Which of the following is a vector quantity?
speed
distance
acceleration

Answer:

the distance moved by the box is 70.03 m.

Explanation:

Given;

mass of the box, m = 35 kg

initial velocity of the box, u = 10 m/s

frictional force, F = 25 N

Apply Newton's second law of motion to determine the deceleration of the box;

-F = ma

a = -F / m

a = (-25 ) / 35

a = -0.714 m/s²

The distance moved by the box is calculated as follows;

v² = u² + 2ad

where;

v is the final velocity of the box when it comes to rest = 0

0 = 10² + (2 x - 0.714)d

0 = 100 - 1.428d

1.428d = 100

d = 100 / 1.428

d = 70.03 m

Therefore, the distance moved by the box is 70.03 m.

The question is incomplete. The complete question is :

Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 6 in a one-dimensional box 34.0 pm in length.

Solution :  

In an one dimensional box, energy of a particle is given by :

[tex]$E=\frac{n^2h^2}{8ma^2}$[/tex]

Here, h = Planck's constant

         n = level of energy

           = 6

         m = mass of particle

         a = box length

For n = 6, the energy associated is :

[tex]$\Delta E = E_6 - E_1 $[/tex]

[tex]$\Delta E = \left( \frac{n_6^2h_2}{8ma^2}\right) - \left( \frac{n_1^2h_2}{8ma^2}\right) $[/tex]

     [tex]$=\frac{h^2(n_6^2 - n_1^2)}{8ma^2}$[/tex]

We know that,

[tex]$E = \frac{hc}{\lambda} $[/tex]

Here, λ = wavelength

         h =  Plank's constant

         c = velocity of light

So the wavelength,

 [tex]$= \frac{hc}{E}$[/tex]

 [tex]$=\frac{hc}{\frac{h^2(n_6^2 - n_1^2)}{8ma^2}}$[/tex]

[tex]$=\frac{8ma^2c}{h(n_6^2 - n_1^2)}$[/tex]

[tex]$=\frac{8 \times 9.109 \times 10^{-31}(0.34 \times 10^{-10})^2 (3 \times 10^8)}{6.626 \times 10^{-34} \times (36-1)}$[/tex]

[tex]$= \frac{ 8 \times 9.109 \times 0.34 \times 0.34 \times 3 \times 10^{-43}}{6.626 \times 35 \times 10^{-34}}$[/tex]

[tex]$=\frac{25.27 \times 10^{-43}}{231.91 \times 10^{-34}}$[/tex]

[tex]$= 0.108 \times 10^{-9}$[/tex]  m

= 108 pm

Answer:

The second ball must be thrown at 30.01 m/s.

Explanation:

First, we need to find the maximum height (H) reached by the ball 1:

[tex] v_{f_{1}}^{2} = v_{0_{1}}^{2} - 2gH [/tex]  

Where:

[tex]v_{f_{1}}[/tex]: is the final speed of ball 1 = 0 (at the maximum height)

[tex]v_{0_{1}}[/tex]: is the initial speed of ball 1        

g: is the gravity = 9.81 m/s²    

We need to find the initial speed, by using the following equation:

[tex] v_{f_{1}} = v_{0_{1}} - gt [/tex]

Where t is the time = 1.71 s (when it reaches the maximum height)

[tex] v_{0_{1}} = gt = 9.81 m/s^{2}*1.71 s = 16.78 m/s [/tex]

So, the maximum height is:                  

[tex] H = \frac{v_{0_{1}}^{2}}{2g} = \frac{(16.78 m/s)^{2}}{2*9.81 m/s^{2}} = 14.35 m [/tex]  

Finally, the speed at which ball 2 must be thrown is:

[tex]v_{f_{2y}}^{2} = (v_{0_{2y}}sin(\theta)})^{2} - 2gH[/tex]      

[tex]v_{0_{2y}}= \frac{\sqrt{2gH}}{sin(\theta)} = \frac{\sqrt{2*9.81 m/s^{2}*14.35 m}}{sin(34)} = 30.01 m/s[/tex]                    

Therefore, the second ball must be thrown at 30.01 m/s.

I hope it helps you!                                                                                    

Answer:

10kN

Explanation:

Given data

m1= 50kg

u1= 3m/s

m2= 100kg

u2= 6m/s

v1= 2m/s

time= 0.04s

let us find the final velocity of Bruce v1

from the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

50*3+100*6= 50*v1+100*2

150+600=50v1+200

750-200=50v1

550= 50v1

divide both sides by 50

v1= 550/50

v1=11 m/s

From

F= mΔv/t

for Bruce

F=50*(11-3)/0.04

F=50*8/0.04

F=400/0.04

F=10000

F=10kN

for Max

F=100*(6-2)/0.04

F=100*4/0.04

F=400/0.04

F=10000

F=10kN

Answer:

A. 72000 C

B. 1100 W

C. 26.4 cents.

Explanation:

From the question given above, the following data were obtained:

Current (I) = 10 A

Voltage (V) = 110 V

Time (t) = 2 h

A. Determination of the charge.

We'll begin by converting 2 h to seconds. This can be obtained as follow:

1 h = 3600 s

Therefore,

2 h = 2 h × 3600 s / 1 h

2 h = 7200 s

Thus, 2 h is equivalent to 7200 s.

Finally, we shall determine the charge. This can be obtained as follow:

Current (I) = 10 A

Time (t) = 7200 s

Charge (Q) =?

Q = It

Q = 10 × 7200

Q = 72000 C

B. Determination of the power.

Current (I) = 10 A

Voltage (V) = 110 V

Power (P) =?

P = IV

P = 10 × 110

P = 1100 W

C. Determination of the cost of operation.

We'll begin by converting 1100 W to KW. This can be obtained as follow:

1000 W = 1 KW

Therefore,

1100 W = 1100 W × 1 KW / 1000 W

1100 W = 1.1 KW

Thus, 1100 W is equivalent to 1.1 KW

Next, we shall determine the energy consumption of the range. This can be obtained as follow:

Power (P) = 1.1 KW

Time (t) = 2 h

Energy (E) =?

E = Pt

E = 1.1 × 2

E = 2.2 KWh

Finally, we shall determine the cost of operation. This can be obtained as follow:

1 KWh cost 12 cents.

Therefore, 2.2 KWh will cost = 2.2 × 12

= 26.4 cents.

Thus, the cost of operating the range for 2 h is 26.4 cents.

Answer:

[tex]V=1.86m/s[/tex]

Explanation:

Mass of bullet [tex]M_B=6.11g[/tex]

Velocity of bullet [tex]V_B=366m/s[/tex]

Mass of first block [tex]M_b_1=1206g[/tex]

Velocity of block [tex]V_b=0.662m/s[/tex]

Mass of second block [tex]M_b_2=1550g[/tex]

Generally the total momentum before collision is mathematically given as

[tex]P_1=0.006kg*366+0+0[/tex]

[tex]P_1=2.196kg\ m/s[/tex]

Generally the total momentum after collision is mathematically given as

[tex]P_2=(1.206kg*0.633)+(1.550+0.00611)V[/tex]

[tex]P_2=0.763398+1.55611V[/tex]

Generally the total momentum is mathematically given as

[tex]P_1=P_2[/tex]

[tex]2.196=0.763398+1.55611V[/tex]

[tex]V=\frac{2.196+0.763398}{1.55611}[/tex]

[tex]V=1.86m/s[/tex]

Answer:

Yes

Explanation:

0.8 kilograms is equal to 800 grams

Answer:

Yes, 0.8 kilograms is greater than 80 grams

Explanation:

0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.

Sorry if I'm wrong, correct me.

The  charge on the inner surface of the shell is -Q

The  charge on the outer surface of the shell is Q

After this contact, the charge on the tack is  0

The charge on the inner surface of the shell now is  0

The charge on the outer surface of the shell now is Q

The charge on a shell depends on the situation and the conditions of the shell. If the shell is an electrically neutral object, such as a metallic spherical shell, it has no net charge, meaning that the total positive charge is equal to the total negative charge. However, if the shell has an excess or deficit of electrons, it will have a net charge, either positive or negative, depending on whether it has an excess of electrons or a deficit of electrons.

Read more on charges here:https://brainly.com/question/18102056

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Answer:

the initial speed of the arrow before joining the block is 89.85 m/s

Explanation:

Given;

mass of the arrow, m₁ = 49 g = 0.049 kg

mass of block, m₂ = 1.45 kg

height reached by the arrow and the block, h = 0.44 m

The gravitational potential energy of the block and arrow system;

P.E = mgh

P.E = (1.45 + 0.049) x 9.8 x 0.44

P.E = 6.464 J

The final velocity of the system after collision is calculated as;

K.E = ¹/₂mv²

6.464 = ¹/₂(1.45 + 0.049)v²

6.464 = 0.7495v²

v² = 6.464 / 0.7495

v² = 8.6244

v = √8.6244

v = 2.937 m/s

Apply principle of conservation of linear momentum to determine the initial speed of the arrow;

[tex]P_{initial} = P_{final}\\\\mv_{arrow} + mv_{block} = (m_1 + m_2)V\\\\0.049(v) + 1.45(0) = (0.049 + 1.45)2.937\\\\0.049v = 4.4026\\\\v = \frac{4.4026}{0.049} \\\\v = 89.85 \ m/s[/tex]

Therefore, the initial speed of the arrow before joining the block is 89.85 m/s

The arrow moving as the speed of "76.36 m/s".

According to the question,

By using the conservation of energy, we have

→                [tex]K.E=P.E[/tex]

→ [tex]\frac{1}{2} (m_1+m_2)v_2^2= (m_1+m_2)gh[/tex]

or,

→                    [tex]v_2 = \sqrt{2mgh}[/tex]

By substituting the values, we have

→                         [tex]= \sqrt{2\times 9.8\times 0.44}[/tex]

→                         [tex]=2.469 \ m/s[/tex]

Now,

By using the conservation of momentum, we get

→ [tex]m_1 v_1 = (m_1+m_2) v_2[/tex]

or,

→      [tex]v_1 = \frac{(m_1+m_2)v_2}{m_1}[/tex]

            [tex]= \frac{1.45+0.049}{0.049}\times 2.469[/tex]

            [tex]= 30.6\times 2.496[/tex]

            [tex]= 76.36 \ m/s[/tex]

Thus the above approach is correct.  

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Answer:

1.76 m

Explanation:

Height from which the object is dropped = 45 m

Time (t) remaining for the ball to land = 0.6 s

Height (h) in the remaining =?

The height to which the object falls in the remaining time can be obtained as follow:

Time (t) remaining for the ball to land = 0.6 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) in the remaining =?

h = ½gt²

h = ½ × 9.8 × 0.6²

h = 4.9 × 0.36

h = 1.76 m

Thus, the distance travelled in the last 0.6 s is 1.76 m

Answer:

a) F_net = 30.47 N ,   θ = 10.6º

b)  Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

          F₁₂ = k [tex]k \frac{ q_{1} \ q_{2} }{ r^{2} }[/tex]

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

        Fₓ = [tex]F_{bc x}[/tex]

Y axis  

       [tex]F_{y}[/tex]Fy = [tex]F_{ab} - F_{bc y}[/tex]

let's find the magnitude of each force

     [tex]F_{ab}[/tex] = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

      F_{ab} = 2.82 10¹ N

      F_{ab} = 28.2 N

      [tex]F_{bc}[/tex] = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

      F_{bc} = 3.75 10¹  N

       F_{bc} = 37.5 N

let's use trigonometry to decompose this force

      tan θ = y / x

      θ = tan⁻¹ and x

       θ= tan⁻¹ ¾

      θ = 37º

let's break down the force

      sin 37 = F_{bcy} / F_{bc}

      F_{bcy} = F_{bc} sin 37

      F_{bcy} = 37.5 sin 37

      F_{bcy} = 22.57 N

      cos 37 = F_{bcx} /F_{bc}

      F_{bcx} = F_{bc} cos 37

      F_{bcx} = 37.5 cos 37

      F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

        Fₓ = 29.95 N

Axis y

        Fy = 28.2 -22.57

        Fy = 5.63 N

we can give the result in two ways

a)  F_net = Fₓ i ^ + [tex]F_{y}[/tex] j ^

    F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

    F_net = [tex]\sqrt{ F_{x}^2 + F_{y}^2 }[/tex]

    F_net = √(29.95² + 5.63²)

     F_net = 30.47 N

we use trigonometry for the direction

      tan θ= [tex]\frac{ F_{y} }{ F_{x} }[/tex]

      θ = tan⁻¹ \frac{ F_{y}  }{  F_{x} }

      θ = tan⁻¹ (5.63 / 29.95)

      θ = 10.6º

Answer:

F = 2.40 × [tex]10^{-6}[/tex]  N

Explanation:

given data

charge q1 = 3.95 nC

x= 0.198 m

charge q2 = 4.96 nC

x= -0.297 m

solution

force on a point charge kept in electric field F = E × q       ................1

here E is the magnitude of electric field and q is the magnitude of charge

and

first we will get here electric field at origin

So net field at origin is

E = (Kq2÷r2²) - (kq1÷r1²)           ...............2

put here value

E = 9[(4.96÷0.297²)-(3.95÷0.198²)]

E = 400.72 N/C        ( negative x direction )

so that force will be

F = 6 × [tex]10^{-9}[/tex] × 400.72

F = 2.40 × [tex]10^{-6}[/tex]  N

The net force on the third charge is 2.404 x 10⁻⁶ N.

The given parameters:

The force on the third charge due to first charge is calculated as follows;

[tex]F_{13} = \frac{kq_1 q_3}{r^2} \\\\F_{13} = \frac{9\times 10^9 \times 3.95 \times 10^{-9} \times 6 \times 10^{-9} }{(0.198)^2} (+i)= 5.44 \times 10^{-6} \ N \ (+i)[/tex]

The force on the third charge due to second charge is calculated as follows;

[tex]F_{23} = \frac{kq_2q_3}{r^2} \\\\F_{23} = \frac{9\times 10^9 \times 4.96 \times 10^{-9}\times 6 \times 10^{-9} }{(0.297)^2} (-i)\\\\F_{23} = (3.036 \times 10^{-6} ) \ N \ (-i)[/tex]

The net force on the third charge is calculated as follows;

[tex]F_{net} = 5.44 \times 10^{-6} - 3.036 \times 10^{-6} \\\\F_{net} = 2.404 \times 10^{-6} \ N[/tex]

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Answer:

Explanation:

See the figure attached

F is electrostatic force .

T cos20 = mg

T sin20 = F

Tan20 = F / mg

F = mg tan 20 = .025 x 9.8 tan20

= .09 N

Distance between bob and balloon

= 15 sin20 = 5.1 cm = .051 m

If q be the charge on balloon

F = 9 x 10⁹ x q² / .051²

= 3460 x 10⁹ q² = .09

q² =  26 x 10⁻⁶ x 10⁻⁹

q = 16.12 x 10⁻⁸ C .

Answer:

15.68 m/s

Explanation:

Given that,

She catches the ball 3.2 s later at the same height from which it was thrown.

When it reaches the maximum height, its height is equal to 0.

It will move under the action of gravity.

[tex]t=\dfrac{2u}{g}[/tex]

2 here comes for the time of ascent and descent.

So,

[tex]u=\dfrac{tg}{2}\\\\u=\dfrac{3.2\times 9.8}{2}\\\\u=15.68\ m/s[/tex]

So, the initial upward speed of the ball is 15.68 m/s.

Answer:

1.9kHz

Explanation:

Given data

wavelength [tex]\lambda= 0.32m[/tex]

velocity [tex]v= 622 m/s[/tex]

We know that

[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]

substitute

[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]

Hence the frequency is 1.9kHz

Answer:

971.2

Explanation:

It was right on acellus :)

Answer:

V= -3.6*10⁻¹¹ V

Explanation:

      [tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]

       [tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]

      [tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]

       [tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]

The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]

We can arrive at this answer as follows:

[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]

[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]

[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]

More information about Gauss' law at the link:

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Answer:

Here its right but its also better than Barney's response

Explanation:

W, Y, Z, X or C

Answer:

W, Y, Z, X

Explanation:

I am not sure but I think it is false

Explanation:

Yes, the law of conservation of energy still applies even if there is waste energy.

The waste energy are the transformation products of energy from one form to another.

According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".

But of then times, energy is lost as heat or sound within a system.

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

Answer:

a)

*The charges of which and the cube are of the same sign..

          Q_exterior = 3 Q

* the charge of the sphere has a different sign than the charge of the cube,

         Q_exterior =  Q

b)   Q = 0

Explanation:

To correctly describe this experiment, you must remember that in metals charges are mobile and that charges of the same sign repel and of different signs attract.

Let's analyze each situation

a) Suppose that the charges of which and the cube are of the same sign.

When the ball is introduced without touching the walls, its charge Q attracts a charge of equal magnitude and different sign to the internal wall. If we create a Gaussian surface around the inner wall of the sphere the net charge between the ball and the inner wall is zero. Consequently, a face Q should have been generated in the outer wall, therefore in this wall it has a total load of

               Q_exterior = 3 Q

   Now suppose that the charge of the sphere has a different sign than the charge of the cube, for simplicity let's say that the charge of the sphere is -Q and the cube + 2Q,

     Again we create a Gaussian surface outside the inner wall, now the charge on the ball attracts a charge of value + Q to neutralize the charge between the ball and the inner wall. Therefore a load remains on the outer wall

              Q_exterior = + Q

b) The cube is connected to earth and it is touched with the ball, in this case the charge of the two bodies is neutralized by the Earth, therefore the bodies have zero charge

             Q = 0

Complete Question

The question image is in the first uploaded image

Answer:

[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]

Explanation:

From the question we are told that

Distance b/w Q mid point and P is given as x

Generally the equation for magnitude of the electric field at the point P is given as

[tex]E=\frac{kQ}{d^2}[/tex]

where

[tex]k=\frac{1}{4\pi e_0}[/tex]

[tex]d=x^2-a^2[/tex]

Therefore

[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]

[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]

Therefore equation for magnitude of the electric field at the point P is

[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]

Answer:

The magnitude and direction of the rocket acceleration is 68.89 m/s² upward.

The speed of the rocket at the given elevation is 186 m/s.

Explanation:

Given;

time to reach the given height, t = 2.7 s

height reached, h = 93 m

initial velocity of the rocket, u = 0

The magnitude and direction of the rocket acceleration is calculated as;

h = ut + ¹/₂at²

h = 0 + ¹/₂at²

h = ¹/₂at²

a = 2h / t²

a = (2 x 93) / 2.7

a = 68.89 m/s²

the direction of the acceleration is upward.

The speed at this elevation, V = u + at

V = at

V = 68.89 x 2.7

V = 186 m/s

Answer:I think it’s self monitoring sorry if wrong

Explanation:

Answer:

It self monitoring

Explanation:

I took the test

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

[tex]H_m=1.65m[/tex]

[tex]H_E=1.16307m[/tex]

Explanation:

From the question we are told that

Mass of ball [tex]M=2kg[/tex]

Length of string [tex]L= 2m[/tex]

Wind force [tex]F=13.2N[/tex]

Generally the equation for [tex]\angle \theta[/tex] is mathematically given as

[tex]tan\theta=\frac{F}{mg}[/tex]

[tex]\theta=tan^-^1\frac{F}{mg}[/tex]

[tex]\theta=tan^-^1\frac{13.2}{2*2}[/tex]

[tex]\theta=73.14\textdegree[/tex]

Max angle =[tex]2*\theta= 2*73.14=>146.28\textdegree[/tex]

Generally the equation for max Height [tex]H_m[/tex] is mathematically given as

[tex]H_m=L(1-cos146.28)[/tex]

[tex]H_m=0.9(1+0.8318)[/tex]

[tex]H_m=1.65m[/tex]

Generally the equation for Equilibrium Height [tex]H_E[/tex] is mathematically given as

[tex]H_E=L(1-cos73.14)[/tex]

[tex]H_E=0.9(1+0.2923)[/tex]

[tex]H_E=1.16307m[/tex]

Solution :

1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).

2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).

3. Non-selective scatter takes place when particle size in greater than the wavelength  (λ).

We have the sizes of different particles :

[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]

Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]

Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]

Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]

Wavelength           [tex]$ O_2 $[/tex]         Smoke particles    Cloud droplets     Rain droplets

                            [tex]$10^{-10} \ m$[/tex]        [tex]$ 3 \times 10^{-7} \ m$[/tex]           [tex]$ 2 \times 10^{-5} \ m$[/tex]              [tex]$ 3 \times 10^{-3} \ m$[/tex]

[tex]$5500 \times 10^{-4} \ m$[/tex]      Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$11 \times 10^{-6} \ m $[/tex]         Rayleigh    Rayleigh            Non-selective      Non-selective

[tex]$1600 \times 10^{-10} \ m $[/tex]    Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$10^{-2} \ m $[/tex]                 Rayleigh      Rayleigh               Rayleigh          Mie

Answer:

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Answer:

True

Explanation:

Bohr proposed an atomic model in which;