The index of refraction of a certain material is 1.25. If I send red light (700 nm) through the material, what will the frequency of the light be in the material

Answer:

Electric field strengh is a measure of the strength of an electric field at a given point in space, equal to the field would induce on a unit electric charge at that point.

Electric field strength is also known as Electric Field Intensity .

Explanation:

Electric Field is also defined as force per charge. The unit will be force unit divided by charge unit. In this case, it will be Newton/Coulomb or N/C.

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Answer:

The speed of the police car is 294 m/s

Explanation:

Given;

frequency of the siren in air, f = 280 Hz

speed of sound in air, v = 343 m/s

Determine the wavelength of the sound in air to the stationary car:

v = fλ

where;

λ is wavelength of the sound

λ = v/f

λ = 343 / 280

λ = 1.225 m

Now, determine the speed at which the police car is approaching the stationary car;

The actual frequency of the police car, F = 240 Hz

V = Fλ

Where;

V is speed of the police car

λ is the distance between the police car and the stationary car, (wavelength)

V = 240 x 1.225

V = 294 m/s

Therefore, the speed of the police car is 294 m/s

Answer: W.D = 1/2mv^2

Explanation:

If an external force or a single force is acting on a body. Just like the first law of thermodynamics, the force acting on the body will cause work done on the system.

Work done = force × distance

And the work done on the body will cause the molecules of the body to experience motion and thereby producing kinetic energy.

The work done will be converted to kinetic energy.

W.D = 1/2mv^2

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

λ = 5.2 x 10⁻⁷ m = 520 nm

Answer:

The magnitude of the magnetic field is 8.333 x 10⁻⁷ T

Explanation:

Given;

charge on the lightening bolt, C = 15.0 C

time the charge passes by, t = 1.5 x 10⁻³ s

Current, I is calculated as;

I = q / t

I = 15 / 1.5 x 10⁻³

I = 10,000 A

Magnetic field at a distance from the bolt is calculated as;

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

where;

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

I is the current in the bolt

r is the distance of the magnetic field from the bolt

[tex]B = \frac{\mu_o I}{2\pi r} \\\\B = \frac{4\pi *10^{-7} 10000}{2\pi *24} \\\\B = 8.333 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field is 8.333 x 10⁻⁷ T

Answer:

Option (A)

Explanation:

A 20 kg boy chases the butterfly with a speed of 2 meter per second.

Angle at which he runs is 70° North of West.

Therefore, Horizontal component (Vx) directing towards West will be,

Vx = v(Cos70°)

Vy = v(Sin70°)

Since momentum of a body is defined by,

Momentum = Mass × Velocity

Therefore, Westerly component of the momentum will be,

Momentum = 20 × (v)(Cos70°)

                   = 20 × 2Cos70°

                   = 13.68

                   ≈ 13.7 kg-meter per second

Therefore, Option (A) will be the answer.

Answer:

Anti-Particle

A wedge of glass of refractive index 1.64 has a silver coating on the bottom, as shown in the image attached below.

Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.

Answer:

the smallest distance x  = 2.74 × 10⁻³ m or 2.74 mm

Explanation:

From the given information:

The net phase change is zero because both the light ray reflecting from the air-glass surface and silver plate undergo a phase change of [tex]\dfrac{\lambda}{2}[/tex] , as such the condition for the  constructive interference is:

nΔy = mλ

where;

n = refractive index

Δy = path length (inside the glass)

So, from the diagram;

[tex]\dfrac{y}{x}=\dfrac{10^{-5} \ m}{0.2 \ m}[/tex]

[tex]\dfrac{y}{x} = 5 \times 10^{-5}[/tex]

[tex]y = 5 \times 10^{-5} x[/tex]

Now;

Δy can now be = 2 ( 5 × 10⁻⁵ [tex]x[/tex])

Δy =1 ×  10⁻⁴[tex]x[/tex]

From nΔy = mλ

n( 1 ×  10⁻⁴[tex]x[/tex] ) = mλ

[tex]x = \dfrac{m \lambda}{n \times 1 \times 10^{-4} }[/tex]

when the thickness is minimum then m = 1

Thus;

[tex]x = \dfrac{1 \times 450 \times 10^{-9} \ m}{1.64 \times 1 \times 10^{-4} }[/tex]

x =  0.00274 m

x = 2.74 × 10⁻³ m or 2.74 mm

Answer: B. The surface of the coating is rough, so light that shines on it gets scattered in many directions.

Explanation: On Edge!!!!!!!!!!!!!!!!!!!!

Answer:

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

Answer:

Explanation:

When the path difference is half the wave length or λ /2 , destructive interference takes place which results in reduced or zero intensity in case equal intensity waves interfere as in Young's double slit experiment

Hence dark fringe is formed at that place where intensity is zero .

Hence the right option is D

Answer:

9.82 rads/sec

Explanation:

We are given;

Time taken; t = 1.6 secs

Number of somersaults = 2

Now, we know that,

1 revolution = 2π radians

And number of somersaults is the same thing as number of revolutions

So,

Total radians = 2π × 2 = 5π

Angular velocity = total number of revolutions/time period = 5π/1.6 = 9.82 rads/sec

Answer:

120000    kgxm/s

Explanation:

momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000  kgxm/s

Explanation:

The statement "our sun is an 'average' sun" is  true when it is used to describe or characterize some unique physical properties of stars generally in the universe. 'Average' in this sense is used to define a typical sun such as, "stars should glow like our sun an average star."

The statement is used wrongly when used to in quantifying other stars in the universe, based on calculated values from our sun. In this case, we cannot truly say if our sun is a true representative average of other stars in the universe.

Yes! it is useful to characterize the milky way by simply citing the average sun. Properties like their ability to glow and radiate heat can be defined by citing an average star like our sun, so long as we don't translate it into citing quantitative properties of the sun as an average of our Milky Way Galaxy like the mass, temperature, etc.

Answer:

The power required by the pump is nearly 230.588 kW

Explanation:

Flow rate of the pump Q = 1 m^3/s

the head flow H = 20 m

specific weight of water γ = 9800 N/m^3

efficiency of the pump η = 85%

First note that specific gravity of water is the product of the density of water and acceleration due to gravity.

γ = ρg

where ρ is density. For water its value is 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

The power to lift this water at this rate will be gotten from the equation

P = ρgQH

but ρg = γ

therefore,

P = γQH

imputing values, we'll have

P = 9800 x 1 x 20 = 196000 W

But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.

we can say that

196000 W is 85% of the power of the pump power needed, therefore

196000 = 85% of [tex]P_{p}[/tex]

where [tex]P_{p}[/tex] is the power of the pump needed

85% = 0.85

196000 = 0.85[tex]P_{p}[/tex]

[tex]P_{p}[/tex] = 196000/0.85 = 230588.24 W

Pump power = 230.588 kW

Answer:

[tex]500\text{N} (490\text{N}) (490.5\text{N})[/tex]

Explanation:

The reaction force is the force that is in the perpendicular direction to the wall.

We have an angle and a hypotenuse, we need to find the adjacent angle - so we can just use cos:

[tex]cos(\theta)=\frac{\text{adj}}{\text{hyp}}\\\text{hyp}*cos(\theta)=\text{adj}\\100*cos(60)=100*0.5=50\text{kg}[/tex]

However, we would like a force and not a mass.

[tex]W=mg\\W=50g\\W=500\text{N} (490\text{N}) (490.5\text{N})[/tex]

Answer 1 if you use g as 10, answer 2 if you're studying mechanics in maths, answer 3 if you're studying mechanics in physics.

Answer:

C. The magnitude of the electron's acceleration inside the field

D. The radius of the circular path the electron travels

Explanation:

The radius of the electron's motion in a uniform magnetic field is given by

[tex]R = \frac{MV}{qB}[/tex]

where;

m is the mass of the electron

q is the charge of the electron

B is the magnitude of the magnetic field

V is speed of the electron

R is the radius of the electron's

Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.

The magnitude of the electron's acceleration inside the field  is given by;

[tex]a_c = \frac{V^2}{R}[/tex]

where;

[tex]a_c[/tex] is centripetal acceleration of electron

Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.

The time the electron is in the magnetic field is given by;

[tex]T = \frac{2\pi M}{qB}[/tex]

The time of electron motion will not change

The magnitude of the net force acting on the electron inside the field will not change;

[tex]qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R} = 0[/tex]

Therefore, the correct options are "C" and "D"

Answer:

x  =  0.27 cm

Explanation:

given data

mass = 150 g

velocity = 10 cm/s = 0.1 m/s

spring constnat = 200 N/m

solution

as we know that here ball is moving with constant speed

so

0.5 × m × v² = 0.5 × k × x²      .......................1

here x is compression in spring

so put here value and we get

0.5 × 150 × (0.1)² = 0.5 × 200 × x²

solve it we get

x  =  0.27 cm

Answer:

2

Explanation:

since the second object was in it stationary, it velocity is 0 m/s

Answer:

Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.

a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of  [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)

Construct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.

The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:

[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].

The question described another equation for the net force on this mass. This equation is the sum of two parts:

Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:

[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].

Combine the two equations for the net force on this mass to obtain:

[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].

From the question:

Hence, the equation will become:

[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].

Rearrange to obtain:

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:

[tex]m^2 + 11\, m + 18 = 0[/tex].

The two roots are:

Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:

[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].

For this particular ODE, that general solution would be:

[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].

Note, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time

For section [tex]\rm a.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm a.[/tex] will be:

[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

Similarly, for section [tex]\rm b.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm b.[/tex] will be:

[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let the first wave is :

[tex]y_1=A\sin\omega t[/tex]

And another wave is :

[tex]y_2=A\sin (\omega t+\phi)[/tex]

[tex]\phi[/tex] is phase difference between waves

Let y is the resultant of these two waves. So,

[tex]y =y_1+y_2[/tex]

The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,

[tex]\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi[/tex]

So, the phase difference between the two waves is [tex]\pi[/tex].

Answer:

D &B

Explanation:

Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force

Answer:

Explanation:

We shall apply the theory of

heat lost = heat gained .

heat lost by water = mass x specific heat x temperature diff

= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J  

heat gained by ice to attain temperature of zero

= m x 2100 x 22.8 = 47880 m

heat gained by ice in melting = latent heat x mass

= 334000m

heat gained by water at zero to become warm at 32 degree

= m x 4190 x 32 = 134080 m

Total heat gained = 515960 m

So

515960 m = 51587.28

m = .1 kg

= 100 gm

Answer:

The  velocity is [tex]v = 37 .46 \ m/s[/tex]

Explanation:

From the question we are told that

    The start distance above the ground is  [tex]h = 71.6 \ m[/tex]

Generally according to the  law of energy conservation we have that

     [tex]PE_{top} = KE_{bottom }[/tex]

Where [tex]PE_{top}[/tex] is potential energy at the top which is mathematically represented as

      [tex]PE_{top} = m * g * h[/tex]

And  [tex]KE_{bottom }[/tex] is the kinetic energy at the bottom which is mathematically represented as

     [tex]KE_{bottom } = \frac{1}{2} * m * v^2[/tex]

Therefore  

       [tex]m * g * h = \frac{1}{2} * m * v^ 2[/tex]

=>    [tex]v = \sqrt{2 * g * h }[/tex]

substituting value

     [tex]v = \sqrt{2 * 9.8 * 71.6 }[/tex]

     [tex]v = 37 .46 \ m/s[/tex]

Answer:

20J

Explanation:

The computation of the change in kinetic energy is shown below:

As we know that

Work was done on system =  change in potential energy + change in kinetic energy + change in thermal energy

-430J = -510J + change in kinetic energy + 100J

-430J = -410J + change in kinetic energy

So, the change in kinetic energy is 20J

We simply applied the above formula to find out the change in kinetic energy

Answer:

fully describes the concave mirror is + f

Explanation:

A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f

This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di

Answer:

+f

Explanation:

because you have to be really dumb to get an -f

Answer:

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = [tex]\frac{\delta m \lambda d}{a}[/tex] where;

[tex]\delta m[/tex] is the first two diffraction minima = 1

[tex]\lambda[/tex] is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given [tex]\lambda[/tex] = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

Answer:

Plane spacing, [tex]d=1.4\times 10^{-10}\ m[/tex]

Explanation:

It is given that,

Wavelength of x-ray, [tex]\lambda=1.4\times 10^{-10}\ m[/tex]

Angle the x-ray made with a set of planes in a crystal causing first order constructive interference is 30 degrees

We need to find the plane spacing. It is based on Bragg's law such that,

[tex]2d\sin\theta=n\lambda[/tex]

d is plane spacing

n = 1 here

[tex]d=\dfrac{\lambda}{2\sin\theta}\\\\d=\dfrac{1.4\times 10^{-10}}{2\times \sin (30)}\\\\d=1.4\times 10^{-10}\ m[/tex]

So, the plane spacing is [tex]1.4\times 10^{-10}\ m[/tex].

Answer:

a) The potential difference should be applied to the d dimension face.

b) The maximum current density j = V/3ρd

c) 3. to the faces that are a distance 3d apart

Explanation:

a) Current density is the ratio of current flowing through a conductor, and cross-sectional area of the conductor. mathematically, it is written as

j = I/A

where I is the electric current, and

A is the area of the conductor.

From the equation, we can see that reducing the area of the conductor will increase the current density for a given amount of current passing through the conductor. The face d wide will give the least cross-sectional area of current flow.

b) current density can be gotten from

j = σE    ....equ 1

where σ is the conductivity of the conductor which is the inverse of resistivity ρ. this means that

σ = 1/ρ    ....equ 2

where ρ is the resistivity of the conductor

E is the electric field and is the volt through the conductor per unit length of the conductor

in this case, the maximum current density will be when the length is length 3d, and the volt is the potential difference V

therefore,

E = V/3d    ....equ 3

substituting equ 2 and equ 3 in equ1, we'll have

the maximum current density j = V/3ρd

c) To get the maximum current, the potential difference should be applied to the faces that are 3d wide apart because the resistance of a conductor varies inversely as the cross-sectional area. The maximum current will be gotten when the resistance is at its minimum, and the minimum resistance will be gotten with the most cross-sectional area. The 3d wide face will give the maximum cross-sectional area.

Answer:

Pencil is to pen

Step by step explanation:

They are similar items, as they are both tools, but are different as to how they function.