If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?

Answer:

0.05cos10t

Explanation:

X(t) = Acos(wt+φ)

The oscillation angular frequency can be calculated using below formula

w = √(k/M)

Where K is the spring constant

But we were given body mass of 5.0 kg

We know acceleration due to gravity as 9.8m)s^2

The lenghth of spring which stretches =10 cm

Then we can calculate the value of K

k = (5.0kg*9.8 m/s^2)/0.10 m

K= 490 N/m

Then if we substitute these values into the formula above we have

w = √(k/M)

w = √(490/5)

= 9.90 rad/s=10rads/s(approximately)

Its position as a function of time can be calculated using the below expresion

X(t) = Acos(wt+φ)

We were given amplitude of 5 cm , if we convert to metre = 0.05m

w=10rads/s

Then if we substitute we have

X(t)=0.05cos(10×t)

X(t)= 0.05cos10t

Therefore,Its position as a function of time=

X(t)= 0.05cos10t

Answer:

Pressure, P = 1250 Pa

Explanation:

Given that,

Weight of a brick, F = 50 N

Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm

We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.

A = 40 cm × 10 cm = 400 cm² = 0.04 m²

So,

Pressure,

[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]

So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.

The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.

The given data in the problem is;

W is the weight of a brick = 50 N

The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm

A is the area,

The area is found as;

A=40 cm × 10 cm = 400 cm² = 0.04 m²

The pressure is the ratio of the force and area

[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]

Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

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Answer:

Explanation:

The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

θ is the angle that the velocity of the electron make with the magnetic field.

Given parameters

F =  1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

Required

Angle that the velocity of the electron make with the magnetic field

Substituting the given parameters into the formula:

1.40*10⁻¹⁶ =  1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

θ₁ = 10.4⁰

Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁

θ₂ = 180-10.4

θ₂ = 169.6⁰

Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰

Answer:

The pressure increase inside the automobile engine block is 1782.18 ATM

Explanation:

Given;

the change in volume of water, ΔV = 9%

the bulk modulus of ice, K = 2 x 10⁹ N/m²

Bulk modulus is given by;

[tex]K = -V\frac{dP}{dV}[/tex]

for pressure increase in the automobile engine block, when the water in there froze;

[tex]dP = K(\frac{dV}{V} )\\\\dP = K(\frac{0.09V}{V} )\\\\dP = 0.09K\\\\dP = 0.09 (2*10^9)\\\\dP = 1.8 *10^{8} \ N/m^2\\\\dP = 1782.18 \ ATM[/tex]

Therefore,  the pressure increase inside the automobile engine block is 1782.18 ATM

The pressure increase inside your automobile engine block will be 1782.18 atm. The force involved vertical to the surface of an object per unit area is pressure.

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

The given data in the problem is;

ΔV is the  change in volume of water = 9%

K is the bulk modulus of ice = 2 x 10⁹ N/m²

dP is the change in the presure=?

The bulk modulus is found as;

[tex]\rm K=-V \frac{dp}{dv} \\\\[/tex]

The change in the presure is obtained as;

[tex]\rm dP = K\frac{dV}{V} \\\\ \rm dP = K\frac{0.09V}{V} \\\\ \rm dP = 0.09 K \\\\ \rm dP = 0.09 \times 2 \times 10^9 \\\\ \rm dP = 1.8 \times 10^8 \\\\ \rm dP =1782.18 \ atm[/tex]

Hence the pressure increase inside your automobile engine block will be 1782.18 atm.

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Answer:

the answer to your question us c honey

Answer:

C

Explanation:

This is so because different materials vary in resistance and conductance of current, heat. Metals are good conductors while none metals like rubber, plastic, glass etc are good insulators or resistors.

Answer:

T = 80√3 N ≈ 139 N

W = 160 N

Explanation:

Sum of forces on B in the x direction:

∑F = ma

80 N sin 60° − T sin 30° = 0

T = 80 N sin 60° / sin 30°

T = 80√3 N

T ≈ 139 N

Sum of forces on B in the y direction:

∑F = ma

80 N cos 60° + T cos 30° − W = 0

W = 80 N cos 60° + T cos 30°

W = 40 N + 120 N

W = 160 N

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

[tex]E=\frac{F}{Q}[/tex]

But the force F

[tex]F= \frac{kQ1Q2}{r^2}[/tex]

But the electric field intensity due to a point charge Q at a distance r meters away is given by

[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Answer:

The separation of the two slits is 0.456 mm.

Explanation:

Given the wavelength of light = 519 nm

The indifference pattern = 4.6 m

Adjacent bright fringes = 5.2 mm

In the interference, the equation required is Y = mLR/d

Here, d sin theta = mL

L = wavelgnth

For bright bands, m is the  order = 1,2,3,4  

For dark bands,  m = 1.5, 2.5, 3.5, 4.5

R = Distance from slit to screen (The indifference pattern)

Y = Distance from central spot to the nth  order fringe or fringe width

Thus,  here d = mLR/Y

d = 1× 519nm × 4.6 / 5.2mm

d = 0.459 mm

Answer:

96 m

Explanation:

Given:

v₀ = 0 m/s

a = 3 m/s²

t = 8 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (8 s) + ½ (3 m/s²) (8 s)²

Δx = 96 m

Complete Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?

Answer:

The velocity is [tex]v = 80.82 \ m/s[/tex]

Explanation:

From  the question we are told that

   The magnitude of charge one is  [tex]q_1 = 3.25 nC = 3.25 *10^{-9} \ C[/tex]

   The  magnitude of charge two  [tex]q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C[/tex]

   The distance of separation is   [tex]d = 58.0 \ cm = 0.58 \ m[/tex]

Generally the electric potential of the electron at the midway point is mathematically represented as

         [tex]V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }[/tex]

substituting values

         [tex]V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }[/tex]

          [tex]V = 1.8103 *10^{-8} \ V[/tex]

Now when the electron is 10 cm   =  0.10 m  from charge 1 , it is  (0.58 - 0.10 =  0.48 m ) m from charge two

Now the electric potential  at that point is mathematically represented as

       [tex]V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}[/tex]

 substituting values

      [tex]V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}[/tex]

      [tex]V_1 = 3.67*10^{-8} \ V[/tex]

Now the law of energy conservation ,

   The  kinetic energy of the electron  =  potential energy of the electron

i.e     [tex]\frac{1}{2} * m * v^2 = [V_1 - V]* q[/tex]

where q is the magnitude of the charge on the electron with value

      [tex]q = 1.60 *10^{-19} \ C[/tex]

While m is the mass of the electron with value  [tex]m = 9.11*10^{-31} \ kg[/tex]

        [tex]\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}[/tex]

         [tex]v = \sqrt{6532.4}[/tex]

        [tex]v = 80.82 \ m/s[/tex]

Answer:

1)   q₁ = 12.987 cm , b)       L = 17.987 cm , c)      m = 179.87

Explanation:

We can solve the geometric optics exercises with the equation of the constructor

         1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and the image respectively.

Let's apply this equation to our case

1) f = 5mm = 0.5 cm

    p₁ = 5.2 mm = 0.52 cm

    h = 0.1 mm = 0.01 cm

    1 / q₁ = 1 / f- 1 / p

    1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923

    1 / q₁ = 0.077

    q₁ = 12.987 cm

2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece

            p₂ = 5 cm

The absolute thing that goes through the two lenses is

           L = q₁ + p₂

           L = 12.987 +5

           L = 17.987 cm

3) This lens configuration forms the so-called microscope, whose expression for the magnifications

           m = -L / f_target 25 cm / f_ocular

           m = - 17.987 / 0.5 25 / 5.0

            m = 179.87

Answer:

The resultant amplitude of the two waves is 2.65.

Explanation:

Given;

amplitude of the first wave, A₁ = 1

amplitude of the second wave, A₂ = 2

phase difference of the two amplitudes, θ = 60.0°.

The resultant amplitude of two waves after interference is given by;

[tex]A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2Cos \theta} \\\\A = \sqrt{1^2 + 2^2 + 2(1)(2)Cos 60} \\\\A= 2.65[/tex]

Therefore, the resultant amplitude of the two waves is 2.65.

Answer:

a) Connect a series resistance of 8,47 ohms

b)14,16 [A]

c) r = 10,96 ohms

Explanation:

My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower

In America

P = V*I

1700 (w) = 120*I

I = 1700/120 [A]

I = 14,16 [A]        current needed for the blower

In Europe

120 (v)  (the drop of voltage I need) when a current of 14,16 passes through to series  resistor is

V = I*R          120 = 14,16* R         R = 8,47 ohms

c) P = I*r²

1700 (w) = 14,16 (A) * r²

r² = 120,06

r = 10,96 ohms

Answer:

it is either constant or zero

Explanation:

Answer:

The current in the ammeter is zero.

(1) is correct option.

Explanation:

Given that,

The capacitor is charged.

We need find the current after closed switched

We know that,

When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.

Then, the ammeter reads zero.

Hence, The current in the ammeter is zero.

(1) is correct option.

Answer:

 Δy = 1 10⁻⁴ m

Explanation:

In double-slit experiments the constructive interference pattern is described by the equation

           d sin θ = m λ

In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle

         tan θ = y / L

as the angles are small,

         tan θ = sin θ / cos θ = sin θ

substituting

         sin θ = y / L

         d y / L = m λ

         y = m λ / d L

let's apply this formula for each wavelength

λ = 450 nm = 450 10⁻⁹ m

m = 5

d = 5.0 mm = 5.0 10⁻³ m

      y₁ = 5 450 10⁻⁹ / (5 10⁻³  1.4)

      y₁ = 3.21 10⁻⁴ m

we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m

      y₂ = 5 590 10⁻⁹ / (5 10⁻³  1.4)

      y₂=  4.21 10⁻⁴ m

the separation of these two lines is

        Δy = y₂ - y₁

        Δy = (4.21 - 3.21) 10⁻⁴ m

        Δy = 1 10⁻⁴ m

Answer:

The distance is [tex]y = 0.00162 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 486 \ nm = 486 *10^{-9} \ m[/tex]

   The  slit spacing is  [tex]d = 0.600 \ mm = 0.60 *10^{-3} \ m[/tex]

    The distance of the screen is  [tex]D = 2.0 \ m[/tex]

 Generally the distance along the screen between adjacent bright fringes is mathematically represented as  

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

substituting values

         [tex]y = \frac{ 486 *10^{-9} * 2}{0.6*10^{-3}}[/tex]

         [tex]y = 0.00162 \ m[/tex]

Answer:

The actual distance is  [tex]d_a = 0.3233\ m[/tex]

Explanation:

From the question we are told that

     The  distance of the can is d =  43 cm  =  0.43 m

Generally the actual distance is mathematically represented as

         [tex]d_a = [\frac{ n_a }{n_w} ]* d[/tex]

Where  [tex]n_a , n_w[/tex] are the refractive index of air and water and their value is  

         [tex]n_a = 1 , \ \ \ n_w = 1.33[/tex]

So

       [tex]d_a = [\frac{ 1 }{1.33} ]* 0.43[/tex]

       [tex]d_a = 0.3233\ m[/tex]

Answer:

     λ = 52.5 nm

Explanation:

De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression

          p = h / λ

where the moment

          p = mv

          λ = h / mv

let's calculate

        λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54

        λ = 5.25 10⁻⁸ m

Let's reduce anm

       λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)

       λ = 52.5 nm

Answer:

34°

Using the relation

θᶜ = sin^-1(n₂/n₁),

where n1= the refractive index of light is propagating from a medium

And n2 = refractive index of medium into which light is entering

So we know that

refractive index of diamond at 589nm = 2.41= n₁

refractive index of ethanol at 589nm and 20°C = 1.36= n₂

Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°

Explanation:

Answer:

d = 6.38 x 10⁻³ m = 6.38 mm

Explanation:

Since, the no. of bright fringes is 34 in a centimeter, therefore, the fringe spacing must be equal to:

Fringe Spacing = Δx = 1 cm/34

Δx = 0.0294 cm = 2.94 x 10⁻⁴ m

But, the formula for fringe spacing in a double slit experiment is:

Δx = λL/d

where,

λ = wavelength of light = 605 nm = 6.05 x 10⁻⁷ m

L = Distance between screen and slits = 3.1 m

d = slit separation = ?

Therefore,

2.94 x 10⁻⁴ m = (6.05 x 10⁻⁷ m)(3.1 m)/d

d = (18.755 x 10⁻⁷ m²)/(2.94 x 10⁻⁴ m)

d = 6.38 x 10⁻³ m = 6.38 mm

Explanation:

1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.

2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.

Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.

We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.

These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.

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Answer:

m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J

Answer:

The magnitude of the charge is 54.9 nC.

Explanation:

The charge on each bead can be found using Coulomb's law:

[tex] F_{e} = \frac{k*q_{1}q_{2}}{r^{2}} [/tex]

Where:

q₁ and q₂ are the charges, q₁ = q₂  

r: is the distance of spring stretching = 4.8x10⁻² m

[tex]F_{e}[/tex]: is the electrostatic force

[tex] F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r [/tex]    

Now, we need to find [tex]F_{e}[/tex]. To do that we have that Fe is equal to the spring force ([tex]F_{k}[/tex]):

[tex] F_{e} = F_{k} = -kx [/tex]

Where:

k is the spring constant

x is the distance of the spring = 4.8 - 4.0 = 0.8 cm

The spring constant can be found by equaling the sping force and the weight force:

[tex] F_{k} = -W [/tex]

[tex] -k*x = -m*g [/tex]

where x is 5.2 - 4.0 = 1.2 cm, m = 1.8 g and g = 9.81 m/s²

[tex] k = \frac{mg}{x} = \frac{1.8 \cdot 10^{-3} kg*9.81 m/s^{2}}{1.2 \cdot 10^{-2} m} = 1.47 N/m [/tex]      

Now, we can find the electrostatic force:

[tex] F_{e} = F_{k} = -kx = -1.47 N/m*0.8 \cdot 10^{-2} m = -0.0118 N [/tex]

And with the magnitude of the electrostatic force we can find the charge:

[tex]q = \sqrt{\frac{F_{e}}{k}}*r = \sqrt{\frac{0.0118 N}{9 \cdot 10^{9} Nm^{2}/C^{2}}}*4.8 \cdot 10^{-2} m = 54.9 \cdot 10^{-9} C = 54.9 nC[/tex]

Therefore, the magnitude of the charge is 54.9 nC.

I hope it helps you!  

The magnitude of the charge (in nC) on each bead is equal to 55.21 nC.

Given the following data:

Extension, e = [tex]0.052 - 0.048[/tex] = 0.012 m

Scientific data:

To calculate the magnitude of the charge (in nC) on each bead, we would apply Coulomb's law:

First of all, we would determine the spring constant of this lightweight spring by using this formula:

[tex]W = mg = Ke \\\\K=\frac{mg}{e} \\\\K=\frac{0.0018 \times 9.8}{0.012} \\\\K=\frac{0.01764}{0.012}[/tex]

Spring constant, K = 1.47 N/m.

For the electrostatic force:

[tex]F = ke\\\\F = 1.47 \times 0.08[/tex]

F = 0.01176 Newton.

Coulomb's law of electrostatic force.

Mathematically, the charge in an electric field is given by this formula:

[tex]q = \sqrt{\frac{F}{k} } \times r[/tex]

Substituting the given parameters into the formula, we have;

[tex]q = \sqrt{\frac{0.01176 }{8.99 \times 10^9} } \times 0.048\\\\q=\sqrt{1.3228 \times 10^{-12}} \times 0.048\\\\q=1.1502 \times 10^{-6} \times 0.048\\\\q= 5.521 \times 10^{-8}\;C[/tex]

Note: 1 nC = [tex]1 \times 10^{-9}\;C[/tex]

Charge, q = 55.21 nC.

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Complete Question

The image of this question  is shown on the first uploaded image

Answer:

a

   [tex]d =0.161 \ m[/tex]

b

  [tex]v = - 0.054 \ m/s[/tex]

c

  [tex]a = 6.12 \ m/s^2[/tex]

Explanation:

From the question we are told that

      The maximum  displacement is  A =  0.17  m  

      The  time considered is  [tex]t = 3.1 \ s[/tex]

     The spring constant is  [tex]k = 137 \ N \cdot m[/tex]

      The mass is  [tex]m = 3.8 \ kg[/tex]

Generally given that the motion which the cylinder is undergoing is a simple harmonic motion , then the displacement is mathematically represented as

             [tex]d = A cos (w t )[/tex]

Where [tex]w[/tex] is the angular frequency which is mathematically evaluated as

        [tex]w = \sqrt{\frac{k}{m} }[/tex]

substituting values

       [tex]w = \sqrt{\frac{137}{ 3.8} }[/tex]

        [tex]w =6[/tex]

So the displacement is at  t

      [tex]d = 0.17 cos (6 * 3.1 )[/tex]

       [tex]d =0.161 \ m[/tex]

Generally the velocity of a  SHM(simple harmonic motion) is mathematically represented as

         [tex]v = - Asin (wt)[/tex]

substituting values

         [tex]v = - 0.17 sin ( 6 * 3.1 )[/tex]

          [tex]v = - 0.054 \ m/s[/tex]

Generally the maximum acceleration is  mathematically represented as

         [tex]a = w^2 * A[/tex]

substituting values

         [tex]a_{max} = 6^2 * (0.17)[/tex]

substituting values

         [tex]a = 6^2 * (0.17)[/tex]

        [tex]a = 6.12 \ m/s^2[/tex]

Answer:

The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.

Explanation:

Answer:

Decrease in entropy

Explanation:

Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.

In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.

On the other hand, in open system both the matter and energy move into and out of the system.

Answer:

4.5x 10^ -9m

Explanation:

See attached file

Answer:

The radius is  [tex]r_{min} = 0.00226 \ m[/tex]

Explanation:

   From the question we are told that

      The  elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]

      The length is  [tex]L = 4.0 \ m[/tex]

      The weight of the commercial sign is    [tex]F_s = 8000 \ N[/tex]

       The maximum extension of the wire is  [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]

Generally the elastic limit of an alloy (stress) is is mathematically represented as

            [tex]\sigma = \frac{ F_s }{ A }[/tex]

Where A is the cross-sectional area of the wire which is mathematically represented as

         [tex]A = \pi r^2[/tex]

here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign

So

          [tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]

=>       [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]

substituting values

             [tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]

           [tex]r_{min} = 0.00226 \ m[/tex]

Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.

              Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m

Explanation: A speed of wave on a string under a tension force can be calculated as:

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

[tex]F_{T}[/tex] is tension force (N)

μ is linear density (kg/m)

Determining velocity:

[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{0.00874 }[/tex]

[tex]|v| =[/tex] 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]

Δx = 11.5×[tex]10^{-6}[/tex]

With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex]  m.

Doubling Tension:

[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{2.0.00874 }[/tex]

[tex]|v| = \sqrt{0.01568}[/tex]

|v| = 0.1252 m/s

Displacement for same time:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]

[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]

With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m

Answer:

1. A

2. C

3. D

Explanation:

Answer:

Answers a) and b) should be marked as correct.

Explanation:

Recall that the resonance in an LRC circuit occurs when the current through the circuit is at its maximum, and such takes place when the impedance  (Z) of the circuit reaches its maximum. This means that the impedance (see formula below) is at its minimum value:

[tex]Z=\sqrt{R^2+(\omega\,L-\frac{1}{\omega\,C})^2 }[/tex]

as per the impedance expression above, such happens when the term in parenthesis inside the root which contains the inductive reactance ([tex]\omega\,L[/tex]) and the capacitive reactance ([tex]1/\omega\,C[/tex]) have the same value.

Therefore, answers:

a) "The impedance of the circuit has its minimum value."

and

b) "The inductive reactance and the capacitive reactance are exactly equal to each other."

are correct answers.

(a) The impedance of the circuit has its minimum value.

(b) The inductive reactance and the capacitive reactance are exactly equal to each other

LRC series circuit consists of inductor, resistor and capacitor is series.

The impedance of the circuit is calculated as follows;

[tex]Z = \sqrt{R^2 + (X_C -X_L)^2}[/tex]

where;

The impedance of the circuit is minimum when the capacitive reactance is equal to the inductive reactance.

[tex]X_C = X_L \\\\Z = \sqrt{R^2 \ + (0)^2} \\\\Z = R[/tex]

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