What explains why a prism separates white light into a light spectrum?
A. The white light, on encountering the prism, undergoes both reflection and refraction; some of the reflected rays re-enter the prism merging with refracted rays changing their frequencies.
B. The white light, on entering a prism, undergoes several internal reflections, forming different colors.
C. The different colors that make up a white light have different refractive indexes in glass.
D. The different colors that make up a white light are wavelengths that are invisible to the human eye until they pass through the prism.
E. The different rays of white light interfere in the prism, forming various colors.

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

Answer:

The Independent variable in this experiment is the time taken by the ball to roll down each distance.

The dependent variable is the distance  through which the ball rolls

The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.

Explanation:

The complete question is

Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?

Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.

A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.

The relationship between the dependent and independent variables in an experiment is given as

y = f(x)

where y is the output or the dependent variable,

and x is the independent variable.

Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

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

angular frequency ω = 2π / T

= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 (  ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ  = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]

= 4.7476 m/sec

= 4.75 m/s

Answer:

8.33` m/s^2 and 8333.3 N

Explanation:

a) acceleration:

ā=v^2/r

ā=(15m/s)^2/27m

ā=225/27 m/s^2

ā=8.333 m/s^2

force:

F=mā. where the is equal to v^2/r

F=1000kg*8.3 m/s^2

F=8333.3 N

Answer:

8.33` m/s^2 and 8333.3 N

Explanation:

Answer:

Explanation:

distance of third dark fringe

= 2.5 x λ D / d

where λ is wavelength of light , D is screen distance and d is slit separation

putting the given values

required distance = 2.5  x 739 x 10⁻⁹  x 6.3 / .49 x 10⁻³

= 23753.57 x 10⁻⁶

= 23.754 x 10⁻³ m

= 23.754 mm .

Answer:

(a) 0.59 ms

(b) 0.15 ms

(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.

Explanation:

(a) distance between ears = 20 cm = 0.2 m

speed of sound generated = 340 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                        = [tex]\frac{0.2}{340}[/tex]

                        = 5.8824 × [tex]10^{-4}[/tex]

                        = 0.59 ms

The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.

(b) distance between ears = 20 cm = 0.2 m

speed of sound in water = 1530 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                         = [tex]\frac{0.2}{1530}[/tex]

                         = 1.4815 × [tex]10^{-4}[/tex]

                         = 0.15 ms

The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.

(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.

A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds

B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds

C) The significance about the time interval of the two situations is that;

Transmission of sound varies with different mediums.

A) We are given;

Distance between the two ears; d = 20 cm = 0.2 m

Speed of sound; v = 340 m/s

Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;

t = d/v

t = 0.2/340

t = 0.588 × 10⁻³ seconds

B) We are now told that the speed of sound in water is 1530 m/s. Thus;

t = 0.2/1530

t = 0.131 × 10⁻⁵ seconds

C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.

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Answer:

a)  A = 449526  J,  b) 449526 J

Explanation:

In this exercise they do not ask for the work of different elements.

Note that as the box rises at constant speed, the sum of forces is chorus, therefore

           T-W = 0

           T = W

           T = m g

           T = 1,390 9.8

           T = 13622 N

Now that we have the strength we can use the definition of work

           W = F .d

            W = f d cos tea

a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel

              A = A  x

              A = 13622  33

               A = 449526  J

b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180

               W = T x cos 180

               W = - 13622 33

               W = - 449526 J

Answer:

vpg = 0.064 N

Explanation:

Upthrust = Volume of fluid displaced

upthrust liquid on the cube g=10ms−2

vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N

vpg = 0.064 N

hope it helps.

Answer:

The wave is traveling in the y axis direction

Explanation:

Because the wave will always travel in a direction 90° to the magnetic and electric components

Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

The equivalent resistance of this combination of resistors is 2.9Ω.

The combined resistance in such arrangement of resistors is provided by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

here.

Req means  the equivalent resistance and R1, R2, R3

.Rn means the resistance of individual resistors interlinked in parallel.

Also,

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

So,

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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Explanation:

It is given that,

Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.

Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]

As the buses are moving in opposite direction, then the concept of relative velocity is used. So,

Distance, [tex]d=v\times t[/tex]

v is relative velocity, v = 108 + 36 = 144 km/h

So,

[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]

So, the distance between them is 48 km after 20 minutes.

Answer:

f = 500 x 10^12Hz

Explanation:

E=hc/wavelength

E=hf

hc/wavelength =hf

c/wavelength =f

f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength

Answer:

The  frequency is  [tex]F = 325 Hz[/tex]

Explanation:

From the question we are told that

    The frequency for the first note is  [tex]F_1 = 330 Hz[/tex]

     The  beat frequency of the first note is  [tex]f_b = 5 \ Hz[/tex]

     The  frequency for the second note is  [tex]F_2 = 350 \ H_z[/tex]

      The  beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]

Generally beat frequency is mathematically represented as

        [tex]F_{beat} = | F_a - F_b |[/tex]

Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source

  Now in the case of this question

For the first note

     [tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]

Where  F is the frequency of the string note

For the second note  

      [tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]

Adding  equation 1 from 2

      [tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]

      [tex]f_b + f_a = F_1 + F_2 -2F[/tex]

substituting values

       [tex]5 +25 = 330 + 350 -2F[/tex]

=>     [tex]F = 325 Hz[/tex]

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

Answer:

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

Answer:

Yes. Something similar occurs here on Earth.

Explanation:

Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.

Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine

Explanation:

Work = force × distance

W = (400 N) (10 m)

W = 4000 J

Answer:

5.33333... seconds

Explanation:

800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....

Answer:

1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.

2)If the angle is zero we see a bright light called undispersed light

For different angles we see the colors of the spectrum

3) must be able to see the well-collimated light emission source

Explanation:

1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.

These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.

The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.

The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.

2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.

If the angle is zero we see a bright light called undispersed light

For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.

3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.

The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.

Answer:

Trial 1: 2 Volts, 0 %

Trial 2: 2.8 Volts, 0%

Trial 3: 4 Volts, 0 %

Explanation:

Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:

V = IR

TRIAL 1:

V = IR

V = (0.1 A)(20 Ω)

V = 2 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(2 - 2)/2| x 100%

% Difference = 0 %

TRIAL 2:

V = IR

V = (0.14 A)(20 Ω)

V = 2.8 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(2.8 - 2.8)/2.8| x 100%

% Difference = 0 %

TRIAL 3:

V = IR

V = (0.2 A)(20 Ω)

V = 4 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(4 - 4)/4| x 100%

% Difference = 0 %

Answer:

[tex]3.1\times 10^{5}m/s[/tex]

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]

[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]

Simply we applied the above formulas

Answer:

Explanation:

The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point

Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima  is π radian .

Answer:

B) the particle's momentum.

Explanation:

We know that

The centripetal force  on the particle when its moving in the radius R and velocity V

[tex]F_c=\dfrac{m\times V^2}{R}[/tex]

The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q

[tex]F_m=q\times V\times B[/tex]

At the equilibrium condition

[tex]F_m=F_c[/tex]

[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]

[tex]R=\dfrac{m\times V}{q\times B}[/tex]

Momentum = m V

Therefore we can say that the radius of curvature is directly proportional to the particle momentum.

B) the particle's momentum.

Answer:

A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.

Explanation:

Answer:

Option A

Explanation:

Acceleration will be obviously zero when Force = 0

That is how:

Force = Mass * Acceleration

So, If force = 0

0 = Mass * Acceleration.

Dividing both sides by Mass

Acceleration = 0/Mass

Acceleration = 0 m/s²

Answer:

[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]

Explanation:

[tex]\mathrm{force=mass \times acceleration}[/tex]

The force is given 0 newtons.

[tex]\mathrm{force=0 \: N}[/tex]

Plug force as 0.

[tex]\mathrm{0=mass \times acceleration}[/tex]

Divide both sides by mass.

[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]

[tex]\mathrm{0 =acceleration}[/tex]

[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]

Answer:

Explanation:

area of the coil  A = .08 x .08 = 64 x 10⁻⁴ m ²

flux through the coil Φ = area of coil x no of turns x magnetic field

= 64 x 10⁻⁴ x 50 x B where B is magnetic field

emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2

= 1.6 B

current induced = emf induced / resistance

12 x 10⁻³ = 1.6 B / 15

B = 112.5 x 10⁻³ T .

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

[tex]distance=v\,*\, t[/tex]

[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

A) The free-body diagram of the forces acting on the brick is attached.

B) The free-body diagram of the forces acting on the rubber cushion is attached.

C) The action and reaction forces of the entire brick–cushion–planet system has been enumerated below.

A) The brick has a Mass M placed on top of a rubber cushion of mass m.

This means that there will be a normal force acting acting upwards on the brick and also a gravitational force acting downward. These forces are denoted as;

Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]

Gravitational force acting on brick = Mg

Find attached the free body diagram.

B) The forces acting on the cushion will be;

Normal force of parking lot pavement on rubber cushion  = [tex]n_{pc}[/tex]

Gravitational force of earth acting on cushion = mg

Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]

C)  The action pairs of forces are;

i) Force; Normal force of rubber cushion acting on brick  = [tex]n_{cb}[/tex]

Reaction Force; Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]

ii) Action Force; Gravitational force acting on brick = Mg

Reaction; Gravitational force of brick acting on the earth

iii) Action Force; Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]

Reaction; Force of rubber cushion on parking lot pavement

iv) Action Force; Gravitational force of earth acting on rubber cushion = mg

Reaction Force; Gravitational force of rubber cushion on the earth.

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