Find the rms (a) electric and (b)magnetic fields at a point 2.00 m from a lightbulb that radiates 90.0 W of light uniformly in all directions.

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

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.

Read more at; https://brainly.com/question/18451537

Answer and Explanation:

When an electronic appliance such as electric heater, electric stove e.t.c is rated, the rating actually specifies the ideal working properties of the appliance. For example if it is rated 200W, 220V, it shows that the power the appliance will consume at a voltage of 220V is 200W.

Therefore, for the electronic stove mentioned with a rating of 1000W, 240V, the stove will consume or draw a power of 1000 watts at a voltage of 240volts.

Ratings can also help determine some other properties of the appliance such as current consumption and resistance in the appliance. For the given electronic stove, the current consumed can be found by using the following relation:

P = IV                 -------------(i)

Where;

P = Power rating = 1000W

I = Current used

V = Voltage rating = 240V

Substituting these values into equation (i) gives;

1000 = I x 240

I = [tex]\frac{1000}{240}[/tex] = 4.17A

Therefore, the current used by the stove is 4.17A

To get the resistance R of the stove, we can use the relation;

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

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

R = [tex]\frac{240^2}{1000}[/tex]

R = 57.6Ω

Therefore, the resistance of the stove is 57.6Ω

Answer:

a)    a = 17.1 m / s², b)    F = 3.04 N

Explanation:

This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities

* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components

* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components

We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is

                θ = 10 -30 = -20º

The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force

              sin (-20) = F_{y} / F

              cos (-20) = Fₓ / F

              F_{y} = F sin (-20)

              Fₓ = F cos (-20)

              F_y = 18 sin (-20) = -6.16 N

              Fₓ = 18 cos (-20) = 16.9 N

The decomposition of the weight is the customary

               sin 30 = Wₓ / W

               cos 30 = W_y / W

               Wₓ = W sin 30 = mg sin 30

                W_y = W cos 30 = m g cos 30

                Wₓ = 0.8 9.8 sin 30 = 3.92 N

                 W_y = 0.8 9.8 cos 30 = 6.79 N

Notice that in the case  the angle is measured with respect to the axis y perpendicular to the plane

Now we can write Newton's second law for each axis

X axis

      Fₓ - fr = m a

Y Axis  

      N - [tex]F_{y}[/tex] - Wy = 0

      N =F_{y} + Wy

      N = 6.16 + 6.79

They both go to the negative side of the axis and

      N = 12.95 N

The friction force has the formula

        fr = μ N

we substitute

        Fₓ - μ N = m a

        a = (Fₓ - μ N) / m

we calculate

       a = (16.9 - 0.25 12.95) / 0.8

       a = 17.1 m / s²

b) now the block slides down with constant speed, therefore the acceleration is zero

ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced

       Newton's law for the x axis

              Fₓ -fr = 0

              Fₓ = fr

              F cos 20 = μ N

              F = μ N / cos 20

we calculate

              F = 0.25 12.95 / cos 20

              F = 3.04 N

this is the force applied at an angle of 10º to the horizontal

Answer:

1.32 is the turns ratio

Explanation:

Note that The transformer steps up the voltage from 12000 V to 390000V

12000 V is the primary and in the secondary it is 390000 V in old transformer

If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils

the formula is

n₂ / n₁ = voltage in secondry / voltage in primary

n₂ / n₁ = 390000 / 12000

ratio of turns in old transformer is 32.5

ratio of turns in new transformer

n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )

= 42.9

T he ratio of turns in the new secondary compared with the old secondary

n₃ / n₂ = 42.9 / 32.5

= 1.32

Answer:

The  wavelength is  [tex]\lambda = 1029 nm[/tex]

Explanation:

From the question we are told that

    The  wavelength of the left light is  [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]

      The  voltage across A  and  B is  [tex]V_{AB } = 1.2775 \ V[/tex]

Let the stopping potential  at A  be [tex]V_A[/tex] and the electric potential at B  be  [tex]V_B[/tex]

The voltage across A and B is mathematically represented as

      [tex]V_{AB} = V_A - V_B[/tex]

Now  According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side  in terms of electron volt is mathematically represented as

        [tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]

Where  W is the work function of the metal

             h is the Planck constant with values  [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]

             c  is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

And  the stopping potential at B for the ejected electron from the right side  in terms of electron volt is mathematically represented as

          [tex]eV_B = \frac{h * c}{\lambda } - W[/tex]

So  

      [tex]eV_{AB} = eV_A - eV_B[/tex]

=>    [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]

=>   [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]

=>   [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]

=>  [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]

Where e is the charge on an electron with the value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>   [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]      

=>  [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]  

=>   [tex]\lambda = 1.029 *10^{-6} \ m[/tex]

=>   [tex]\lambda = 1029 nm[/tex]

Answer:

    f= 4,186  10²  Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

             w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El  momento de inercia de indican que giran un eje que pasa                 por enronqueces

           I= ½ M R2  

reduzcamos las cantidades al sistema SI

         R= 1,4 cm = 0,014  m

         M= 430 g = 0,430 kg

substituimos

           w= √ (2 k/M R2)

calculemos  

           w = RA ( 2 370 / (0,430  0,014 2)

           w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

            w =2pi f

            f= w/2π

            f= 2,963 10³/ (2π)

            f= 4,186  10²  Hz

Answer:

The angle between the polarizing axes of the two polarizers is 54°

Explanation:

Given;

intensity of unpolarized light, I₀ = 54.0 W/m²

intensity of light that emerges from second ideal polarizer, I₁ = 19.0 W/m²

The angle between the polarizing axes of the two polarizers is dtermined by applying Malus' law for intensity of a linearly polarized light passing through a polarizer.

I₁ = I₀Cos²θ

Cos²θ = I₁ / I₀

Cos²θ = 19 / 54

Cos²θ =0.3519

Cos θ = √0.3519

Cos θ = 0.5932

θ = Cos⁻¹(0.5932)

θ = 53.6°

θ = 54°

Therefore, the angle between the polarizing axes of the two polarizers is 54°

Answer:

The time interval is [tex]t = 5.48 *10^{-3} \ s[/tex]

Explanation:

From the question we are told that

   The length of the string is  [tex]l = 3.00 \ m[/tex]

    The  mass of the string is [tex]m = 5.00 \ g = 5.0 *10^{-3}\ kg[/tex]

     The  tension on the string is  [tex]T = 500 \ N[/tex]

The  velocity of the pulse is mathematically represented as

      [tex]v = \sqrt{ \frac{T}{\mu } }[/tex]

Where [tex]\mu[/tex] is the linear density which is mathematically evaluated as

       [tex]\mu = \frac{m}{l}[/tex]

substituting values

     [tex]\mu = \frac{5.0 *10^{-3}}{3}[/tex]

     [tex]\mu = 1.67 *10^{-3} \ kg /m[/tex]

Thus  

     [tex]v = \sqrt{\frac{500}{1.67 *10^{-3}} }[/tex]

    [tex]v = 547.7 m/s[/tex]

The time taken is evaluated as

    [tex]t = \frac{d}{v}[/tex]

substituting values

      [tex]t = \frac{3}{547.7}[/tex]

      [tex]t = 5.48 *10^{-3} \ s[/tex]

Answer:

a) 2.278 x 10^-5 volts

b) 1.139 x 10^-6 Ampere

c) 2.59 x 10^-11 W

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = [tex]\pi r^{2}[/tex]

==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

[tex]I[/tex] = E/R

where R is the resistor

[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere

c) power delivered to the resistor is given as

P = [tex]I[/tex]E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W

Answer:

Infrared telescope and camera

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

Explanation:

ω = 45 rev/min × (2π rad/rev) × (1 min / 60 s) = 4.71 rad/s

r = 10 cm = 0.10 m

1) The linear speed is:

v = ωr

v = (4.71 rad/s) (0.10 m)

v = 0.471 m/s

2) The linear acceleration in the tangential direction is 0.

The linear acceleration in the radial direction is:

a = v² / r

a = (0.471 m/s)² / (0.10 m)

a = 2.22 m/s²

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:

Electric potential at position, x = -20 m, = -100 V

Electric potential at position, x = 0 m, = 0

Electric potential at position, x = 10 m, = 50 V

Electric potential at position, x = 11 m, = 55 V

Electric potential at position, x = 99 m, 495 V

Explanation:

Given;

magnitude of electric field, E = 5.0 V/m

at position x = 10 m, electric potential = 50 V

Electric potential at position, x = -20 m

V = Ex

V = 5 (-20)

V = -100 V

Electric potential at position, x = 0 m

V = Ex

V = 5(0)

V = 0

Electric potential at position, x = 10 m

V = Ex

V = 5(10)

V = 50 V

Electric potential at position, x = 11 m

V = Ex

V = 5(11)

V = 55 V

Electric potential at position, x = 99 m

V = Ex

V = 5(99)

V = 495 V

Answer:

Δx = 9 x 10⁻³ m = 9 mm

Explanation:

The formula for fringe spacing in Young's Double Slit Experiment is given as follows:

Δx = λL/d

where,

Δx = fringe spacing = ?

λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m

L = Distance between slits and screen = 2 m

d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m

Therefore,

Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)

Δx = 9 x 10⁻³ m = 9 mm

Answer:

P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.

Explanation:

The force, F on a conductor of length, L in a magnetic field of magnetic field strength, B and current, I flowing through it is given by

F = BIL.

Now, if the conductor has a velocity, v, the energy dissipated by this force is P

P = Fv = BIL × v = BILv.

Now, we know that the induced e.m.f due to the motion of the conductor is given by ε = BLv.

From P above, P = BILv = I(BLv)

substituting ε = BLv into P, we have

P = Iε

Thus, P is the electrical energy dissipated due to the motion of the conductor.

Now since P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.

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:

F = 213.75 N

Explanation:

First we need to calculate the angular acceleration of merry-go-round. For that purpose we use 1st equation of motion in angular form.

ωf = ωi + αt

where,

ωf=final angular velocity=(5.5 rev/min)(2π rad/1 rev)(1 min/60 s)=0.58 rad/s

ωi =initial angular velocity = 0 rad/s

t = time = 12 s

α = angular acceleration = ?

Therefore,

0.58 rad/s = 0 rad/s + α(12 s)

α = (0.58 rad/s)/(12 s)

α = 0.05 rad/s²

Now, we shall find the linear acceleration of the merry-go-round:

a = rα

where,

a = linear acceleration = ?

r = radius = 4.5 m

Therefore,

a = (4.5 m)(0.05 rad/s²)

a = 0.225 m/s²

Now, the force is given by Newton;s 2nd Law:

F = ma

where,

F = Force = ?

m = mass pf merry-go-round = 950 kg

Therefore,

F = (950 kg)(0.225 m/s²)

F = 213.75 N

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

[tex]m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v[/tex]

Where:

[tex]m_{M}[/tex], [tex]m_{S}[/tex] - Masses of the small meteorite and the communication satellite, measured in kilograms.

[tex]v_{M}[/tex], [tex]v_{S}[/tex] - Speeds of the small meteorite and the communication satellite, measured in meters per second.

[tex]v[/tex] - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

[tex]v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}[/tex]

If [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex] and [tex]v_{S} = 0\,\frac{m}{s}[/tex], the final speed is now calculated:

[tex]v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}[/tex]

[tex]v = 0.1\,\frac{m}{s}[/tex]

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

[tex]K_{S} + K_{M} - K - Q_{disp} = 0[/tex]

[tex]Q_{disp} = K_{S}+K_{M}-K[/tex]

Where:

[tex]K_{S}[/tex], [tex]K_{M}[/tex] - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

[tex]K[/tex] - Kinetic energy of the satellite-meteorite system, measured in joules.

[tex]Q_{disp}[/tex] - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

[tex]Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}][/tex]

Given that [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex], [tex]v_{S} = 0\,\frac{m}{s}[/tex] and [tex]v = 0.1\,\frac{m}{s}[/tex], the dissipated heat is:

[tex]Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right][/tex][tex]Q_{disp} = 200000\,J[/tex]

[tex]Q_{disp} = 200\,kJ[/tex]

The energy coverted to heat is 200 kilojoules.

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:

The refractive index of fluid 2 is 1.78

Explanation:

Refractive index , n = real depth/apparent depth

For the first fluid, n = 1.37 and apparent depth = 9.00 cm.

The real depth of the container is thus

real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm

To find the refractive index of fluid index of fluid 2, we use the relation  

Refractive index , n = real depth/apparent depth.

Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.

So, n = 12.33 cm/6.86 cm = 1.78

So the refractive index of fluid 2 is 1.78

Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8

Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,

Then,

Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]

Real depth = Index of refraction x apparent depth

Since the same container is used, we can make an assumption that;

real depth of fluid 1 = real depth of fluid 2

That is,

1.37 x 9 = n x 6.86

Where n = Index of refraction for the second fluid.

make n the subject of formula

n = 12.33 / 6.86

n = 1.79

Therefore, the index of refraction of the second fluid is 1.8 approximately.

Learn more about refraction here: https://brainly.com/question/10729741

Answer:

d. if any of the above occurs

Explanation:

That is The number of fringes per unit length on the screen can be doubled if

if the distance between the slits is doubled.

if the wavelength is changed to λ = λ/2.

And if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

Answer:

C. The spinning of an object on its axis.

Explanation:

The definition of rotation answers this question.

A good way to remember is that when you rotate, you turn, and when you revolve, you move around.

Answer: A) [tex]P_{x}[/tex] = 564.4 N

              B) [tex]P_{y}[/tex] = 374 N

Explanation: The ladder forms with the wall a right triangle, with one unknown side. To find it, use Pythagorean Theorem:

[tex]hypotenuse^{2} = side^{2} + side^{2}[/tex]

[tex]side = \sqrt{hypotenuse^{2} - side^{2}}[/tex]

side = [tex]\sqrt{3^{2} - 2.5^{2}}[/tex]

side = 1.65

Tom's weight is a vector pointing downwards. Since he is at an angle to the floor, the gravitational force has two components: one that is parallel to the floor ([tex]P_{x}[/tex]) and othe that is perpendicular ([tex]P_{y}[/tex]). These two vectors and weight, which is gravitational force, forms a right triangle with the same angle the ladder creates with the floor.

The image in the attachment illustrates the described above.

A) [tex]P_{x}[/tex] = P sen θ

[tex]P_{x} = P.\frac{oppositeside}{hypotenuse}[/tex]

[tex]P_{x}[/tex] = 680.[tex]\frac{1.65}{3}[/tex]

[tex]P_{x}[/tex] = 564.4 N

B) [tex]P_{y}[/tex] = P cos θ

[tex]P_{y} = P.\frac{adjacentside}{hypotenuse}[/tex]

[tex]P_{y}[/tex] = 680. [tex]\frac{1.65}{3}[/tex]

[tex]P_{y}[/tex] = 374 N

Answer:

The fraction of light that escapes the water surface as the water moves deeper into the water will decrease.

Explanation:

The speed of light in water is small compared to the speed of light in air, and a larger part of the light energy is absorbed in water than in air. When the bulb is immersed in water, some of the light energy is absorbed by the mass of water. When the light bulb is further moved deeper into the water, the fraction of light that escapes decreases, because more mass of water is made available to absorb more of the light energy from the bulb.

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:

5.95 A

Explanation:

From the question

R = ρL/A..................... Equation 1

Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.

Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m

Substitute these values into equation 1

R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶

R = 0.084 Ω.

Finally, using Ohm law,

V = IR

Where V = Voltage, I = current

Make I the subject of the equation

I = V/R............... Equation 2

I = 0.5/0.084

I = 5.95 A

Explanation:

Work = force × distance

W = (400 N) (10 m)

W = 4000 J