Given 1ft = 12in, how many feet are in 36 inches?

Answer:

633nm

Explanation:

Given the following :

Number of lines per centimeter(N) = 7000

Angle θ = 62.4°

Order (n) = 2

If grating element = d

Wavelength (λ) = (d* SinΘ) / 2

If number of lines = 7000 per cm

Converting to metre :

100 cm = 1m

7000 lines per 1 cm

Number of lines per m:

7000 lines * 100 = 700,000 lines per meter

Recall :

d = reciprocal of N

d = 1 / 700,000

d = 0.00000142857

Substituting into (λ) = (d* SinΘ) / 2

λ = (0.00000142857 * Sin 62.4°) / 2

λ = 0.00000126600 / 2

λ = 0.000000633002

λ = 0.000000633

λ = 633 × 10^-9 m = 633nm

Answer:

Explanation:

If the mass of the probe is 500kg, its weight W = mass  acceleration due to gravity.

Weight of the probe = 500*9.81

Weight of the probe = 4,905N

If its average density =  1400kg/m³

Volume = Mass/Density

Volume = 500/1400

Volume = 0.3571m³

According to the floatation principle, the volume of the probe is equal to the volume of liquid displaced. Hence the volume of water displaced is 0.357m³.

Since density of water is 1000kg/m³, we can find the mass of the water using the formula;

Mass of water = Density of water * Volume of water

Mass of water = 1000*0.3571

Mass of water = 357.1kg

Weight of water displaced = 3571 * 9.81 = 3503.15N

The tension in the cable will be the difference between the weight of the probe and weight of the displaced fluid.

Tension in the cable = 4,905N -  3503.15N

Tension in the cable = 1,401.85N

Hence the tension in the cable is 1,401.85N

Answer and Explanation: When tossing a ball up in the air, the forces acting on the ball are due to Gravity, which is defined by gravitational acceleration on that location on Earth (approximately 9.8 m/s²) multiplied by mass of the ball; Force of thrown, i.e., the force you threw the ball and air resistance force, which is proportional to the square of the ball's through the air and the ball's cross section area. To facilite calculations, air resistance force is normally ignored.

Answer:

weight and drag

Explanation:

Answer:

a) A = 4.0 m , b)   w = 3.0 rad / s , c)  f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

          x = A cos (wt + fi)

In the exercise we are told that the expression is

          x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

         A = 4.0 m

b) the frequency or angular velocity

         w = 3.0 rad / s

c) angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 3 / 2π

           f = 0.477 Hz

d) the period

frequency and period are related

           T = 1 / f

           T = 1 / 0.477

           T = 20.94 s

e) the phase constant

          Ф = 0.10 rad

f) velocity is defined by

          v = dx / dt

         v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

         v = A w

          v = 4 3

          v = 12 m / s

g) the angular velocity is

          w² = k / m

          k = m w²

          k = 1.2 3²

          k = 10.8 N / m

h) the total energy of the oscillator is

          Em = ½ k A²

           Em = ½ 10.8 4²

          Em = 43.2 J

i) the potential energy is

           Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

               x = 3.98 m

j) kinetic energy

           K = ½ m v²

for t = 00.1 ²

    v = A w sin 0.10

    v = 4 3 sin 0.10

    v = 1.98 m / s

Answer:

[tex] \boxed{\sf Angle \: of \: refraction \: (r) = {sin}^{ - 1} ( \frac{1}{2.8} )} [/tex]

Given:

Refractive index of air ( [tex] \sf \mu_{air} [/tex] )= 1

Refractive index of glass slab ( [tex] \sf \mu_{glass} [/tex]) = 1.40

Angle of incidence (i) = 30.0°

To Find:

Angle of refraction (r)

Explanation:

From Snell's Law:

[tex] \boxed{ \bold{ \sf \mu_{air}sin \ i = \mu_{glass}sin \: r}}[/tex]

[tex] \sf \implies 1 \times sin \: 30 ^ \circ = 1.4sin \:r[/tex]

[tex] \sf sin \:30^ \circ = \frac{1}{2} : [/tex]

[tex] \sf \implies \frac{1}{2} = 1.4 sin \: r[/tex]

[tex] \sf \frac{1}{2} = 1.4 sin \: r \: is \: equivalent \: to \: 1.4 sin \: r = \frac{1}{2} : [/tex]

[tex] \sf \implies 1.4 sin \: r = \frac{1}{2} [/tex]

Dividing both sides by 1.4:

[tex] \sf \implies \frac{\cancel{1.4} sin \: r}{\cancel{1.4}} = \frac{1}{2 \times 1.4} [/tex]

[tex] \sf \implies sin \: r = \frac{1}{2 \times 1.4} [/tex]

[tex] \sf \implies sin \: r = \frac{1}{2.8} [/tex]

[tex] \sf \implies r = {sin}^{ - 1} ( \frac{1}{2.8} )[/tex]

[tex] \therefore[/tex]

[tex] \sf Angle \: of \: refraction \: (r) = {sin}^{ - 1} ( \frac{1}{2.8} )[/tex]

Answer:

The work done to lift the counterweight equals the potential energy acquired

Explanation:

since this is vertically applied force on the counterweight, and the distance the force is displacing the counterweight is in the same direction as the applied force, it equals the gained potential energy

Answer:

The velocity is  [tex]v_x = 0.356 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass is  [tex]m = 1.10 \ kg[/tex]

   The  spring constant is  [tex]k = 18 \ N/m[/tex]

   The  speed is [tex]v = 40 \ cm / s = 0.4 m/s[/tex]

    The  position considered is  x =  0.45 A  

Here A is the amplitude which is mathematically represented as

     [tex]A = v * \sqrt{\frac{m}{k} }[/tex]

=>     [tex]A = 0.4 * \sqrt{\frac{1.10}{18 } }[/tex]

=>     [tex]A = 0.0989 \ m[/tex]

So     [tex]x = 0.45 * 0.0989[/tex]

=>     [tex]x = 0.045 \ m[/tex]

Generally the speed at  x  is mathematically represented as

      [tex]v_x = \sqrt{ \frac{k}{m} * [A^2 - x^2 ]}[/tex]

=>    [tex]v_x = \sqrt{ \frac{18}{ 1.10} * [0.0989^2 - 0.045^2 ]}[/tex]

=>    [tex]v_x = 0.356 \ m/s[/tex]

Answer:

The angular separation is  [tex]k = 0.8594^o[/tex]

Explanation:

From the question we are told that

   The  wavelength of the light is [tex]\lambda = 600 \ nm = 600*10^{-9} \ m[/tex]

   The  distance of separation between the slit is  [tex]d = 0.080 \ mm = 0.080 *10^{-3} \ m[/tex]

    The distance from the screen is

Generally the condition for  constructive interference is mathematically represented as

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

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

    here [tex]\theta[/tex] is the angular separation between the central maxima and one side of the first order maxima

given that we are considering the first order of maxima n =  1  

        =>   [tex]\theta = sin ^{-1} [ \frac{1 * 600*10^{-9} }{ 2.0 } ][/tex]

        =>    [tex]\theta = sin ^{-1} [ 0.0075 ][/tex]

        =>   [tex]\theta = 0.4297^o[/tex]

So the angular separation of the two first order maxima  is  

     [tex]k = 2 * \theta[/tex]

     [tex]k = 2 * 0.4297[/tex]

      [tex]k = 0.8594^o[/tex]

This question is incomplete, the complete question is;

When an auditorium has a solid wall, sound waves will tend to perfectly reflect off the wall (i.e. with a 180o phase change). If listening to music, as from an orchestra, the incoming and reflected waves will interfere with each other. For a listener sitting 0.5 m from the wall, what is the lowest frequency which gets suppressed by this interference? Use vsound=330 m/s.

Answer: f = 165 Hz

the lowest frequency which gets suppressed by this interference is 165 Hz

Explanation:

For a reflected wave (out of phase), the path difference between the incoming and reflected wave should be equal to the half integral multiple of wavelength.

r₂ - r₁ = ( m + 1/2) λ/2

r₂ is the distance from the source to observer via reflection

r₁ is distance from source to observer

here r₂ would travel an additional distance of 0.5 m due to reflection that straight approaching wave.

Therefor to have minimum/lowest possible frequency, we say m = 0

we substitute

0.5 = ( 0 + 1/2 ) λ/2

λ = 2m

The frequency would be

f = Vsound / λ

f = 330 / 2

f = 165 Hz

Therefore the lowest frequency which gets suppressed by this interference is 165 Hz

Answer:

(a) 5.0 kg

(b) 10 kg

Explanation:

Draw a free body diagram for each block.  There are 4 forces on block A:

Weight force mAg pulling down,

Normal force N pushing up,

Tension force T pulling right,

and friction force Nμ pushing left.

There are 2 forces on block B:

Weight force mBg pulling down,

and tension force T pulling up.

Whether the system is just starting to move, or moving at constant speed, the acceleration is 0.

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = 0

mBg = T

Sum of forces on A in the +y direction:

∑F = ma

N − mAg = 0

N = mAg

Sum of forces on A in the +x direction:

∑F = ma

T − Nμ = 0

T = Nμ

Substitute:

mBg = mAg μ

mA = mB / μ

(a) When the system is just starting to move, μ = 0.40.

mA = 2.0 kg / 0.40

mA = 5.0 kg

(b) When the system is moving at constant speed, μ = 0.20.

mA = 2.0 kg / 0.20

mA = 10 kg

m_1=5kg

m=10kg

From the question we are told

the coefficient of static friction between mass mA  and the table is 0.40, where as the coefficient of kinetic friction  is 0.20.

a)  

Generally the equation for the Tension  is mathematically given as

T=mg

Where

[tex]m_1g=m_2g[/tex]

Therefore

[tex]m_1=\frac{2.0}{0.4}\\\\m_1=5kg[/tex]

b

Generally the equation for the Tension  is mathematically given as

[tex]T=f\\\\T=u_km_1g\\\\\m_1=\frac{m_2}{u}\\\\m_1=\frac{2}{0.2}[/tex]

m=10kg

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Answer:

F = 2009.64 N

Explanation:

It is given that,

Mass of a baseball, m = 0.145 kg

Initial speed if the baseball, u = 33 m/s

It hit on a horizontal line drive straight back at the pitcher at 46.0 m/s, final velocity, v = -46 m/s

Time of contact between the bat and the ball is [tex]t=5.7\times 10^{-3}\ s[/tex]

We need to find the magnitude of the force exerted by the ball on the bat. It can be calculated using impulse-momentum theorem. So,

[tex]Ft=m(v-u)\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.145\times (-46-33)}{5.7\times 10^{-3}}\\\\F=-2009.64\ N[/tex]

So, the magnitude of force exerted on the ball by the bat is 2009.64 N.

Answer:

a. [tex]3.04\times 10^{-8}m[/tex]

b. = 40 km/ million years

Explanation:

The computation is shown below;

According to the question, Data provided in the question

Average speed = 4.0 cm / per year

Distance move in 1 second at this speed

Based on the above information

a. For distance move in 1 second is

As we know that

[tex]d_1 = v_g \times t\\\\ = 4\ cm \times \frac{1}{100\times 365.25\times 3,600} \times 1\s\\\\= 3.04\times 10^{-8}m[/tex]

b. For speed per kilometer per million years is

[tex]v_1 = 4\times \frac{10^6}{10^5} \\\\[/tex]

= 40 km/ million years

Answer:

d) y(x,t) = (10.5 mm)cos(172rad?m?1 x ?2730rad?s?1t)

Answer:B

Explanation: I just did it on a p e x

273 K gives the object's temperature in the SI unit therefore the correct answer is option B

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

The International System of Units, sometimes known as the SI system of units, is the most frequently used and acknowledged system of units in use nowadays. There are three additional units and 7 SI basic units in this system of SI units.

The three supplemental SI units are radian, steradian, and becquerel, whereas the base SI units are meter, kilogram, second, kelvin, ampere, candela, and mole. These base units can be used to create all other SI units.

Thus,273 K gives the object's temperature in the SI unit therefore the correct answer is option B

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Answer:

   k = -3.1450 10⁻⁴  s⁻¹

Explanation:

In this exercise we are given the equation that describes the cooling process

        dT / dt = k (T -)

Let's solve is this equation,

        dT / (T-T_ {e}) = k dt

change of variable for integration

       T -T_{e} = T ’

       dT = dT '

       ∫ dT ’/ T’ = k  ∫ dt

we integrate

        ln T ’= k t

we change to the initial variables

        ln (T - T_{e}) = k t

Let's evaluate from the lower limit T = T for t = 0 to the upper limit T = T₀ for time t

       ln (T₀ -T_{e}) - ln (T -T_{e}) = k (t-0)

we simplify

       ln (T₀ -T_{e} / T -T_{e}) = k t

       k = ln (T₀ -T_{e})  / (T-Te) / t

In the exercise they indicate that the temperature T = 205 F, the ambient temperature is T_{e} = 70F, the temperature to which T₀ = 200 F falls in a time t = 2 min = 120 s

Let's calculate

           k = ln [(200- 70) / (205 -70)] / 120

           k = -0.0377403 / 120

           k = -3.1450 10⁻⁴  s⁻¹

Answer:

 λ =365.4 nm

Explanation:

Boh's atomic model of the Hydrogen atom the energy of each level is

        Eₙ = - 13.606 / n²

where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.

An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.

                 ED = 13.606 (1 / n₀² - 1 /[tex]n_{f}^{2}[/tex])

we are going to apply this relationship to answer slash.

At the beginning of the studies of atomic transitions, each group did not consider having a different name

name        Initial state

Lymman         1

Balmer           2

the final state is any other state sta the continuum that corresponds to n = inf

Let's look for the highest energy of the Balmer series

              ΔE = 13.606 (1/2² - 1 /∞)

              ΔE = 3.4015 eV

Let's use the Planck relation for the energy

                E = h f = h c /λ

                λ = h c / E

Let's reduce the energy to J

              E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹

            λ = 6.63 10⁻³⁴  3 10⁸ / 5.4424 10⁻¹⁹

            λ = 3.654 10⁻⁷ m

            λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)

            λ =365.4 nm

this eta radiation in the ultraviolet range

Answer:

its 45 over 6

Explanation:the answer is in  the question

Answer: Only the melted cube's shape changed.

Explanation:

Answer:

Infrared light

Explanation:

Infrared light is the spectrum of electromagnetic wave given off by a body possessing thermal energy. Infrared light is preferred over visible light in this region of space because visible light is easily scattered in the presence of fine particles. Infrared ray makes it easy for us to observe Cold, dark molecular clouds of gas and dust in our galaxy that glows when irradiated by the stars . Infrared can also be used to detect young forming stars, even before they begin to emit visible light. Stars emit a smaller portion of their energy in the infrared spectrum, so nearby cool objects such as planets can be more readily detected with infrared light which won't be possible with an ultraviolet or visible light.

Answer:

The  work function is  [tex]\phi = 2.46 \ eV[/tex]

Explanation:

From the question we are told that

    The light energy is  [tex]E = 3.56 eV[/tex]

     The  stopping voltage is  [tex]V = 1.10 \ V[/tex]

Generally work function is mathematically represented as

      [tex]\phi = E - KE[/tex]

Where KE is the kinetic energy of the ejected electron and it is mathematically represented as

         [tex]KE = V * e[/tex]

Where  e is the charge on the electron

So  

        [tex]KE = 1.10eV[/tex]

Thus  

        [tex]\phi = 3.56eV - 1.10 eV[/tex]

=>      [tex]\phi = 2.46 \ eV[/tex]

Answer:

(a) the current will flow in opposite direction

(b) the current needed is 33.25 A

Explanation:

(a) At the center of the two parallel wires, the two wires will have the same magnitude of magnetic field. In order to have a non a zero value of magnetic field at the center, the field must be in the same direction and the current will flow in opposite direction according to right hand rule.

(b) How much current is needed

Given;

distance between the two parallel wires, d = 9.5 cm = 0.095 m

magnitude of magnetic field at a point halfway between the wires, [tex]B_c[/tex] = 280 μT (This unit was corrected to obtain feasible current)

The magnetic field at distance R due to an infinite wire is given by;

[tex]B = \frac{\mu_o I}{2\pi R}[/tex]

At the center of the wire, [tex]B_c = 2B[/tex]

[tex]B_c = 2(\frac{\mu_o I}{2\pi R} )\\\\B_c = \frac{\mu_o I}{\pi R}[/tex]

where;

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

R is the center point between the wires, R = d/2 = 0.095m / 2 = 0.0475 m

I is the current needed

[tex]B_c = \frac{\mu_o I}{\pi R} \\\\I = \frac{B_c \pi R}{\mu_o} \\\\I = \frac{280* 10^{-6}*\pi *0.0475}{4\pi *10^{-7}} \\\\I = 33.25 \ A[/tex]

Answer:

Explanation:

Given the simultaneous equation,

3C+4D=5  .............. 1

2C+5D=2 ............... 2

Solving for the value of C and D using substitution method.

From equation 1;

3C = 5-4D

Divide both sides by 3

3C/3 =  (5-4D)/3

C = (5-4D)/3 .... 3

From equation 2:

2C+5D=2

5D = 2-2C

Divide both sides by 5;

5D/5 = 2-2C/5

D = (2-2C)/5 ..... 4

Substitute equation 4 into 3;

C = 5-4{(2-2C)/5}/3

C = [5 - (8-8C/5)]/3

C = [25-(8-8C)/5]/3

C = (17+8C)/15

15C = 17+8C

15C-8C = 17

7C = 17

C = 17/7

Substitute C = 17/7 into equation 4 to get the value of D

D = (2-2(17/7))/5

D = (2-34/7)/5

D = 14-34/35

D = -20/35

D = -4/7

Hence the value of C = 17/7, D = -4/7

Answer:

The speed of the car just before braking is 23.66 m/s.

Explanation:

Given;

mark of the skid, d = 80 m

deceleration of the car, a = 3.5 m/s²

To determine the speed of the car just before braking, we apply the following kinematic equation;

[tex]v^2 = u^2 + 2ad\\\\v^2 = 0 + 2(3.5)(80)\\\\v^2 = 560\\\\v= \sqrt{560}\\\\v = 23.66 \ m/s[/tex]

Therefore, the speed of the car just before braking is 23.66 m/s.

Explanation:

time taken fir 50 oscillations is 6 seconds

time taken for 1 oscillation is 6/50

convert it into a decimal

Answer:

1.44*10^-3m

Explanation:

Given that distance BTW two bright fringes is

DetaY = lambda* L/d

So for second wavelength

Deta Y2= Lambda 2* L/d

=lambda 2 x deta y1/ lambda1

So substituting

= 360 x 10^-9 x (1.6*10^-3/640*10^-9)

1.44*10^ -3m

Answer:

a

 [tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]

b

The  direction of the electric field is opposite that of the current              

Explanation:

From the question we are told that

   The current is  [tex]I = 17\ A[/tex]

   The diameter of the ring is  [tex]d = 3.0 \ cm = 0.03 \ m[/tex]

Generally the  radius is mathematically represented as

       [tex]r = \frac{d}{2}[/tex]

       [tex]r = \frac{0.03}{2}[/tex]

       [tex]r = 0.015 \ m[/tex]

The  cross-sectional area is mathematically represented as

       [tex]A = \pi r^2[/tex]

=>     [tex]A = 3.142 * (0.015^2)[/tex]

=>    [tex]A = 7.07 *10^{-4 } \ m^ 2[/tex]

Generally  according to ampere -Maxwell equation we have that

      [tex]\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }[/tex]

Now given that [tex]\= B = 0[/tex] it implies that

     [tex]\oint \= B \cdot \= ds = 0[/tex]

So

    [tex]\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0[/tex]

Where  [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

            [tex]\mu_o[/tex] is the permeability of free space with value  

[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

      [tex]\phi[/tex] is magnetic flux which is mathematically represented as

       [tex]\phi = E * A[/tex]

Where E is the electric field strength

  So  

       [tex]\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0[/tex]

=>   [tex]\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }[/tex]

=>   [tex]\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }[/tex]

=>   [tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]

The  negative  sign shows that the  direction  of  the electric field is opposite that of the current

Answer:

The initial speed of the block is 1.09 m/s

Explanation:

Given;

mass of block, m = 1.7 kg

force constant of the spring, k = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

From principle of conservation of energy

kinetic energy of the block = elastic potential energy of the spring

¹/₂mv² = ¹/₂kx²

mv²  = kx²

[tex]v = \sqrt{\frac{kx^2}{m} }[/tex]

where;

v is the initial speed of the block

x is the compression of the spring

[tex]v = \sqrt{\frac{955*(0.046)^2}{1.7} } \\\\v = 1.09 \ m/s[/tex]

Therefore, the initial speed of the block is 1.09 m/s

This question is incomplete, the complete question is;

In a simple model of a potassium iodide (KI) molecule, we assume the K and I atoms bond ionically by the transfer of one electron from K to I.

(a) The ionization energy of K is 4.34 eV, and the electron affinity of I is 3.06 eV. What energy is needed to transfer an electron from K to I, to form K+ and I- ions from neutral atoms? This quantity is sometimes called the activation energy Ea.eV

(b) A model potential energy function for the KI molecule is the Lennard - jones potential:

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

where r is the internuclear separation distance and α and ∈ are adjustable parameters (constants) . The Ea term is added to ensure the correct asymptotic behavior at large r and is activation energy calculated in a. At the equilibrium separation distance, r=r₀=0.305 nm, U(r) is a minimum, and dU/dr=0. In addition, U(r₀)=-3.37 eV.

Us the experimental values for the equilibrium sepeartion and dissociation energy of KI to determine/find 'α' and '∈'.

(c) calculate the force needed to break the KI molecule in nN

Answer:

a) energy is needed to transfer an electron from K to I, to form K+ and I- ions from neutral atoms is 1.28 eV

b) α = 0.272, ∈ = 4.65 eV

c) the force needed to break the KI molecule in nN 65.6 nN

Explanation:

a) The ionization energy of K is 4.34 ev ( energy needed to remove the outer most electrons)

And the electron affinity of I is 3.06 ev ( which is energy released when electron is added)

Now the energy that is need to transfer an electron from K to I,

i.e the ionization energy of K(4.34 ev) and the electron affinity of I (3.06 ev)

RE = 4.34 - 3.06 = 1.28 eV

b)

from the question we have

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

now taking d/drU(r₀)=0  (at r = r₀)

= 4∈d/dr [ (α/r)¹² - (α/r)⁶ ] = 0

= ( -12(α¹²/r¹³)) - (-6 (α⁶/r⁷)) = 0

12(α¹²/r¹³) = 6 (α⁶/r⁷)

α⁶ = r⁶/2

α = r/(2)^1/6

at equilibrium r = r₀ = 0.305 nm

α = 0.305 nm / (2)^1/6

C = 0.0305/1.1246

α = 0.272

Now substituting the values of U(r₀), α, Eₐ in the initial expression

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

we have

- 3.37eV = 4∈ [ (0.272 nm / 0.305 nm)¹² - (0.272 nm / 0.305 nm )⁶ ] + 1.28

- 1.65 eV = ∈(0.25 - 0.5)

∈ = 4.65 eV

c)

Now to break the molecule then the potential energy should be zero(0)

and we know r = 0.272 nm

therefore force needed to break the molecule is

F = -dU/dR_r-α

F = -4∈ (-12/α  + 6/α)

F = -4(4.65eV) ( -12/0.272nm + 6/0.272nm)

F = 65.6 nN

Answer:

n₁ > n₂.

prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50

Explanation:

Total internal reflection occurs when the refractive index of the incident medium the light is greater than the medium to which the light is refracted, let's use the refraction equation

                 n₁ sin θ₁ = n₂ sin  θ₂

the incident medium is 1, at the limit point where refraction occurs is when the angle in the refracted medium is 90º, so sin θ₂ = 1

                 n₁ sin θ₁ = n₂

                 sin θ₁ = n₂ / n₁

We mean that this equation is defined only for n₁ > n₂.

In our case, for the total internal reflection to occur, the refractive incidence of the medium must be greater than the index of refraction of the prism.

In general, prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50

Calculate the acceleration ,a .

s = ut + 1/2 at²

According to the second equation of motion, we have

s = ut + 1/2 at²

★Substituting the values in the above formula,we get:

⇒ 16 = 8 × 4 + 1/2 × a × 4

⇒ 16 = 32 + 2a

⇒ 2a = 16 - 32

⇒ 2a = -16

⇒ a = -16/2

⇒ a = -8 m/s²

Hence,the acceleration is -8 m/s² .

Answer:

Therefore, electron will have a longer de Broglie Wavelength.

Explanation:

The de Broglie wavelength is given by the following formula:

λ = h/mv

where.

λ = de Broglie wavelength

h = Plank's Constant

m = mass of the particle.

v = speed of the particle

Since, the speed of both electron and proton is same and Plank's constant is also a constant. Therefore, the de Broglie wavelength depends solely upon the mass of electron and proton, as follows:

λ ∝ 1/m

It shows that wavelength is inversely proportional to the mass of particle.

Since, the mass of electron is less than the mass of proton.

Therefore, electron will have a longer de Broglie Wavelength.