beam of white light goes from air into water at an incident angle of 58.0°. At what angles are the red (660 nm) and blue (470 nm) parts of the light refracted? (Enter your answer to at least one decimal place.) red ° blue °

Answer:

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

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Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

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

Where;

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

[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

Answer:

The discharge rate is [tex]Q = 0.0192 \ m^3 /s[/tex]

Explanation:

From the question we are told that

   The  diameter is  [tex]d = 60 \ mm = 0.06 \ m[/tex]

    The  head is  [tex]h = 6 \ m[/tex]

     The  coefficient of contraction is  [tex]Cc = 0.68[/tex]

     The  coefficient of  velocity is  [tex]Cv = 0.92[/tex]

The radius is mathematically evaluated as

         [tex]r = \frac{d}{2}[/tex]

substituting values

        [tex]r = \frac{ 0.06 }{2}[/tex]

        [tex]r = 0.03 \ m[/tex]

The  area is mathematically represented as

      [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * (0.03)^2[/tex]

      [tex]A = 0.00283 \ m^2[/tex]

 The  discharge rate is mathematically represented as

        [tex]Q = Cv *Cc * A * \sqrt{ 2 * g * h}[/tex]

substituting values

       [tex]Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}[/tex]

       [tex]Q = 0.0192 \ m^3 /s[/tex]

Answer:

I know the answer

Explanation:

We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.

You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.

Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.

So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.

Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).

Answer:

The acceleration in the block is 2.1 m/s²

Explanation:

Given that,

Mass = 50 kg

Angle = 54°

Force = 40 N

Coefficient of friction = 0.33

We need to calculate the acceleration in the block

Using balance equation

[tex]F_{net}=F_{f}-F\cos\theta[/tex]

[tex]ma=\mu mg\sin\theta-F\cos\theta[/tex]

[tex]a=\dfrac{\mu mg\sin\theta-F\cos\theta}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{0.33\times50\times9.8\sin54-40\cos54}{50}[/tex]

[tex]a=2.1\ m/s^2[/tex]

Hence, The acceleration in the block is 2.1 m/s²

Answer:

3. the intermolecular bonding of water

Explanation:

Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.

Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.

Answer:

D. The tendency for water to freeze from the bottom up.

Explanation:

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature =  Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

ΔL = 0.66 m

Answer:

[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]

Given:

Mass of oxygen (m) = 32.0 g = 0.032 kg

Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³

Speed of sound in oxygen (v) = 317 m/s

To Find:

Bulk modulus of oxygen

Explanation:

[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]

[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]

[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]

[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]

[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]

[tex]\sf \implies B = 100489 \times 1.428[/tex]

[tex]\sf \implies B = 143498.292 \: Pa[/tex]

[tex]\sf \implies B \approx 143.5 \: kPa[/tex]

Answer:

The rms current is 0.3112 A.

Explanation:

Given that,

Suppose, The capacitance is 170 μF and the inductance is 2.94 mH. The resistance in the top branch is 278 Ohms, and in the bottom branch is 151 Ohms. The potential of the power supply is 47 V .

We know that,

When the frequency is very large then the capacitance can be treated as a short circuit and inductance as open circuit.

So,

We need to calculate the rms current

Using formula of current

[tex]I=\dfrac{V}{R}[/tex]

Where, V = voltage

R = resistance

Put the value into the formula

[tex]I=\dfrac{47}{151}[/tex]

[tex]I= 0.3112 \ A[/tex]

Hence, The rms current is 0.3112 A.

Answer:

The maximum wavelength of the e-m wave is 6.9 x 10^-7 m

Explanation:

Energy required to trigger a response = 1.8 eV

we convert to energy in Joules.

1 eV = 1.602 x 10^-19 J

1.8 eV = [tex]x[/tex] J

[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J

The energy of an electromagnetic wave is gotten as

E = hf

where

h is the Planck's constant = 6.63 x 10^-34 J-s

and f is the frequency of the wave.

substituting values, we have

2.88 x 10^-19 = 6.63 x 10^-34 x f

f = (2.88 x 10^-19)/(6.63 x 10^-34)

f = 4.34 x 10^14 Hz

We know that the frequency of an e-m wave is given as

f = c/λ

where

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength of the e-m wave

From this we can say that

λ = c/f

λ = (3 x 10^8)/(4.34 x 10^14)

λ = 6.9 x 10^-7 m

Answer:

the atom had a dense, positively charged nucleus.​

Explanation:

Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.

According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.

Answer:

C. planetary accretion

Explanation:

Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.

Answer:

[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]

Explanation:

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.

Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.

Answer:

space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

Answer:

It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J

But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/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]

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

Answer:

they must be affordable because they have to pay for it or they wont get the stuff they are bying.

Explanation:

need a brainliest please.

Answer: B, they must be affordable.

Explanation:

The researchers need to compare those who contracted the disease to those who did not.

Answer:

D Is 430m

Explanation:

See attached file

Answer: i think it is d. none of them.

Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

Answer:a) 4+8+24=36

B) 1/4+1/8+1/24=10

C) yu will connect them in parallel connection.

D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.

Explanation:

(a)The maximum resistance that can be produced using all three resistors will be 36 ohms.

(b)The minimum resistance that can be produced using all three resistors will be 10 ohms.

(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.

(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The maximum resistance that can be produced using all three resistors is obtained by adding all the given resistance;

[tex]\rm R_{max}=(4 +8+24 )\ ohms \\\\ R_{max}=36 \ ohms[/tex]

The minimum resistance that can be produced using all three resistors is obtained when connected in the parallel.

[tex]\rm R_{min}=\frac{1}{4} +\frac{1}{8} +\frac{1}{24} \\\\ R_{min}=10 \ ohm[/tex]

(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.

(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.

Hence,the maximum resistance that can be produced using all three resistors will be 36 ohms.

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Answer:

1. 0.2m

1. 66m

Explanation:

See attached file

The expressions of fluid mechanics allows to find the result for the maximum height that the water leaves through the two points are;

1) The maximum height when the water leaves the hose is: Δy = 0.20 m

2) The maximum height of the water leaves the nozzle is: Δy = 68.6m

Given parameters

To find

   1. Maximum height of water in hose

  2. Maximum height of water at the nozzle

Fluid mechanics studies the movement of fluids, liquids and gases in different systems, for this it uses two expressions:

           A₁v₁ = A₂.v₂

          P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂

Where the subscripts 1 and 2 represent two points of interest, P is the pressure, ρ the density, v the velocity, g the acceleration of gravity and y the height.

1, Let's find the exit velocity of the water in the hose.

Let's use subscript 1 for the nozzle and subscript 2 for the hose.

The continuity equation of the flow value that must be constant throughout the system.

      Q = A₁ v₁

      v₁ = [tex]\frac{Q}{A_1 }[/tex]  

The area of ​​a circle is:

     A = π r²

Let's calculate the velocity in the hose.

    A₁ = π (0.94 10⁻²) ²

    A₁ = 2.78 10⁻⁴ m²

    v₁ = [tex]\frac{0.55 \ 10^{-3}}{2.78 \ 10^{-4}}[/tex]

    v₁ = 1.98 m / s

Let's use Bernoulli's equation.

When the water leaves the hose the pressure is atmospheric and when it reaches the highest point it has not changed P1 = P2

      ½ ρ v₁² + ρ g y₁ = ½ ρ v₂² + ρ g v₂

      y₂-y₁ = ½  [tex]\frac{v_i^2 - v_2^2}{g}[/tex]  

At the highest point of the trajectory the velocity must be zero.

     y₂- y₁ = [tex]\frac{v_1^2}{2g}[/tex]

Let's calculate

     y₂-y₁ =  [tex]\frac{1.98^2}{2 \ 9.8}[/tex]  

     Δy = 0.2 m

2.  Let's find the exit velocity of the water at the nozzle

          A₁ = π r²

          A₁ = π (0.22 10⁻²) ²

          A₁ = 0.152 10⁻⁴ m / s

With the continuity and flow equation.

           Q = A v

            v₁ = [tex]\frac{Q}{A}[/tex]  

             v₁ = [tex]\frac{0.55 \ 10{-3} }{0.152 \ 10^{-4} }[/tex]  

             v₁ = 36.67 m / s

Using Bernoulli's equation, where the speed of the water at the highest point is zero.

           y₂- y₁ =  [tex]\frac{v^1^2}{g}[/tex]  

Let's calculate.

           Δy =  [tex]\frac{36.67^2 }{2 \ 9.8 }[/tex]  

           Δy = 68.6m

In conclusion using the expressions of fluid mechanics we can find the results the maximum height that the water leaves through the two cases are:

      1) The maximum height when the water leaves the hose is:

          Δy = 0.20 m

      2) The maximum height of the water when it leaves the nozzle is:

          Δy = 68.6 m

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Answer:

θ = 45.15°

Explanation:

We need to use the grating equation in this question. The grating equation is given as follows:

mλ = d Sin θ

where,

m = order number = 3

λ = wavelength of light = 520 nm = 5.2 x 10⁻⁷ m

d = slit spacing = 2.2 μm = 2.2 x 10⁻⁶ m

θ = angle from the axis = ?

Therefore,

(3)(5.2 x 10⁻⁷ m) = (2.2 x 10⁻⁶ m) Sin θ

Sin θ = (3)(5.2 x 10⁻⁷ m)/(2.2 x 10⁻⁶ m)

Sin θ = 0.709

θ = Sin⁻¹(0.709)

θ = 45.15°

Answer:

Explanation:

1 )

An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm.

Since the object is placed at a distance more than twice the focal length , its image will be inverted , real  and will be of the size less than the size of object . So option B is applied .

B)  real, inverted and smaller than the object.

2 )

An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm.

The object is placed at a point beyond its radius of curvature, its image will be formed at a point between f and C   or between focal point and centre of curvature . Its size will be smaller than size of object and it will be real and inverted .

B)  real, inverted and smaller than the object.

Answer:

Repulsion

Explanation:

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

[tex]E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A[/tex]

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

[tex]E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV[/tex]

Therefore, the average emf induced in the coil is 175 mV

Answer:

No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.

Explanation:

If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)

For a body under this type of equilibrium,

ΣF = 0   ...1

where ΣF is the resultant force (total effective force due to all the forces acting on the body)

For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as

ΣF = ma   ...2

where m is the mass of the body

a is the acceleration of the body

Substituting equation 2 into equation 1, we have

0 = ma

therefore,

a = 0

this means that if the resultant force acting on a rigid body is zero, then there won't be any force available to produce acceleration on the body.