What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.630 m long and has a mass of 5.69 g.

Answers

Answer 1

Answer:

220.698Hz

Explanation:

The fundamental frequency f₀ is expressed as f₀ =V/2L where;

V is the speed of the string = [tex]\sqrt{\frac{T}{M} }[/tex]

m is the mass of the string

L is the length of the string

T is the tension in the string

f₀ = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

Given datas

m = 5.69g = 0.00569 kg

T = 440N

L = 0.630 m

Required

Fundamental frequency of the steel piano wire f₀

[tex]f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz[/tex]

Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz


Related Questions

(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.


pF

(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV

Answers

Explanation:

(a) Given that,

Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]

The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]

The dielectric constant of, k = 2.1

When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :

[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]

Putting all the values we get :

[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]

(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]

The voltage difference between the plates at this critical voltage is given by :

[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]

or

V = 0.6 kV

We have that the Capacitance and potential difference is mathematically given as

[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF



Capacitance &potential difference

Question Parameters:

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.

a)

Generally the equation for the Capacitance  is mathematically given as

[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]

C=334.68pF

b)

Generally the equation for the Capacitance  is mathematically given as

[tex]Vmax=\frac{Q}{C}[/tex]

Where

Q is the charge on the plates, and hence not given

Therefore, maximum potential difference is

[tex]Vmax=\frac{Q}{334.68pF}[/tex]

For more information on potential difference visit

https://brainly.com/question/14883923

A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What happens after he pulls his arms inwards

Answers

Answer:

His angular velocity will increase.

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = [tex]I[/tex]'ω'

where

[tex]I[/tex]' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = [tex]mr'^{2}[/tex]

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

since [tex]I[/tex] is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

Two narrow slits are illuminated by a laser with a wavelength of 593 nm. The interference pattern on a screen located x = 4.80 m away shows that the fourth-order bright fringe is located y = 8.20 cm away from the central bright fringe. Calculate the distance between the two slits.

Answers

Answer:

The distance is  [tex]d = 1.39 *10^{-4} \ m[/tex]

Explanation:

From the question we are told that

   The wavelength is  [tex]\lambda = 593 \ nm = 593 *10^{-9} \ m[/tex]

    The distance of the screen is   x  =  4.80 m

    The  location of the fourth order bright fringe is  y  =  8.20 cm = 0.082 m

    The order of the fringe is  n  =  4

   

Generally the position of a fringe with respect to the central fringe is mathematically represented as

           [tex]y = \frac{ n * x * \lambda }{d}[/tex]

Where d is the distance between the slits, so making d the subject

          [tex]d = \frac{\lambda * x * n }{ y }[/tex]

substituting values

          [tex]d = \frac{ 593 *10^{-9} * 4.80 * 4 }{ 0.082 }[/tex]

           [tex]d = 1.39 *10^{-4} \ m[/tex]

A bicycle has wheels that are 60 cm in diameter. What is the angular speed of these wheels when it is moving at 4.0 m/s

Answers

Answer:

13.33 rad/s

Explanation:

Applying,

v = ωr......................... Equation 1

Where v = linear speed, ω = angular speed and r = radius.

Note that,

r = d/2................. Equation 2

Where d = diameter of the wheel.

Substitute equation 2 into equation 1

v = ωd/2............... Equation 3

make ω the subject of the equation

ω = 2v/d................ Equation 4

Given: v = 4 m/s, d = 60 cm = 0.6 m

Substitute these values into equation 4

ω = 2(4)/0.6

ω = 13.33 rad/s

Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion​

Answers

Last point, Newton’s first law of motion is the correct answer.

Hope this helps ya

Why was Bohr's atomic model replaced by the
modern atomic model?

Answers

Answer:

Explanation:

Bohr's atomic model was replaced by the  modern atomic model because of its limitations, which included :

(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )

(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).

(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.

For a beam of light in air (n = 1) reflecting off glass (n = 1.5), what is Brewster's angle to the nearest degree?

Answers

Answer: 56°

Explanation:

Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.

The formula is;

[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]

[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]

= 56.30993247

= 56°

A 1.5 V battery is connected to a 1000 ohm resistor and a 500 ohm resistor in series. The voltage across the 1000 ohm resistor is _____ V.

Answers

Answer:

1 volt and 0.5 volt

Explanation:

Given data

voltage supplied Vs= 1.5 volts

resistance R1= 1000 ohms

resistance R2= 500 ohms

The total resistance is

Rt= 1000+ 500

Rt= 1500 ohms

The current I is given as

[tex]I= \frac{Vs}{Rt} \\\\ I= \frac{1.5}{1500} = 0.001mA[/tex]

Voltage across R1

[tex]VR1= Vs(\frac{R1}{R1+R2} )=1.5(\frac{1000}{1000+500} )= 1.5(\frac{1000}{1500} )\\ \\\ VR1= 1v[/tex]

Voltage across R2

[tex]VR2= Vs(\frac{R2}{R1+R2} )=1.5(\frac{500}{1000+500} )= 1.5(\frac{500}{1500} ) \\\ VR2=0.5v[/tex]

In series connection the current is the same for all components while the voltage divides across all components,the voltages consumed by each individual resistance is equal to the source voltage.

At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m

Answers

Answer:

a) The value of g at such location is:

[tex]g=9.8005171\,\frac{m}{s^2}[/tex]

b) the period of the pendulum with the length is 0.970 m is:

[tex]T=1.9767 sec[/tex]

Explanation:

Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under  an acceleration of gravity g:

[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]

a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:

[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]

b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:

?a wire is stretched 30% what is the percentage change in resistance ​

Answers

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]

So that;

[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])

    = (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])

    = [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]

   = 1.69R               (∵ R = (ρl) ÷ A)

[tex]R_{o}[/tex] = 1.69R

Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.

The change in resistance of the wire = [tex]R_{o}[/tex] - R

                                      = 1.69R  - 1R

                                      = 0.69R

The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100

                                                               = 0.69 × 100

                                                              = 69%

The percentage change in resistance of the wire is 69%.

a ball is kicked on level ground with a speed of 30 m/s at angle of 40 degrees above horizontal g Find the minimum velocity of the ball during its flight

Answers

Answer:

The minimum velocity of the ball during its flight is 22.98 m/s.

Explanation:

The velocity of the ball v = 30 m/s

The angle it makes with the horizontal ∅ = 40°

The minimum velocity of the ball during flight will be the horizontal axis component of the velocity, as acceleration is zero on this axis.

[tex]V_{x}[/tex] = v cos ∅

[tex]V_{x}[/tex]  = 30 cos 40°

[tex]V_{x}[/tex] = 30 x 0.766 = 22.98 m/s

Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form _______. A. pure elements B. molecules C. metals D. the periodic table

Answers

Answer:

Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form pure elements

The answer is option A

Answer:

its molecues

Explanation:

In the summer of 2010 a huge piece of ice roughly four times the area of Manhattan and 500 m thick caved off the Greenland mainland.

Required:
a. How much heat would be required to melt this iceberg (assumed to be at 0°C) into liquid water at 0°C?
b. The annual U.S. energy consumption is 1.2 x 10^20 J. If all the U.S. energy was used to melt the ice, how many days would it take to do so?

Answers

Answer:

a

  [tex]Q = 5.34 *10^{19} \ J[/tex]

b

   [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Explanation:

From the question we are told that

     The  area of  Manhattan is  [tex]a_k = 87.46 *10^{6} \ m^2[/tex]

      The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]

        The  thickness is  [tex]t = 500 \ m \\[/tex]

       

Generally the volume of the ice is mathematically represented is

         [tex]V = a_i * t[/tex]

substituting value

         [tex]V = 500 * 3.498*10^{8}[/tex]

         [tex]V = 1.75 *10^{11}\ m^3[/tex]

Generally the mass of the ice is

       [tex]m_i = \rho_i * V[/tex]

Here [tex]\rho_i[/tex] is the density of ice the value is  [tex]\rho _i = 916.7 \ kg/m^3[/tex]

=>   [tex]m_i = 916.7 * 1.75*10^{11}[/tex]

=>    [tex]m_i = 1.60 *10^{14} \ kg[/tex]

Generally the energy needed for the ice to melt is mathematically represented as

        [tex]Q = m _i * H_f[/tex]

Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is  [tex]H_f = 3.33*10^{5} \ J/kg[/tex]

=>    [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]

=>    [tex]Q = 5.34 *10^{19} \ J[/tex]

Considering part b

  We are told that the annual energy consumption is  [tex]G = 1.2*10^{20 } \ J / year[/tex]

So  the time taken to melt the ice is

      [tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]

        [tex]T = 0.445 \ years[/tex]

converting to days

      [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sphere is then grounded as shown. Then the ground wire is removed. A is the inner surface and B is the outer surface. Which statement is correct

Answers

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

Bob and Lily are riding on a merry-go-round. Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform. When the merry-go-round is rotating at a constant angular speed, Bob's angular speed is:_____.
a. the same as Lily's.
b. larger than Lily's.
c. exactly half as much as Lily's.
d. exactly twice as much as Lily's.
e. smaller than Lily's.

Answers

Answer: the same as Lily's

Explanation:

Angular velocity has to do with the speed at which an object will be able to rotate. We are informed that Bob and Lily are riding on a merry-go-round.

Since we are further told that Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform and that he merry-go-round is rotating at a constant angular speed.

Based on the above analysis, Bob's angular speed will be thesame as that of Lily.

¿Cómo podrías utilizar el sistema de posicionamiento global para proponer recorridos alternativos para llegar a un lugar específico (centro educativo, supermercado, el hogar, el parque, entre otros)?

Answers

Answer:

El sistema de posicionamiento global (conocido mundialmente como GPS) podría utilizarse para proponer recorridos alternativos para llegar a un lugar específico, como un parque, a través de la creación de un recorrido guiado por una aplicación móvil con diferentes rutas de acceso al lugar.

Así, por ejemplo, se crearían diferentes rutas de acceso desde un punto A hasta un punto B, teniendo en cuenta factores como: rapidez, congestión vehicular, pago o no de peajes, posibilidad de acceso a pie y determinados factores extra que influyan en la forma de llegar al lugar. Todo ello plasmado en un mapa interactivo en el cual se señalen las rutas disponibles mediante el marcado del mapa en cuestión.

what conventions are used in SI to indicate units

Answers

Answer:

Conventions used in SI to indicate units are as follows:

Only singular form of units are used. for example: use kg and not kgs.Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.Units named after a scientist should be written in small letters. for example: newton, henry.Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k

A diffraction grating with 200 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At what angle from the beam axis will the first order peak occur if the tube emits light with wavelength of 617.3 nm

Answers

Answer

123.5 x 10 ^-3 radian

Explanation:

Given the Width of slit a = 1 x 10⁻³ / 200

a = 5x 10⁻⁶ m .

angle at which first order peak is formed

= λ / a (where λ is wavelength and a is width of slit)

given λ = 617.3 x 10⁻⁹ m

a = 5 x 10⁻⁶

θ = 617.3 x 10⁻⁹ / 5 x 10⁻⁶

= 123.5x 10⁻³ radian .

first order peak is formed at an angle of 123.5 x 10⁻³ radian .

Explanation:

If an electromagnetic wave has components Ey = E0 sin(kx - ωt) and Bz = B0 sin(kx - ωt), in what direction is it traveling?

Answers

Answer:

Its traveling in the +x direction

Explanation:

The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.

The electromagnetic wave is travelling in the +x direction.

Electromagnetic waves are waves formed as a result of vibrations between an electric field and a magnetic field.

Given that:

Ey = E0 sin(kx - ωt)

Hence the electric field is moving in the +y direction.

Also, Bz = B0 sin(kx - ωt)

Hence the magnetic field is moving in the +z direction

Electric fields and magnetic fields (E and B) in an EM wave are perpendicular to each other and are also perpendicular to the direction of propagation of the wave.

Therefore the direction of the wave is travelling in the +x direction.

Find out more at: https://brainly.com/question/25559554

A 300 MWe (electrical power output) Power Plant having a thermal efficiency of 40% is cooled by sea water. Due to environmental regulations the seawater can only increase temperature by 5 C during the process. How much sea water (minimum) must be used in kg/s for cooling if the plant operates at it's rated capacity?

Answers

Answer:

m = 22,877 kg / s

Explanation:

Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat

The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%

           Q = 300 60% / 40%

           Q = 450 MW

having the amount of heat generated we can use the calorimeter equation,

           Q = m [tex]c_{e}[/tex] [tex](T_{f} - T_{o})[/tex]

            m = Q / c_{e} (T_{f} - T_{o})

let's use the maximum temperature change allowed

           (T_{f} - T_{o}) = 5

the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water

     

power and energy are related

              W = Q / t

               Q = W t

             

let's calculate

             m = 450 10⁶ / (3934 5)

             m = 22,877 kg / s

A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.

Answers

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

Please help!
Much appreciated!​

Answers

Answer:

your question answer is 22°

I’m pretty sure the answer is 22

The same force is applied to two hoops. The hoops have the same mass, but the larger hoop has twice the radius. How are the angular accelerations of the hoops related

Answers

Answer:

The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].

Explanation:

Net force on the hoop is given by;

[tex]F_{net} = ma[/tex]

where;

a is linear acceleration

m is the mass

Net torque on the hoop is given by;

[tex]\tau_{net} =I\alpha[/tex]

where;

I is moment of inertia

α is the angular acceleration

But, τ = Fr

[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]

[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]

let the angular acceleration of the smaller hoop = α₁

let the radius of the smaller hoop = r₁

then, the radius of the larger loop, r₂ = 2r₁

let the angular acceleration of the larger hoop = α₂

[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]

Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]

A generator consists of 1000 turns of wire around a coil of area 10cm2 (0.0010m 2 ). Calculate the peak output voltage of this generator when operating at 100 cycles/second in a 0.50 tesla field.

Answers

Answer:

The peak output voltage of this generator is 314.2 V.

Explanation:

Given;

the number of turns of the coil, N = 1000 turns

area of the coil, A = 0.001 m²

angular frequency of the coil, f = 100 cycles/seconds

magnitude of the magnetic field, B = 0.5 T

The peak output voltage of this generator is given by;

E = NBAω

Where;

ω is the angular velocity = 2πf

E = NBA(2πf)

E = 1000 x 0.5 x 0.001(2 x π x 100)

E = 314.2 V

Therefore, the peak output voltage of this generator is 314.2 V.

How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m

Answers

Answer:

Explanation:

volume of water being lifted

= π r² h , where r is radius of cylinder and h is height of cylinder

= 3.14 x5² x 10

= 785 m³

mass of water = 785 x 10³ kg

mass of this much of water is lifted so that its centre of mass is lifted by height

10 / 2 = 5m .

So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity

= 785 x 10³ x 9.8 x 5

= 38.465 x 10⁶ J  

A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]

Answers

Answer:

The  apparent depth is  [tex]D' = 0.00376 \ m[/tex]

Explanation:

From the question we are told that

     The  depth of the water is  [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]

      The  refractive index of water is  [tex]n = 1.33[/tex]

     

Generally the apparent depth of the coin is mathematically represented as

          [tex]D' = D * [\frac{ n_a}{n} ][/tex]

Here  [tex]n_a[/tex]  is the refractive index of  air the value is  [tex]n_a = 1[/tex]

So

        [tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]

        [tex]D' = 0.00376 \ m[/tex]

The apparent depth will be 0.00376 m.

What is an index of refraction?

The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.

d is the depth of the water =5.0 mm =5.0 ×10⁻³

n is the refractive index of water =1.33

[tex]\rm n_a[/tex] is the refractive index of wire=1

The apparent depth of the coin is given as;

[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]

Hence the apparent depth will be 0.00376 m.

To learn more about the index of refraction refer to the link;

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A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that apply.If the container is heated, the gas particles will lose kinetic energy and temperature will increase.

Answers

Answer:

B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.

C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.

E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.

#FreeMelvin

Just wondering if I did this right

Answers

Yeah

All they are all correct

An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.440 mm.
1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
2. What would be the angular position of the second-order, two-slit, interference maxima in this case?
3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?
4. What is the intensity at the angular position of θ2?

Answers

Answer:

1) θ = 0.00118 rad, 2)  θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4)  I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

          d sin θ = m λ

1) the first order maximum occurs for m = 1

           sin θ = λ  / d

           θ = sin⁻¹ λ  / d

let's reduce the magnitudes to the SI system

           λ  = 520 nm = 520 10⁻⁹  θ = 0.00118 radm

           d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

           θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

            θ = sin⁻¹ (1.18 10⁻³)

            θ = 0.00118 rad

2) the second order maximum occurs for m = 2

            θ = sin⁻¹ (m λ  / d)

            θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

            θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

          I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

             x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

             x = 2.21

            I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

            I / I₀ = cos² (3.0945) [0.363] 2

            I / I₀ = 0.9978 0.1318

            I / I₀ = 0.1738

4) the maximum second intensity is

            I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

            x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

            x = 4.41

            I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

            I / Io = cos² 6.273    0.216

            I / Io = 0.216

.

The concentration of sodium and potassium ions in the blood and body fluid is regulated by :

Answers

Answer:

Kidney

Explanation:

One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.

Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.

Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.

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