what are the applications of pascal's principle​

Answer:

C if it is a negitive ion it has more electrons because protons determine what element it is

The answer is 3. Observation

Explanation:

The sentence "The popcorn kernels popped twice as fast as the last batch" is the result of observing or measuring the time popcorn kernels require to pop. In this context, the sentence best matches the word "observation" which the term used in the Scientific method to refer to statements that are the result of studying a phenomenon, either through the senses such as sight or through precise instruments that allow scientists to understand numerically variables such as time, speed, temperature, etc.

Answer:

1) θ = 0.00118 rad, 2)  θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4)  I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

          d sin θ = m λ

1) the first order maximum occurs for m = 1

           sin θ = λ  / d

           θ = sin⁻¹ λ  / d

let's reduce the magnitudes to the SI system

           λ  = 520 nm = 520 10⁻⁹  θ = 0.00118 radm

           d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

           θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

            θ = sin⁻¹ (1.18 10⁻³)

            θ = 0.00118 rad

2) the second order maximum occurs for m = 2

            θ = sin⁻¹ (m λ  / d)

            θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

            θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

          I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

             x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

             x = 2.21

            I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

            I / I₀ = cos² (3.0945) [0.363] 2

            I / I₀ = 0.9978 0.1318

            I / I₀ = 0.1738

4) the maximum second intensity is

            I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

            x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

            x = 4.41

            I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

            I / Io = cos² 6.273    0.216

            I / Io = 0.216

.

Answer:

A. 990v/m

B.330x10^-8T

C.2.19x10^-6J/m³

D.1.45x10^-11J

Explanation:

See attached file

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω²     .................1

and kinetic energy is  directly proportional to square of the amplitude.

so

[tex]\frac{KE2}{KE1} = \frac{A2^2}{A1^2}[/tex]      .............2

[tex]\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}[/tex]

[tex]\frac{KE2}{KE1}[/tex] = 0.52284

so factor that bug maximum KE change is 0.52284

The factor does the bug's maximum KE change should be considered as the 0.52284.

Since

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

So here we apply the given formula

KE = (0.5) × m × A² × ω²     .................1

here,

kinetic energy is directly proportional to square of the amplitude.

So,

= 4.7^2/ 6.5^2

= 0.52284

hence, The factor does the bug's maximum KE change should be considered as the 0.52284.

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Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

[tex]a_t = ar[/tex]

where;

a is the angular acceleration and

r is the radius of the circular path

[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]

Determine time of the rotation;

[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]

The magnitude of total linear acceleration is given by;

[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Answer:

Explanation:

C. The temperature of the coffee will decrease while the temperature of the table increases.

The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more-approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms. In common usage, the mass of an object is often referred to as its weight, though these are in fact different concepts and quantities. In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons. Weight and mass are considered to be the same quantities. But many people tend to misuse these terms in their daily conversations. The main difference between weight and mass is that weight is the force of gravity by which the earth attracts towards it whereas mass is the amount of matter in an object.

Answer:

y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Explanation:

The diffraction of a slit is given by the expressions

          a sin θ = m λ

where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.

          sin θ = m λ / a

If this equation

          a ’= 2 a

we substitute

          2 a sin θ'= m λ

          sin θ'= (m λ / a)  1/2

          sin θ ’= sin θ / 2

We can use trigonometry to find the width

         tan θ = y / L

as the angle is small

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L  

we substitute

        y ’/ L = y/L   1/2

        y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Answer:

velocity and acceleration

Answer:

Hey there!

Centripetal (Circular Motion) and Oscillating Motion.

Let me know if this helps :)

Answer:

The power of the mirror in diopters is 58.8 D

Explanation:

Given;

radius of curvature of the spoon, R = 3.4 cm = 0.034 m

The focal length of a mirror is given by;

[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]

The focal length of the mirror is 0.017 m

(b) The power of the mirror is given by;

[tex]P = \frac{1}{f}[/tex]

where;

P is the power of the mirror

f is the focal length

[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]

Thus, the power of the mirror in diopters is 58.8 D

Answer: the same as Lily's

Explanation:

Angular velocity has to do with the speed at which an object will be able to rotate. We are informed that Bob and Lily are riding on a merry-go-round.

Since we are further told that Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform and that he merry-go-round is rotating at a constant angular speed.

Based on the above analysis, Bob's angular speed will be thesame as that of Lily.

Answer:

The number of turns of the inductor is 2000 turns.

Explanation:

Given;

emf of the inductor, E = 0.8 V

the rate of change of current with time, dI/dt = 10 A/s

steady current in the solenoid, I = 0.2 A

flux per turn, Ф = 8.0 μWb per

Determine the inductance of the solenoid, L

E = L(dI/dt)

L = E / (dI/dt)

L = 0.8 / (10)

L = 0.08 H

The inductance of the solenoid is given by;

[tex]L = \frac{\mu_o N^2 A}{l}[/tex]

Also, the magnetic field of the solenoid is given by;

[tex]B = \frac{\mu_o NI}{l}[/tex]

I is 0.2 A

[tex]B = \frac{\mu_oN(0.2)}{l} = \frac{0.2\mu_o N}{l}[/tex]

[tex]\frac{B}{0.2 } = \frac{\mu_o N}{l}[/tex]

[tex]L = \frac{\mu_o N^2 A}{l} \\\\L = \frac{\mu_o N }{l} (NA)\\\\L = \frac{B}{0.2} (NA)\\\\L = \frac{BA}{0.2} (N)[/tex]

But Ф = BA

[tex]L = \frac{\phi N}{0.2} \\\\\phi N = 0.2 L\\\\N = \frac{0.2 L}{\phi} \\\\N = \frac{0.2 *0.08}{8*10^{-6}}\\\\N = 2000 \ turns[/tex]

Therefore, the number of turns of the inductor is 2000 turns.

This question involves the concepts of magnetic flux, magnetic field, and inductance.

The inductor has "2000" turns.

The magnetic field due to an inductor coil is given as follows:

[tex]B=\frac{\mu_o NI}{L}\\\\[/tex]

where,

B = magnetic field

μ₀ = permeability of free space \

N = No. of turns

I = current = 0.2 A

L = length of inductor

Therefore,

[tex]\frac{\mu_oN}{L}=\frac{B}{0.2\ A}---------- eqn(1)[/tex]

Now, the inductance of a solenoid is given by the following formula:

[tex]E = L\frac{dI}{dt}\\\\L = \frac{E}{\frac{dI}{dt}}[/tex]

The inductance of solenoid can also be given using the following formula:

[tex]L = \frac{\mu_o N^2A}{L}[/tex]

comparing both the formulae, we get:

[tex]\frac{E}{\frac{dI}{dt}}= \frac{\mu_oN^2A}{L}\\\\E=\frac{dI}{dt}\frac{\mu_oN}{l}(NA)\\\\using\ eqn (1):\\\\E=\frac{dI}{dt}\frac{B}{0.2}(NA)\\\\[/tex]

where,

BA = magnetic flux = [tex]\phi[/tex] = 8 μWb/turn = 8 x 10⁻⁶ Wb/turn

N = No. of turns = ?

E = E.M.F = 0.8 volts

[tex]\frac{dI}{dt}[/tex] = rate of change in current = 10 A/s

Therefore,

[tex]0.8=(10)\frac{8\ x\ 10^{-6}}{0.2}N\\\\N=\frac{(0.8)(0.2)}{8\ x\ 10^{-5}}[/tex]

N = 2000 turns

Learn more about magnetic flux here:

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The attached picture shows the magnetic flux.

Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.

Explanation:

Answer:

Light amplification by stimulated emission of radiation

Answer:

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Explanation:

When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.

In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:

[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]

Where:

[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.

If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:

[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]

[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]

And change in heat energy ([tex]Q[/tex]), measured by joules, by:

[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]

Where:

[tex]n[/tex] - Molar quantity, measured in moles.

The final temperature of the monoatomic ideal gas is now cleared:

[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]

Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:

[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]

[tex]T_{2} = 420.279\,K[/tex]

The final pressure of the system is calculated by the following relationship:

[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]

If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:

[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]

[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Answer:

     e% = 0.99%   this value is within the 5% tolerance given by the manufacturer

Explanation:

Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by

          e% = | R_nominal - R_measured | / R_nominal 100

where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors

let's apply this formula to our case

R_nominal = 10 kΩ = 10000 Ω

R_measured = 100 99 Ω

        e% = | 10000 - 10099.1 | / 10000 100

        e% = 0.99%

this value is within the 5% tolerance given by the manufacturer

Answer:

     λ = 52.5 nm

Explanation:

De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression

          p = h / λ

where the moment

          p = mv

          λ = h / mv

let's calculate

        λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54

        λ = 5.25 10⁻⁸ m

Let's reduce anm

       λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)

       λ = 52.5 nm

The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.

To find the angle of a resultant vector, the vector must be resolved into y-component and x-component.

The angle of this resultant vector is also known as the direction of the vector.

Mathematically, the direction of a resultant vector is given as;

[tex]\theta = tan^{-1} (\frac{R_y}{R_x} )\\\\where;\\\\\theta \ is \ the \ direction \ of \ the \ resultant \ vetcor\\\\R_y \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ y - direction\\\\R_x \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ x - direction[/tex]

Therefore, the angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.

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Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

[tex]F_{r}=kv^2[/tex]

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]

Put the value into the formula

[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]

Put the value of k

[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]

[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]

At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]

So, [tex]1600-v^2=0[/tex]

[tex]v=\sqrt{1600}[/tex]

[tex]v=40\ ft/sec[/tex]

Hence, The velocity is 40 ft/sec.

Answer:

The voltage is equal to the batteries terminal voltage

Explanation:

Explanation:

Answer:

The number is  [tex]Z = 216 \ fringes[/tex]

Explanation:

From the question we are told that

      The wavelength is  [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]

       The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]

      The distance between the plates (radius of wire ) =  [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]

   Generally the condition for constructive  interference in a film is mathematically represented as

            [tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]

Where  t is the thickness of the separation between the glass i.e  

    t  = 0 at the edge where the glasses are touching each other and  

     t =  2d at the edge where the glasses are separated by the wire  

   m is the order of the fringe it starts from  0, 1 , 2 ...

So  

       [tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]

=>   [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]

=>    

       [tex]m = 215[/tex]

given that we start counting m from zero

   it means that the number of  bright fringes that would appear is

         [tex]Z = m + 1[/tex]

=>    [tex]Z = 215 +1[/tex]

=>     [tex]Z = 216 \ fringes[/tex]

Answer:

The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]

    The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]

    The  number of slit is  N =  1000 slits

    The order of the maxima is  m =  2

Generally the spacing between the slit is mathematically represented as

         [tex]d = \frac{k}{N}[/tex]

substituting values

        [tex]d = \frac{ 0.01}{1000}[/tex]

       [tex]d = 1.0 *10^{-5} \ m[/tex]

Generally the condition for constructive interference is

       [tex]d\ sin(\theta ) = m * \lambda[/tex]

substituting values

      [tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]

       [tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]

      [tex]\theta = 6.315^o[/tex]

Generally the dispersion is mathematically represented as

           [tex]D = \frac{ m }{d cos(\theta )}[/tex]

substituting values

          [tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]

           [tex]D = 2.01220 *10^{5} \ rad/m[/tex]

Answer:

The wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm

d. 198 mm

Explanation:

Refractive index is given by;

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}}[/tex]

where;

[tex]\lambda_{vacuum}[/tex] is the wavelength of the light beam in vacuum

[tex]\lambda_{medium}[/tex] is the wavelength of the beam in a material

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}} \\\\\lambda_{vacuum} = \mu *\lambda _{medium}\\\\\ the \ wavelength \ of \ the \ light \ beam \ is \ constant \ in \ a \ vacuum\\\\ \mu_1 *\lambda _{medium}_1 = \mu_2 *\lambda _{medium}_2\\\\\lambda _{medium}_2 = \frac{ \mu_1 *\lambda _{medium}_1 }{ \mu_2} \\\\\lambda _{medium}_2 =\frac{1.5*330}{2.5} \\\\\lambda _{medium}_2 = 198 \ mm[/tex]

Therefore, the wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm.

d. 198 mm

Answer:

A)  The distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B)  The electrical force of repulsion among them shows a small effect  in beam stability.

Explanation:

Given that:

A Van de Graaff generator produces a beam of 2.02-MeV deuterons

If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.

[tex]\mathtt{K.E = \dfrac{1}{2}mv^2}[/tex]

2 K.E = mv²

[tex]\mathtt{v^2 = \dfrac{2 K.E }{m}}[/tex]

[tex]\mathtt{v =\sqrt{ \dfrac{2 K.E }{m}}}[/tex]

so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.

[tex]\mathtt{v =\sqrt{ \dfrac{2 (2.02 \ MeV) \times \dfrac{10^6 \ eV}{ 1 \ MeV} \times \dfrac{1.60 \times 10^{-19} \ J }{1 \ eV} }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]

[tex]\mathtt{v =\sqrt{ \dfrac{6.464 \times 10^{-13} \ J }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]

v = 13911611.49  m/s

v = 1.39 × 10⁷ m/s

So, If the beam current is 10.0 μA.

We all know that:

[tex]I = \dfrac{q}{t}[/tex]

[tex]t = \dfrac{q}{I}[/tex]

[tex]\mathtt{ t = \dfrac{1.6 * 10 ^{-19} \ C}{10.0 * 10^{-6} \ A}}[/tex]

t = 1.6 × 10⁻¹⁴ s

Finally, the distance of the deuterons from one another  = v × t

the distance of the deuterons from one another  = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)

the distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.

The electrical force of repulsion among them shows a small effect  in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.

Answer:

Ultraviolet light from the sun.

Explanation:

This is an example of radiation.

Answer:

X-Ray

Explanation:

x-Ray is an example of radiation.

Answer: i dont do physics yet lol

Explanation:

Answer:

T = 80√3 N ≈ 139 N

W = 160 N

Explanation:

Sum of forces on B in the x direction:

∑F = ma

80 N sin 60° − T sin 30° = 0

T = 80 N sin 60° / sin 30°

T = 80√3 N

T ≈ 139 N

Sum of forces on B in the y direction:

∑F = ma

80 N cos 60° + T cos 30° − W = 0

W = 80 N cos 60° + T cos 30°

W = 40 N + 120 N

W = 160 N

Answer: no sorry../

Explanation:

Answer:

Parte B : 31.18º , Parte C: 31.17º

Explanation:

Parte B: The angle between the glass and the water before it enters the water is going to be equal to the value of the angle when it enters the glass , 27.13º.

Using the formula n1 senθ1 = n2 senθ2 , where n1=1.51 , θ1=27.13º, n2=1.33 , it gives us θ2=31.18º.

Parte C: n1= 1 , θ1=43.5º, n2=1.33

Using the same formula : n1 senθ1 = n2 senθ2 , it gives us θ2= 31.17º.

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

[tex]p_{f} = p_{i}[/tex]

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.