation:

Given;

number of turns of the coil, N = 1300 turn

diameter of the coil, d = 2.4 cm = 0.024 m

initial magnetic field, B₁ = 0 T

final magnetic field, B₂ = 0.12 T

change in time, dt = 9.0 ms = 9 x 10⁻³ s

Area of the coil is given by;

A = πr²

radius of the coil, r = 0.024 / 2

radius of the coil, r = 0.012 m

A = π(0.012)²

A = 4.525 x 10⁻⁴ m²

The induced emf in the coil is given by;

E = NA(dB/dt)

E =  NA [(B₂ - B₁) /dt]

E = 1300 x 4.525 x 10⁻⁴ (0.12 - 0) / (9 x 10⁻³)

E = 7.843 V

Therefore, the induced emf in the coil is 7.843 V

Just wondering if I did this right

Answers

Yeah

All they are all correct

You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination

Answers

Answer:

4 / 7

Explanation:

1/total resistance = 1/1 + 1/2 + 1/4

= 1¾

total resistance = 1 ÷ 1¾

= 4/7

Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The resultant vector + is given by

Answers

Answer:

The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

[tex]\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)[/tex]

[tex]\vec A = 5.196\,i + 3\,j[/tex]

4-meter vector with direction: 30º east of north.

[tex]\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)[/tex]

[tex]\vec B = 2\,i + 3.464\,j[/tex]

The resultant vector is obtaining by sum of components:

[tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex]

The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].

When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?

Answers

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?

Answers

Answer:

The separation of the two slits is 0.456 mm.

Explanation:

Given the wavelength of light = 519 nm

The indifference pattern = 4.6 m

Adjacent bright fringes = 5.2 mm

In the interference, the equation required is Y = mLR/d

Here, d sin theta = mL

L = wavelgnth

For bright bands, m is the  order = 1,2,3,4  

For dark bands,  m = 1.5, 2.5, 3.5, 4.5

R = Distance from slit to screen (The indifference pattern)

Y = Distance from central spot to the nth  order fringe or fringe width

Thus,  here d = mLR/Y

d = 1× 519nm × 4.6 / 5.2mm

d = 0.459 mm

A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.

Answers

Answer:

Explanation:

When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .

Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .

hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .

An electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Which of the following are also the same for the two particles?
(A) speed
(B) kinetic energy
(C) frequency
(D) momentum

Answers

Explanation:

The De-Broglie wavelength is given by :

[tex]\lambda=\dfrac{h}{p}[/tex]

h is Planck's constant

p is momentum

In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.

Only momentum is the factor that remains same for both particles i.e. momentum.

Kasek rides his bicycle down a 6.0° hill (incline is
6° with the horizontal) at a steady speed of 4.0
m/s. Assuming a total mass of 75 kg (bicycle and
Kasek), what must be Kasek's power output to
climb the same hill at the same speed? ​

Answers

Answer:

 P = 2923.89 W  

Explanation:

Power is

     P = F v

for which we must calculate the force, let's use Newton's second law, let's set a coordinate system with a flat parallel axis and the other axis (y) perpendicular to the plane

X Axis  

         F - Wₓ = 0

         F = Wₓ

Y Axis

         N -  [tex]W_{y}[/tex] = 0

let's use trigonometry for the components of the weight

         sin 6 = Wₓ / W

         cos 6 = W_{y} / W

         Wₓ = W sin 6

         W_{y} = W cos 6

          F = mg cos 6

          F = 75 9.8 cos 6

          F = 730.97 N

let's calculate the power

        P = F v

        P = 730.97 4.0

        P = 2923.89 W

What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?​

Answers

Answer:

If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.

The simple reason is the desirability for genetic variation using recessive genes.

In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.

Hope this helps

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

Answers

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?

Answers

Answer:

a

   [tex]k = 11600000 N/m[/tex]

b

   [tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]

c

  [tex]F = 3750.28 \ N[/tex]  

Explanation:

From the question we are told that

    The Young modulus is  [tex]E = 1.4 *10^{10} \ N/m^2[/tex]

     The length is  [tex]L = 0.35 \ m[/tex]

      The  area is  [tex]2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2[/tex]

   

Generally the force acting on the tibia is mathematically represented as

       [tex]F = \frac{E * A * \Delta L }{L}[/tex]    derived from young modulus equation

Now this force can also be mathematically represented as

      [tex]F = k * \Delta L[/tex]    

So

     [tex]k = \frac{E * A }{L}[/tex]

substituting values

     [tex]k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}[/tex]

     [tex]k = 11600000 N/m[/tex]

    Since the tibia support half the weight then the force experienced by the tibia is  

        [tex]F_k = \frac{750 }{2} = 375 \ N[/tex]

 From the above equation the extension (compression) is mathematically represented as

          [tex]\Delta L = \frac{ F_k * L }{ A * E }[/tex]        

substituting values

           [tex]\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }[/tex]

           [tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]

From the above equation the maximum force is  

        [tex]F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}[/tex]  

         [tex]F = 3750.28 \ N[/tex]  



48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?

Answers

Answer:

the interation blood veins

Explanation:

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?

Answers

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]

   

How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

Answers

Answer:

[tex]y = 0.0394 \ m[/tex]

Explanation:

From the question we are told that

        The  distance of the screen is  [tex]D = 2.20 \ m[/tex]

       The distance of separation of the slit is  [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]

        The  wavelength of light is  [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]

Generally the condition for constructive interference is

            [tex]dsin\theta = n * \lambda[/tex]

=>        [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]

here n = 1 because we are considering the central diffraction peak

=>        [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]

=>       [tex]\theta = 1.0274 ^o[/tex]

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           [tex]y = D tan (\theta )[/tex]

substituting values

        [tex]y = 2.20 * tan (1.0274)[/tex]

        [tex]y = 0.0394 \ m[/tex]

You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 31° from the edge of the building with an initial velocity of 15 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?

Answers

Answer:

69.58 m tall

Explanation:

Pls see attached file

Consider two parallel wires where the magnitude of the left currentis 2 I0(io) and that of the right current is I0(io). Point A is midway between the wires,and B is an equal distance on the other side of the wires.
The ratio ofthe magnitude of the magnetic field at point A to that at point Bis________

Answers

Answer:

Explanation:

At the point midway between wires

magnetic field due to wire having current 2I₀

= 10⁻⁷ x 2 x2I₀ / r     where 2r is the distance between wires .

magnetic field due to wire having current I₀

= 10⁻⁷ x 4 I₀ / r

magnetic field due to wire having current I₀

= 10⁻⁷ x 2I₀ / r    

= 10⁻⁷ x 2 I₀ / r     where 2r is the distance between wires .

these fields are in opposite direction as direction of current is same in both .

net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r

= 2 x 10⁻⁷ x  I₀ / r

At point A net magnetic field = 2 x 10⁻⁷ x  I₀ / r

At point B , we shall calculate magnetic field

magnetic field due to nearer wire having current  2 I₀ = 10⁻⁷ x 4 I₀ / r

magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r

These magnetic fields act in the same direction so they will add up

net magnetic field = [ (4 I₀ / r)  + (2 I₀ / 3r) ] x 10⁻⁷

= (14 I₀ / 3r ) x 10⁻⁷

Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷

Ratio of field at A and B

= 3 / 7 . Ans

The ratio of the magnitude of the magnetic field at point A to point B is :

3 / 7

Given data :

Magnitude of the left current is  2I₀

Magnitude of the right current is  I₀

First step : Determine the magnetic field at point A  

The magnetic field due to the left current ( 2I₀ )

10⁻⁷ * 2 * 2I₀ / r       ( 2r = distance between wires )

The magnetic field due to the right current ( I₀ )

10⁻⁷ * 2 I₀ / r

From the expressions above the magnetic fields are in  opposite direction

∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r =   2 * 10⁻⁷ *  I₀ / r

Hence The magnetic field at point A = 2 * 10⁻⁷ *  I₀ / r

Next step : determine the magnetic field at point B

Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r

Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r

Since the fields acts in the same directions

The net magnetic field =  (4 I₀ / r)  + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷

Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷

Therefore the ratio of the magnitude of the magnetic field at point A to point B  =  3/ 7

Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B  = 3 / 7

Learn more : https://brainly.com/question/22403676

From a hot air balloon 2 km​ high, a person looks east and sees one town with angle of depression of 16 degrees. He then looks west to see another town with angle of depression of 84 degrees. What is the distance between the two towns?

Answers

Answer:

7km

Explanation:

The angle of depression is congruent to the angle of elevation and can be explained as angle below horizontal in which the person observing an object must view for him/her to view object's that are lower than him/her.

In angle of depression, there is assumption that object is closer to the person observing it, so there is parallel horizontal for both observing and object been observed.

hot air balloon 2 km​ high,

there exist two triangles

From trigonometry

Tanx= opposite/adjacent

Opp= 2km

Adj= X1

first triangle have base length of

Tan(16)=2/X1

X1=2/ tan(16)

X1=6.97

For Second triangle

Tanx= opposite/adjacent

Opp= 2km

Adj= X2

the other with a base length of

X2=2/tan(84)

X2=0.21

Therefore,, the total distance between the two towns is

x1+x2=6.97+0.21=7.18km

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