What would you estimate for the length of a bass clarinet, assuming that it is modeled as a closed tube and that the lowest note that it can play is a D b whose frequency is 69 Hz

Answer:

e. The speed of the object is a constant 6 m/s

Explanation:

Since the net force is towards the centre , hence there is no tangential acceleration . Only centripetal acceleration is there . Hence point mass is moving with uniform speed . Let it be u .

Centripetal force = m v² / r , r is radius of circular path .

Putting the given values

.10 x v² / .36 = 10

v = 6 m /s

Answer:

a. 24 m

Explanation:

Destructive interference occurs when two waves arrive at a point, out of phase. In a completely destructive interference, the two waves cancel out, but in a partially destructive interference, they produce a wave with a time varying amplitude, but maintain a wavelength the wavelength of one of the original waves. Since the two waves does not undergo complete destructive interference, then the possible value of the new wave formed can only be 24 m, from the options given.

Answer:

Final energy = Uf = initial energy × d₂/d₁

Explanation:

Energy is the ability to do work.

capacitor is an electronic device that store charges

where

V is the potential difference

d is the distance of seperation between the two plates

ε₀ is the dielectric constant of the material used in seperating the two plates, e.g., paper, mica, glass etc.

A = cross sectional area

U =¹/₂CV²

C =ε₀A/d

C × d=ε₀A=constant

C₂d₂=C₁d₁

C₂=C₁d₁/d₂

charge will  'q' remains same in the capacitor, if the capacitor was disconnected from the electric potential source (v) before the separation of the plates was replaced

Energy=U =(1/2)q²/C

U₂C₂ = U₁C₁

U₂ =U₁C₁ /C₂

U₂ =U₁d₂/d₁

Final energy = Uf = initial energy × d₂/d₁

Answer:

Explanation:

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

Therefore the potential on the ferric surface is

        V = k Q / r

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

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

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

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

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

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

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

         E = k Q / r²

Answer:

[tex]I=1.19\times 10^{-2}\ W/m^2[/tex]

Explanation:

It is given that,

Maximum value of magnetic field strength, [tex]B=10^{-8}\ T[/tex]

We need to find the intensity of the light at that place.

The formula of the intensity of magnetic field is given by :

[tex]I=\dfrac{c}{2\mu _o}B^2[/tex]

c is speed of light

So,

[tex]I=\dfrac{3\times 10^8}{2\times 4\pi \times 10^{-7}}\times (10^{-8})^2\\\\I=1.19\times 10^{-2}\ W/m^2[/tex]

So, the intensity of the light is [tex]1.19\times 10^{-2}\ W/m^2[/tex].

Answer:

The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].

Explanation:

Let be an uniform rod of length L whose origin is located at one one end and axis is perpendicular to the rod, such that:

[tex]\lambda = \frac{dm}{dr}[/tex]

Where:

[tex]\lambda[/tex] - Linear density, measured in kilograms per meter.

[tex]m[/tex] - Mass of the rod, measured in kilograms.

[tex]r[/tex] - Distance of a point of the rod with respect to origin.

Mass differential can translated as:

[tex]dm = \lambda \cdot dr[/tex]

The equation of the moment of inertia is represented by the integral below:

[tex]I = \int\limits^{L}_{0} {r^{2}} \, dm[/tex]

[tex]I = \lambda \int\limits^{L}_{0} {r^{2}} \, dr[/tex]

[tex]I = \lambda \cdot \left(\frac{1}{3}\cdot L^{3} \right)[/tex]

[tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex] (as [tex]m = \lambda \cdot L[/tex])

The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].

Answer:

  x = -3 cm

Explanation:

The electrical potential is the sum of the potentials of each charge

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

let's apply this to our case where the potential is V = 0 for x = 0

         0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))

in our case

q₁ = + 2.0 10⁻⁶ C

q₂ = - 6.0 10⁻⁶ C

q₃ = + 3.0 10⁻⁶ C

x₁ = -1.0 cm = 1.0 10⁻² m

x₂ = +2.0 cm = 2.0 10⁻² m

we substitute in the equation

          0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + ​​3 10⁻⁶ / x)

          3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴

          3 10⁻⁶ / x = -1 10⁻⁴

           x = - 3 10⁻² m

           x = -3 cm

Answer:

2.9Ω

Explanation:

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

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

Where;

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

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

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

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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

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

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

here.

Req means  the equivalent resistance and R1, R2, R3

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

Also,

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

So,

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

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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Answer:

The horizontal acceleration of the block is 4.05 m/s².

Explanation:

The horizontal acceleration can be found as follows:

[tex] F = m \cdot a [/tex]

[tex] Fcos(\theta) - \mu_{k}N = m\cdot a [/tex]

[tex] Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a [/tex]  

[tex] a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m} [/tex]

Where:

a: is the acceleration

F: is the force exerted by the rope = 28.2 N

θ: is the angle = 30°

[tex]\mu_{k}[/tex]: is the kinetic coefficient = 0.12

m: is the mass = 5 kg

g: is the gravity = 9.81 m/s²

[tex] a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2} [/tex]

Therefore, the horizontal acceleration of the block is 4.05 m/s².

I hope it helps you!

Answer:

151.58 V

Explanation:

From the question,

The work done in a circuit in moving a charge is given as,

W = 1/2QV..................... Equation 1

Where W = Work done in moving the charge, Q = The magnitude of charge, V = potential difference between the plates.

make V the subject of the equation

V = 2W/Q.................. Equation 2

Given: W = 144 J. Q = 1.9 C

Substitute into equation 2

V = 2(144)/1.9

V = 151.58 V

Answer:

    q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

                    ΔV = Δ[tex]V_{C}[/tex]

we assume that the source has a voltage of the form

                    ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

                   C = q / ΔV

                  q = C ΔV sin wt

the current in the circuit is

                    i = dq / dt

                    i = c ΔV₀ w cos wt

if we use

                  cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

                  i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

                     V_{L} + V_{C} = 0

                      L di / dt + q / C = 0

the current is

                      i = dq / dt

they ask us for a solution so that

                    L d²q / dt² + 1 / C q = 0

                     d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

                    q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

                di / dt = Aw cos (wt + Ф)

                d²i / dt²= - A w² sin (wt + Ф)

                 - A w² + 1 /LC  A = 0

                  w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

                 0 = A sin Ф

                  Ф = 0

                  q = q₀ sin (wt)

Answer:

4.1 N/C

Explanation:

First of all, we know from maths that the surface area of a sphere = 4πr²

Charge on inner sphere ..

Q(i) = 40.0*10^-12C/m² x 4π(0.01m)²

Q(i) = 5.03*10^-14 C

Charge on outer sphere

Q(o) = 60*10^-12 x 4π(0.03m)²

Q(o) = 6.79*10^-13 C

Inner sphere has a - 5.03*10^-14C charge (-Qi) on inside of the outer shell. As a result, there is a net zero charge within the outer shell.

For the outer shell to show a NET charge +6.79*10^-13C, it's must have a +ve charge

= +6.79*10^-13C + (+5.03*10^-14C)

= +7.29*10^-13 C

Now again, we have

E = kQ /r²

E = (9.0*10^9)(+7.29*10^-13 C) / (0.04)²

E = 6.561*10^-3 / 1.6*10^-3

E = 4.10 N/C

Thus, the magnitude of the electric field is 4.1 N/C

Answer:

35.6 m

Explanation:

The given ball possesses a projectile motion from its initial height. So, the required launch velocity of the ball is 6.55 m/s.

The horizontal component of velocity during the projection of an object is known as launch velocity. It is obtained when the horizontal range is known.

Given data -

The angle of projection is, [tex]\theta = 10.0 {^\circ}[/tex].

The initial height of the projection is, h = 1.50 m.

The horizontal displacement is, R = 3.00 m.

The mathematical expression for the horizontal displacement (Range) of the projectile is given as,

[tex]R = \dfrac{u^{2} \times sin2 \theta}{g}[/tex]

here,

u is the launch velocity.

g is the gravitational acceleration.

Solving as,

[tex]u =\sqrt{\dfrac{R \times g}{sin2 \theta}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin(2 \times 10)}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin20^\circ}}\\\\\\u=6.55 \;\rm m/s[/tex]

Thus, we can conclude that the required launch velocity of the ball is 6.55 m/s.

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Answer:

500 nm, 530 nm, 650 nm

Explanation:

Let's say that there is diffraction grating observed with a slit spacing of s. Respectively we must determine the angle θ which will help us determine the 3 wavelengths ( λ ) of the light emitted by element X. This can be done applying the following formulas,

s( sin θ ) = m [tex]*[/tex] λ, such that y = L( tan θ ) - where y = positioning, or the distance of the first - order maxima, and L = constant, of 77 cm

Now the grating has a slit spacing of -

s = 1 / N = 1 / 1200 = 0.833 [tex]*[/tex] 10⁻³ mm

The diffraction angles of the " positionings " should thus be -

θ = tan⁻¹ [tex]*[/tex] ( 0.58 / 0.77 ) = 37°,

θ = tan⁻¹ [tex]*[/tex] ( 0.654 / 0.77 ) = 40°,

θ = tan⁻¹ [tex]*[/tex] ( 0.945 / 0.77 ) = 51°

The wavelengths of these three bright fringes should thus be calculated through the formula : λ = s( sin θ ) -

λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 37° ) = ( 500 [tex]*[/tex] 10⁻⁹ m )

λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 40° ) = ( 530 [tex]*[/tex] 10⁻⁹ m )

λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 51° ) = ( 650 [tex]*[/tex] 10⁻⁹ m )

Wavelengths : 500 nm, 530 nm, 650 nm

This question will be solved using the "grating equation".

The wavelengths of the light emitted by element X are:

"1. 6.654 x 10⁻⁷ m = 665.4 nm

2. 6.349 x 10⁻⁷ m = 634.9 nm

3. 5.262 x 10⁻⁷ m = 526.2 nm"

The diffraction grating equation is given as follows:

[tex]m\lambda = d Sin\ \theta[/tex]

where,

m = order of maxima = 1

λ = wavelength of light = ?

d = grating element = [tex]\frac{1}{no.\ of\ slits\ per\ unit\ length} = \frac{1}{1200\ slits/mm}[/tex]

d =  (8.33 x 10⁻⁴ mm/slit)(1 m/ 1000 mm) = 8.33 x 10⁻⁷ m/slit

θ = angle of diffraction = [tex]tan^{-1}(\frac{L}{y})[/tex]

where,

L = distance of grating from the screen = 77 cm

y = distance of maxima from central maxima

Hence, the general equation after substituting constant values becomes:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{y}))[/tex]

FOR y = 58 cm:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{58\ cm}))[/tex]

λ = 6.654 x 10⁻⁷ m = 665.4 nm

FOR y = 65.4 cm:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{65.4\ cm}))[/tex]

λ = 6.349 x 10⁻⁷ m = 634.9 nm

FOR y = 94.5 cm:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{94.5\ cm}))[/tex]

λ = 5.262 x 10⁻⁷ m = 526.2 nm

The attached picture shows the arrangement of the light rays in a diffraction grating.

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Answer:

Explanation:

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

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

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

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

= 1.6 B

current induced = emf induced / resistance

12 x 10⁻³ = 1.6 B / 15

B = 112.5 x 10⁻³ T .

Answer:  (x - 4)² + (y - 7)² = 9

Explanation:

The equation of a circle is: (x - h)² + (y - k)² = r²    where

Given: (h, k) = (4, 7)

Find the intersection of the given equation and the perpendicular passing through (4, 7).

3x - 4y = -1

     -4y = -3x - 1

        [tex]y=\dfrac{3}{4}x-1[/tex]

              [tex]m=\dfrac{3}{4}[/tex]      -->      [tex]m_{\perp}=-\dfrac{4}{3}[/tex]

[tex]y-y_1=m_{\perp}(x-x_1)\\\\y-7=-\dfrac{4}{3}(x-4)\\\\\\y=-\dfrac{4}{3}x+\dfrac{16}{3}+7\\\\\\y=-\dfrac{4}{3}x+\dfrac{37}{3}[/tex]

Use substitution to find the point of intersection:

[tex]x=\dfrac{29}{5}=5.8,\qquad y=\dfrac{23}{5}=4.6[/tex]

Use the distance formula to find the distance from (4, 7) to (5.8, 4.6) = radius

[tex]r=\sqrt{(5.8-4)^2+(4.6-7)^2}\\\\r=\sqrt{3.24+5.76}\\\\r=\sqrt9\\\\r=3[/tex]

Input h = 4, k = 7, and r = 3 into the circle equation:

(x - 4)² + (y - 7)² = 3²

(x - 4)² + (y - 7)² = 9

Answer:

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

Explanation:

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

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

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

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

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

Simply we applied the above formulas

Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r²           [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

Answer:

Explanation:

Here image distance is fixed .

In the first case if v be image distance

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

In the second case

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

So lens must be moved forward by 5.06 - 5.01 =  .05 cm ( away from film )

Answer:

1.97 V

Explanation:

Applying

N1/N2 = V1/V2................... Equation 1

Where N1 = primary turns, N2 = Secondary turns, V1 = primary/input voltage, V2 = secondary/output voltage.

make V2 the subject of the equation

V2 = V1(N2/N1)............. Equation 2

Given: V1 = 118 V, N1 = 480 turns, V2 = 8 turns.

Substitute into equation 2

V2 = 118(8/480)

V2 = 1.97 V.

Answer:

a= 4.9m/s²

Explanation:

Using Fnet= mgsintheta = ma

But a= gsintheta

a= 9.8xsin 30

= 4.9m/s²

Answer:

a) d = 7.62 10⁻⁶ m, b)    l = 3.25 10⁴ m

Explanation:

Resistance is expressed by the formula

     R = ρ l / A                       (1)

density is defined by

     density = m / V

the volume of a wire is the cross section by the length

     V = A l

we substitute

     density = m / A l

     A = m / density l

we substitute in 1

     R = ρ l density l / m

     R =ρ density l² / m

     l = √ (R m /ρ density)

let's calculate the cable length

     l = √(11.7  13.5 10⁻³ / (1.68 10⁻⁸ 8.9 10³))

     l = √(10.56 10⁸)

     l = 3.25 10⁴ m

now we can find the cable diameter with the density equation

      A = m / density l

      A = 13.5 10⁻³ / (8.9 10³ 3.25 10⁴)

      A = 4,557 10⁻¹¹ m²

the area of ​​the circle is

      A = π r² = π d² / 4

      d = √ (4A /π)

      d = √ (4 4,557 10⁻¹¹/π)

      d = 7.62 10⁻⁶ m

The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m

Data;

The resistivity of copper is calculated by

[tex]R = \frac{\rho L}{A}\\[/tex]

Let's calculated the volume of the wire first;

[tex]\rho = \frac{mass}{volume} \\volume = \frac{mass}{density} \\volume = \frac{13.5*10^-^3}{8.9*10^3} \\v = 1.52*10^-6m^3[/tex]

The diameter of the wire will be

[tex]R = \frac{\rho L}{A}\\ R = \frac{\rho LA}{A^2}\\ R = \frac{\rho V}{A^2} \\A^2 = \frac{\rho V}{R}\\ A^2 = \frac{8.9*10^3 * 1.516*10^-^6}{11.7} \\A^2 = 0.0011\\A = \sqrt{0.0011} \\A = 0.034m^2[/tex]

Taking the area

[tex]a = \pi r^2\\0.034 = 3.14 * r^2\\r^2 = \frac{0.034}{3.14} \\r^2 = 0.01083\\r = \sqrt{0.01083}\\ r = 0.104m\\d = 2r\\d = 2 * 0.104\\d = 0.208m[/tex]

Length of the wire can be calculated as

[tex]V = AL\\L = V/A\\L = \frac{1.52*10^-^6}{0.034}\\ L = 4.47*10^-^5m[/tex]

The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m

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Answer:

the right answer is true

Answer:

True

Explanation:

Answer:

The  initial velocity is  [tex]v_h = 8.66 \ m/s[/tex]

Explanation:

From the question we are told that

    The height of the tree is  [tex]h = 3.30\ m[/tex]

    The distance of the position of landing from base  is  [tex]d = 5.30 \ m[/tex]

According to the second equation of motion

    [tex]h = u_o * t + \frac{1}{2} at^2[/tex]

[tex]Where\ u_o[/tex] is the initial velocity in the vertical axis  

           a  is equivalent to acceleration due to gravity which is positive because the tiger is downward

    So

     [tex]3 = 0 + 0.5 * 9.8 *t^2[/tex]

=>    [tex]t = \frac{3 }{9.8 * 0.5}[/tex]

      [tex]t = 0.6122\ s[/tex]

Now the initial velocity in the horizontal direction is mathematically evaluated as

         [tex]v_h = \frac{5.30}{0.6122}[/tex]

        [tex]v_h = 8.66 \ m/s[/tex]

Answer:

5.33333... seconds

Explanation:

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

Answer:

Explanation:

When water droplet condenses on the outer wall of glass of ice , it releases heat equal to mass x latent heat of condensation of water . This heat reaches the ice melting inside  glass . Due to this heat , the melting process is accelerated .

Hence the process of melting gets accelerated when water droplet condenses on the outer wall of glass containing mixture of ice and water .

Answer:

2.8

Explanation:

Using p = m/v; (old density)

p' = m/v (new density)

=m/0.350 V

p'/p = (m/0.350V)/(m/v) = 1/0.350 = 2.86

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

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

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, [tex]q_1=-3\ nC[/tex]

It is placed at a distance of 9 cm at x axis

Charge, [tex]q_2=+4\ nC[/tex]

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]

Here,

[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]

So,

[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]

Squaring both sides,

[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

According to the question,

Charge,

Now,

→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]

or,

→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]

→  [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]

here,

[tex]r_1 = \sqrt{y^2+81}[/tex]

[tex]r^2 = \sqrt{y^2+225}[/tex]

By substituting the values,

→      [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]

By applying cross-multiplication,

[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]

By squaring both sides, we get

→  [tex]9(y^2+225) = 16(y^2+81)[/tex]

    [tex]9y^2+2025 = 16 y^2+1296[/tex]

   [tex]2025-1296=7y^2[/tex]

               [tex]7y^2=729[/tex]

                  [tex]y = 10.2 \ m[/tex]

Thus the solution above is correct.

Learn more about charge here:

https://brainly.com/question/12437696

Answer:

Trial 1: 2 Volts, 0 %

Trial 2: 2.8 Volts, 0%

Trial 3: 4 Volts, 0 %

Explanation:

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

V = IR

TRIAL 1:

V = IR

V = (0.1 A)(20 Ω)

V = 2 volts

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

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

% Difference = 0 %

TRIAL 2:

V = IR

V = (0.14 A)(20 Ω)

V = 2.8 volts

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

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

% Difference = 0 %

TRIAL 3:

V = IR

V = (0.2 A)(20 Ω)

V = 4 volts

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

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

% Difference = 0 %