How do the electrons in the bulb's filament, far from the battery and its emf, "know" that they should flow?

Answer:

Explanation:

This question pertains to resonance in air column.  It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

All objects are described using classical mechanics, with the exception of the buckyball, which requires a quantum mechanical description due to its small mass and high speed that result in a significant de Brogli wavelength.

Whether an object shows particle-wave duality, requiring quantum mechanics to describe, or behaves like everyday objects, described by classical mechanics, is determined by the de Broglie wavelength of the object. Calculating the de Broglie wavelength (λ) can be accomplished using the equation λ = h/mv, where h is Planck's constant and m and v are the object's mass and velocity, respectively.

For the virus, buckyball, mosquito, and turtle, the calculated de Broglie wavelengths are so small compared to their physical dimensions that quantum mechanical effects would be negligible. Hence, all these objects can be described using classical mechanics. The buckyball is the only exception. Owing to its tiny mass and high speed, its de Broglie wavelength becomes significant in relation to its size, requiring quantum mechanical description.

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Objects that are atomic or subatomic in size require quantum mechanics for their description, while larger objects can be adequately explained by classical mechanics. In the provided cases, a virus and a buckyball would require quantum mechanics, while a mosquito and a turtle would be described by classical mechanics.

To determine whether an object would be adequately described by classical mechanics or require quantum mechanics, we need to consider its size, mass, and speed. Under the domain of quantum mechanics are typically objects that are atomic or subatomic in size, meaning they exhibit wave-particle duality due to their small size and significant motion.

1. A virus: Being atomic in size, a virus would require quantum mechanics for its description. Specifically, it would demonstrate both particle and wave properties due to its small size.

2. A buckyball: Given its subatomic mass, size, and significant speed, a buckyball falls into the quantum domain, displaying both particle and wave properties.

3. A mosquito and 4. A turtle : These are larger, macroscopic objects. Their motion and behavior can be adequately described using classical mechanics as they primarily exhibit particle properties, their wave-like characteristics being effectively obscured due to their larger mass and size. Their behaviors conform to the laws defined under classical mechanics and do not exhibit the quantum nature that smaller objects do.

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Answer:

a)  x_{cm} = L / 2 , y_{cm}= L/2, b) x_{cm} = L / 2 , y_{cm}= L/2

Explanation:

The center of mass of a body is the point where all external forces are applied, it is defined by

      [tex]x_{cm}[/tex] = 1 / M ∑  [tex]x_{i} m_{i}[/tex] = 1 /M ∫ x dm

      [tex]y_{cm}[/tex] = 1 / M ∑ [tex]y_{i} m_{i}[/tex] = 1 / M ∫ y dm

where M is the total body mass

Let's calculate the center of mass of our L-shaped body, as formed by two rods one on the x axis and the other on the y axis

a) let's start with the reference zero at the left end of the horizontal rod

let's use the concept of linear density

    λ = M / L = dm / dl

since the rod is on the x axis

     dl = dx

    dm = λ dx

let's calculate

      x_{cm} = M ∫ x λ dx = λ / M ∫ x dx

      x_{cm} = λ / M x² / 2

we evaluate between the lower integration limits x = 0 and upper x = L

      x_{cm} = λ / M (L² / 2 - 0)

  we introduce the value of the density that is cosntnate

     x_{cm} = (M / L) L² / 2M

     x_{cm} = L / 2

We repeat the calculation for verilla verilla

     λ = M / L = dm / dy

     y_{cm} = 1 / M ∫ y λ dy

     y_{cm} = λ M y² / 2

     [tex]y_{cm}[/tex] = M/L  1/M (L² - 0)

     y_{cm}= L/2

b) we repeat the calculation for the origin the reference system is top of the vertical rod

     horizontal rod

        x_{cm} = 1 / M ∫λ x dx = λ/M   x² / 2

we evaluate between the lower limits x = 0 and the upper limit x = -L

      x_{cm} = λ / M [(-L)²/2 - 0] = (M / L) L² / 2M

      x_{cm} = L / 2

vertical rod

      y_{cm} = 1 / M ∫y dm

      y_{cm} = λ / M ∫y dy

      y_{cm} = λ / M y2 / 2

we evaluate between the integration limits x = 0 and higher x = -L

      y_{cm} = (M / L) 1 / M ((-L)²/2 -0)

      y_{cm} = L / 2

To find the center of mass of the 'L' structure, use the formal definition and consider the midpoint of each member. The x-coordinate is halfway between the ends and the y-coordinate is halfway between the top and bottom ends.

To determine the location of the center of mass of the 'L' structure, we can use the formal definition of center of mass. Since the thin vertical and horizontal members have the same length and mass, the center of mass of each member is located at its midpoint. The x-coordinate of the center of mass is halfway between the x-coordinate of the left end and the x-coordinate of the right end of the 'L'. The y-coordinate of the center of mass is halfway between the y-coordinate of the bottom end and the y-coordinate of the top end of the 'L'.

(a) Origin at the lower left:

x = -L/4, y = L/4

(b) Origin at the top of the vertical member:

x = L/4, y = -L/4

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

     [tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]

    [tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]

Explanation:

   From the question we are told that

         The first string has a frequency of   [tex]f_1 = 230 Hz[/tex]

          The period of the beat is  [tex]t_{beat} = 0.99s[/tex]

Generally the frequency of the beat is

             [tex]f_{beat} = \frac{1}{t_{beat}}[/tex]

  Substituting values

            [tex]f_{beat} = \frac{1}{0.99}[/tex]

                   [tex]= 1.01 Hz[/tex]

From the question

        [tex]f_2 - f_1 = f_{beat}[/tex]   for  [tex]f_2[/tex]  having a  higher tension

So

       [tex]f_2 - 230 = 1.01[/tex]

               [tex]f_2 = 231.01Hz[/tex]

 From the question

            [tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]

         [tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]

Substituting values

         [tex]\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}[/tex]

      [tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]

    For [tex]f_2[/tex] having a lower tension

           [tex]f_1 - f_2 = f_{beat}[/tex]

  So

       [tex]230 - f_2 = 1.01[/tex]

            [tex]f_2 = 230 -1.01[/tex]

                  [tex]= 228.99[/tex]

  From the question

            [tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]

         [tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]

    Substituting values

         [tex]\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}[/tex]

      [tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]        

Answer:

1.04m

Explanation:

the distance is determined by the diameter of the trajectory.

you can find the radius of the trajectory by using the following formula:

[tex]r=\frac{m_cv}{qB}[/tex]

mc : mass of the Big C = 0.303kg/mol/(6.02*10^{23}/mol)=5.03*10^{-25}kg

v: velocity of the ion = 50000m/s

B: magnetic constant = 0.3T

q: 1.6*10^{-19}C

By replacing you obtain:

[tex]r=\frac{(5.03*10^{-25}kg)(50000m/s)}{(1.6*10^{-19}C)(0.3T)}=0.52m[/tex]

the diameter wiil be:

d=2r=1.04m

the ion strikes the detector from 1.04m to the entrance of the spectrometer

The Florida Snow ions strike the spectrometer detector after travelling through a semicircle trajectory will be "1.04 m" far.

According to the question,

Big C's mass, [tex]m_c[/tex] = [tex]\frac{0.303}{6.02\times 10^{23}}[/tex]

                             = 5.03 × 10⁻²⁵ kg

Ion's velocity, v = 50000 ms

Magnetic constant, B = 0.3 T

Charge, q = 1.6 × 10⁻¹⁹ C

We know that,

→ r = [tex]\frac{m_c v}{qB}[/tex]

or,

The radius, r = [tex]\frac{5.03\times 10^{-25}\times 50000}{1.6\times 10^{-19}\times 0.3 }[/tex]

                     = 0.52 m or,

Diameter, d = 2 × r

                     = 2 × 0.52

                     = 1.04 m

Thus the above approach is correct.  

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Answer:

false  b) a = v²(radial) / r  and e)

Explanation:

Let's review given statement separately

a) centripetal acceleration

       a = v² / r

linear and angular velocity are related

     v = w r

we substitute

       a = w²r

this acceleration is directed to the center of the circle, so the vector must be negative

        a = - w r2

the bold are vector

True this statement

b) the magnitude is the scalar value

     a = v² / r

where v is the tangential velocity, not the radial velocity, so this statement is

False

c) This is true

      a = v²/ r

this speed is tangential

d) Newton's second law is

         F = m a

if the acceleration is centripetal

         F = m (- v² / r)

we substitute

         F = m (- s² / R)

the statement is true

e) when the car turns to the left, the objects have to follow in a straight line, which is why

you need a force toward the center of the circle to take the curve.

     Consequently there is no outward force

This statement is false

Answer:

The maximum charge around the capacitor is 0.03170189C

Explanation:

See attached file

In an RLC series circuit with given values for resistance, inductance, and capacitance, we can calculate the charge on the capacitor five cycles later and fifty cycles later using the equation q(t) = q(0) * e^(-(R/L)t)

In an RLC series circuit with a resistance of 7.092 ohms, an inductance of 10 mH, and a capacitance of 3.0 µF, the charge on the capacitor can be calculated using the equation:

q(t) = q(0) * e^(-(R/L)t)

Given that the initial charge on the capacitor is 8.0 µC, we can calculate the charge five cycles later and fifty cycles later using this equation.

(a) To find the charge five cycles later, we need to find the time period of one cycle. The time period can be calculated using the formula:

T = 2π√(LC)

Once we have the time period, we can calculate the time it takes for five cycles to pass and substitute it in the equation to find the charge.

(b) To find the charge fifty cycles later, we can use the same approach as in part (a), but substituting the time it takes for fifty cycles to pass.

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Answer:

The tension is  [tex]T= \frac{11}{2\sqrt{3} } Mg[/tex]

The horizontal force provided by hinge   [tex]Fx= \frac{11}{4\sqrt{3} } Mg[/tex]

Explanation:

   From the question we are told that

          The mass of the beam  is   [tex]m_b =M[/tex]

          The length of the beam is  [tex]l = L[/tex]

           The hanging mass is  [tex]m_h = 3M[/tex]

            The length of the hannging mass is [tex]l_h = \frac{3}{4} l[/tex]

            The angle the cable makes with the wall is [tex]\theta = 60^o[/tex]

The free body diagram of this setup is shown on the first uploaded image

The force [tex]F_x \ \ and \ \ F_y[/tex] are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           [tex]\sum F =0[/tex]

Now about the x-axis the moment is

              [tex]F_x -T cos \theta = 0[/tex]

     =>     [tex]F_x = Tcos \theta[/tex]

Substituting values

            [tex]F_x =T cos (60)[/tex]

                 [tex]F_x= \frac{T}{2} ---(1)[/tex]

Now about the y-axis the moment is  

           [tex]F_y + Tsin \theta = M *g + 3M *g ----(2)[/tex]

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          [tex]M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0[/tex]

            [tex]\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0[/tex]

               [tex]\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}[/tex]

               [tex]T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }[/tex]

                   [tex]T= \frac{11}{2\sqrt{3} } Mg[/tex]

The horizontal force provided by the hinge is

             [tex]F_x= \frac{T}{2} ---(1)[/tex]

Now substituting for T

              [tex]F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}[/tex]

                  [tex]Fx= \frac{11}{4\sqrt{3} } Mg[/tex]

The answer is:[tex]\frac{4Mg}{\sqrt{3}}.[/tex]

To solve this problem, we will analyze the forces acting on the uniform beam and apply the conditions for rotational and translational equilibrium. Here's the step-by-step solution:

1. Identify the forces acting on the beam:

- The weight of the beam, [tex]\( W_b = Mg \)[/tex], acting at the center of mass of the beam, which is at [tex]\( \frac{L}{2} \)[/tex] from the wall.

- The weight of the mass suspended from the bar, [tex]\( W_s = 3Mg \), acting at \( \frac{3L}{4} \)[/tex] from the wall.

- The tension in the cable, [tex]\( T \)[/tex], acting at an angle[tex]\( \theta = 60^\circ \)[/tex] with the horizontal.

- The horizontal force provided by the hi-nge,[tex]\( F_h \)[/tex].

2. Break down the tension in the cable into its horizontal and vertical components:

- The horizontal component of the tension,[tex]\( T_h = T \cos(60^\circ) \).[/tex]

- The vertical component of the tension,[tex]\( T_v = T \sin(60^\circ) \)[/tex].

3. Apply the condition for translational equilibrium in the vertical direction:

- The sum of the vertical forces must be zero:[tex]\( T_v + F_h = W_b + W_s \).[/tex]

- Substituting the expressions for the weights, we get: [tex]\( T \sin(60^\circ) = Mg + 3Mg \).[/tex]

- Simplifying, [tex]\( T \sin(60^\circ) = 4Mg \)[/tex].

4. Solve for the tension [tex]\( T \)[/tex]:

[tex]- \( T = \frac{4Mg}{\sin(60^\circ)} \).- Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), we have \( T = \frac{4Mg}{\frac{\sqrt{3}}{2}} = \frac{8Mg}{\sqrt{3}} \).[/tex]

5. Apply the condition for rotational equilibrium about the hi-nge:

- The sum of the torques about the hi-nge must be zero.

- The torque due to the weight of the beam is [tex]\( \tau_b = W_b \left(\frac{L}{2}\right) = Mg \left(\frac{L}{2}\right) \)[/tex].

- The torque due to the weight of the suspended mass is[tex]\( \tau_s = W_s \left(\frac{3L}{4}\right) = 3Mg \left(\frac{3L}{4}\right) \).[/tex]

 - The torque due to the tension in the cable is[tex]\( \tau_T = T_v \left(\frac{L}{2}\right) \)[/tex] (since the tension acts at the end of the beam).

- Setting the sum of the torques to zero: [tex]\( \tau_T = \tau_b + \tau_s \).[/tex]

- Substituting the expressions for the torques, we get: [tex]\( T \sin(60^\circ) \left(\frac{L}{2}\right) = Mg \left(\frac{L}{2}\right) + 3Mg \left(\frac{3L}{4}\right) \)[/tex]

- We already know that [tex]\( T \sin(60^\circ) = 4Mg \)[/tex], so this equation is satisfied, confirming that our value for [tex]\( T \)[/tex] is correct.

6. Solve for the horizontal force [tex]\( F_h \)[/tex]:

- From the vertical equilibrium equation, [tex]\( T \sin(60^\circ) = 4Mg \)[/tex] , we have already found [tex]\( T \)[/tex].

- The horizontal component of the tension is [tex]\( T_h = T \cos(60^\circ) \)[/tex].

- Since there are no other horizontal forces, [tex]\( F_h = T_h \)[/tex].

- Substituting the value of[tex]\( T \), we get \( F_h = \frac{8Mg}{\sqrt{3}} \cos(60^\circ) \).[/tex]

- Since [tex]\( \cos(60^\circ) = \frac{1}{2} \), we have \( F_h = \frac{8Mg}{\sqrt{3}} \cdot \frac{1}{2} = \frac{4Mg}{\sqrt{3}} \).[/tex]

Therefore, the tension in the cable is [tex]\( \boxed{\frac{8Mg}{\sqrt{3}}} \)[/tex] and the horizontal force provided by the hi-nge is[tex]\( \boxed{\frac{4Mg}{\sqrt{3}}} \).[/tex]

Answer:

Explanation:

7 rev /s = 7 x 2π rad /s

angular velocity = 14π rad /s

Angular acceleration α = increase in angular velocity / time

= 14π - 0 / 9

α = 4.8844 rad / s

θ  = 1/2 α x t²   θ is angle of rotation , t is time

= 1/2 x 4.8844  x 9²

= 197.8182 rad

2π n = 197.8182

n = 31.5 rotation

During acceleration , no of rotation made = 31.5

During deceleration : -----

deceleration =

Angular deceleration α = decrease in angular velocity / time

= 14π - 0 / 13

α = 3.3815  rad / s

θ  = 1/2 α x t²   θ is angle of rotation , t is time

= 1/2 x 3.3815  x 13²

= 285.736 rad

2π n = 285.736

n = 45.5 rotation

total rotation

= 45.5 + 31.5

= 77 .

Answer:

4117.65 N

Explanation:

Speed of ball, u = 126 km/h = 35 m/s

Mass of ball, m = 160 g = 0.16 kg

Time interval, t = 1.36 ms = 0.00136 s

We can calculate the force as a measure of the momentum of the ball:

F = P/t

Momentum, P, is given as:

P = mv

Therefore:

F = (mv) / t

F = (0.16 * 35) / (0.00136)

F = 4117.65 N

The force imparted to the two glove d hands of the inexperienced catcher is 4117.65 N.

Answer:

200 nm

Explanation:

Refractive index of gasoline = 1.4

Wavelength = 560 nm

t = Thickness of film

m = Order = 1

Wavelength is given by

[tex]\lambda=\dfrac{560}{1.4}=400\ nm[/tex]

We have the relation

[tex]2t=m\lambda\\\Rightarrow t=\dfrac{m\lambda}{2}\\\Rightarrow t=\dfrac{1\times 400}{2}\\\Rightarrow t=200\ nm[/tex]

The thickness of the film is 200 nm

Answer:

12078.46 V

Explanation:

Applying,

E₀ = BANω.................... Equation 1

Where: E₀ = maximum emf, B = magnetic Field, A = Area of the coil, N = Number of turns of the coil. ω = angular velocity

Given: N = 99 turns, B = 2.96 T, ω = 1200 rev/min = (1200+0.10472) = 125.664 rad/s

A = L×W, where L= Length = 93.9 cm = 0.939 m, W = width = 34.9 cm = 0.349 m

A = (0.939×0.349) = 0.328 m²

Substitute into equation 1

E₀ = 99(2.96)(0.328)(125.664)

E₀ = 12078.46 V

Hence the maximum value of the emf produced = 12078.46 V

Answer:

The maximum emf induced in the loop is 9498.268 V

Explanation:

Given;

number of turns of coil, N = 99 turns

area of the rectangular loop, A = 0.739 m x 0.349 m = 0.2579 m²

magnetic field strength, B = 2.96 T

angular speed of the loop, ω = 1200 rev/min

angular speed of the loop, ω (rad/s) = (2π x 1200) / 60 = 125.68 rad/s

The maximum value of the emf produced is calculated using the formula below;

ξ = NABω

Substitute the given values and calculate the maximum emf induced;

ξ = (99)(0.2579)(2.96)(125.68)

ξ = 9498.268 volts

Therefore, the maximum emf induced in the loop is 9498.268 V

Answer:

1.attraction and repulsion behavior do the north and south poles of a magnet exhibit?A.Like poles repel each other, and opposite poles attract.

2.Explain the left-hand rule for flux direction in conductors. Ans.The left-hand rule is used to determine the direction of rotation of the flux around the conductor

.3.What is the most common source of power for most HVACR systems?Ans.Alternating current is the most common source of power used for most HVACR systems.

4.What is the purpose of transformers?Ans..Transformers are used to increase or decrease incoming voltages to meet the requirements of the load.

5.What is a solenoid valve?Ans.Solenoid valves are electrically operated valves that open or close automatically by energizing or de-energizing a coil of wire.6.Is there an electrical connection between the primary and secondary windings of a transformer?A.There is no electrical connection between the primary and secondary windings.

Explanation:

The poles of a magnet attract and repel based on their polarity. The left-hand rule explains how magnetic fields interact with electric currents. Most HVACR systems use electricity, and transformers convert voltage levels to transmit power efficiently. A solenoid valve controls fluid or gas flow through electromagnetism.

The north and south poles of a magnet exhibit a behavior where like poles repel and opposite poles attract. This means that north pole to north pole or south pole to south pole will repel, while north pole to south pole will attract.

The left-hand rule for flux direction, often used in electromagnetism, states that if you point your thumb in the direction of the current, the direction in which your fingers curl represents the direction of the magnetic field lines. This is significant in understanding the interaction of electric current with magnetic fields in conductors.

The most common source of power for most HVACR (Heating, Ventilation, Air Conditioning, and Refrigeration) systems is electricity. This power is used to run the system's various components including compressors, fans, and controls.

Transformers are used in electric power systems to convert voltages from one level to another. This is mainly used to transmit electricity over long distances with low energy losses and is essential in power grids.

A solenoid valve is an electromechanical device used to control the flow of a fluid or gas. When an electric current is passed through the coil, a magnetic field is created causing the valve to open or close.

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Answer:

is closest to  100*C  temperature  at the aluminum-copper junction

Explanation:

The expression for calculating the resistance of each rod is given by

[tex]R =\frac{ L}{ kA}[/tex]

Now; for Aluminium

[tex]R_{al} =\frac{ 0.20 }{ 205*0.0004}[/tex]

[tex]R_{al}[/tex] = 2.439

For Copper

[tex]R_{Cu}=\frac{0.70}{385*0.0004}[/tex]

[tex]R_{Cu} = 4.545[/tex]

Total Resistance [tex]R = R_{al} + R_{Cu}[/tex]

= 2.439 + 4.545

= 6.9845

Total temperature  difference = 230*C + 30*C

= 200 *C

The Total rate of heat flow is then determined which is  = [tex]\frac{ total \ temp \ difference}{total \ resistance }[/tex]

=[tex]\frac{200}{ 6.9845 }[/tex]

= 28.635 Watts

However. the temperature difference across the aluminium = Heat flow × Resistance of aluminium

= 28.635 × 2.349

= 69.84 *C

Finally. for as much as one end of the aluminium is = 30 *C , the other end is;

=30*C + 69.84*C  

= 99.84  *C

which is closest to  100*C  temperature  at the aluminum-copper junction

Answer:

The magnitude of the change in gravitational potential energy of the earth-athlete system over the course of the race is 643,536Joules

Explanation

Potential energy is one of the form of mechanical energy and it is defined as the energy possessed by a body due to virtue of its position. When the body is under gravity, it possesses an energy called the gravitational potential energy.

Gravitational potential energy is expressed as shown:

GPE = mass × acceleration due to gravity × height

Given mass of athlete = 80kg

height covered = 820m

acceleration due to gravity = 9.81m/s

GPE = 80×9.81×820

GPE = 643,536Joules

Answer:

a) [tex]\Delta \theta = 617.604\,rad[/tex], b) [tex]\Delta t = 10.392\,s[/tex], c) [tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]

Explanation:

a) The final angular speed at the end of the acceleration stage is:

[tex]\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)[/tex]

[tex]\omega = 111.5\,\frac{rad}{s}[/tex]

The angular deceleration is:

[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}[/tex]

[tex]\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}[/tex]

[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]

The change in angular position during the acceleration stage is:

[tex]\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}[/tex]

[tex]\theta = 177.604\,rad[/tex]

Finally, the total change in angular position is:

[tex]\Delta \theta = 440\,rad + 177.604\,rad[/tex]

[tex]\Delta \theta = 617.604\,rad[/tex]

b) The time interval of the deceleration interval is:

[tex]\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }[/tex]

[tex]\Delta t = 7.892\,s[/tex]

The time required for the grinding wheel to stop is:

[tex]\Delta t = 2.50\,s + 7.892\,s[/tex]

[tex]\Delta t = 10.392\,s[/tex]

c) The angular deceleration of the grinding wheel is:

[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]

Answer:

Ф_T =544

The wheel will stop after 10 s.  

α_z = -11.15 rad/s^2

Explanation:

The angular acceleration is constant. Thus, we will apply the equations of rotation with constant angular acceleration model.

(a) In order to calculate the total angle, we will divide the entire interval from t = 0 to the time the wheel stops into two intervals.  

From t = 0 to t = 2 s:  

Ф-Ф_o =1/2(w_0z+w_z)t                       (1)

We will calculate w_z first:  

w_z = w_0x +α_xt

w_z = 24 + (35)(2.5)

w_z = 111.5

Substitute w_x into Eq.1  

Ф-Ф_o = 1/2(24+111.5)(2)

Ф-Ф_o = 136 rad

We can calculate it directly from the following equation:  

Ф-Ф_o = w_0x*t+1/2a_xt^2

Ф-Ф_o = 24*2.5+1/2*35*2.5

Ф-Ф_o =103.75 rad

Thus, the total angle the wheel turned between t = 0 and the time it stopped:  

Ф_T =103.75 +440

Ф_T =544

(b)  

We will take the interval from when the circuit breaker trips until the wheel comes to a stop.  

w_oz = 111.5 rad/s  

wz = 0 the wheel will stop at the end. 1

Ф-Ф_o =1/2(w_0z+w_z)t    

440 = 1/2(111.5+0)*t

t = 8s

Adding t of the first interval to t of the second interval :  

t_T = 2 + 8  =10

The wheel will stop after 10 s.  

(c)  

We will take the interval from when the circuit breaker trips until the wheel comes to a stop.  

w_oz = 111.5 rad/s  

w_z = 0  

 w_x =w_oz+α_z*t

α_z = -11.15 rad/s^2

Answer:

[tex]4.048\times 10^{-7}\ m[/tex]

Explanation:

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

E = Energy = [tex]4.91\times 10^{-19}\ J[/tex]

Wavelength ejected is given by

[tex]\lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{4.91\times 10^{-19}}\\\Rightarrow \lambda=4.048\times 10^{-7}\ m[/tex]

The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is [tex]4.048\times 10^{-7}\ m[/tex]

Answer:

3 Volts

Explanation:

Using Ohm's Law

         Voltage=current x resistance

              V = I x R

              V = 1.5 x 2

              V = 3 V

To solve for the velocity at t=0.40s of a mass in SHM, differentiate the position function with respect to time, and use the resulting velocity function. The spring constant can be calculated using Hooke's law, and the total energy can be found through the energy relationship in SHM.

The student is asking about the characteristics of simple harmonic motion (SHM) related to a mass attached to a spring. To find the velocity at a specific time t for the mass undergoing SHM, we need to differentiate the position function x(t) with respect to time. The provided position function is x(t)=(2.0 cm)cos(10t), implying that the velocity function will be v(t)=-ω A sin(ωt), where ω is the angular frequency, and A is the amplitude. To calculate the spring constant k and total energy E, we will employ both Hooke's law, F = -kx, and the relationship between the total energy in SHM and the displacement from the equilibrium position, given by E = ½ k A².

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The velocity at [tex]t=0.40s[/tex] is [tex]0.1514m/s[/tex].

The spring constant is [tex]6N/m[/tex].

The total energy of the mass is 0.0012J.

Determining Velocity, Spring Constant, and Total Energy in Simple Harmonic Motion

The position of a 60 g oscillating mass is given by the function: [tex]x(t) = (2.0 cm)cos(10t)[/tex], where time t is in seconds. Here's how we solve for the velocity at [tex]t = 0.40 s[/tex], the spring constant k, and the total energy E of the system.

Step-by-Step Solution

Answer:

a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged

Explanation:

When a ray of light passes from one transparent material to another, it must comply with the law of refraction

     n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the incident ray and index 2 for the refracted ray

With this expression let's examine the questions

a) They indicate that the refractive index increases,

      sin θ₂ = n₁ / n₂ sin θ₁

     θ₂ = sin⁻¹ (n₁ /n₂   sin θ₁)

    As m is greater than n1 the quantity on the right is less than one, the whole quantity in parentheses decreases so the Angle te also decreases

Answer is decrease

b) The wave velocity eta related to the wavelength and frequency

      v = λ f

The frequency does not change since the passage from one medium to the other is a process of forced oscillation, resonance whereby the frequency in the two mediums is the same.

The speed decreases with the indicated refraction increases and therefore the wavelength decreases

      λ = λ₀ / n

The answer is decrease

c) from the previous analysis the frequency remains unchanged

d) the refractive index is defined by

       n = c / v

So if n increases, the speed must decrease

The answer is decrease

e) the energy of the photon is given by the Planck equation

      E = h f

Since the frequency does not change, the energy does not change either

Answer remains unchanged

a) The Angle te also decreases, b) Decrease, c) Unchanged, d) The speed decrease, e) Unchanged

When a glimmer of light perishes from one translucent material to another, it must capitulate with the law of refraction

n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the happening ray and index 2 is for the refracted ray With this expression let's discuss.

a) They indicate that the refractive index increases,

sin θ₂ = n₁ / n₂ sin θ₁

θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)

As m is more significant than n1 the quantity on the ownership is less than one, the whole quantity in parentheses diminishes so the Angle te also decreases

Solution is decrease

b) The wave velocity eta connected to the wavelength and frequency

v = λ f

The frequency does not modify since the passage from one medium to the other is a procedure of forced oscillation and resonance whereby the frequency in the two mediums is the same.

The speed decreases with the foreshadowed refraction increases and thus the wavelength decreases

So, λ = λ₀ / n

The answer is to decrease

c) from the earlier analysis the frequency remains unchanged

d) When the refractive index is defined by

n is = c / v

Then, if n increases, the speed must decrease

The solution is to decrease

e) When the energy of the photon is given by the Planck equation

E is = hf. Since When the frequency does not transform, the energy does not change either Solution is remains unchanged

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Answer:

I = 215.76 A  

Explanation:

The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.  

Fm = u_o*I^2*L/2*π*d

where I is the current in each rod, d = 0.0082 m is the distance 27rId  

between each, L = 0.85 m is the length of each rod.

Fm = 4π*10^-7*I^2*1.1/2*π*0.0083

The mass of each rod is m = 0.0276 kg  

F_m = mg

4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8

I = 215.76 A  

note:

mathematical calculation maybe wrong or having little bit error but the method is perfectly fine

The current in the rods is approximately 132.75 A.

To determine the current in the rods, we need to use the formula for the force between two parallel current-carrying conductors, which is given by:

[tex]\[ F = \frac{\mu_0 I^2 L}{2 \pi d} \][/tex]

where:

F is the force between the two rods,

[tex]\( \mu_0 \)[/tex] is the permeability of free space [tex](\( 4\pi \times 10^{-7} \) T\·m/A)[/tex],

I is the current in each rod,

L is the length of the rods (1.1 m),

d is the distance between the rods (8.3 mm or [tex]\( 8.3 \times 10^{-3} \) m[/tex]).

The rods repel each other, and the force of repulsion must equal the gravitational force on the floating rod for it to levitate. The gravitational force is given by:

[tex]\[ F_g = m g \][/tex]

where:

m is the mass of the rod (0.11 kg),

g is the acceleration due to gravity (approximately 9.81 m/s²).

Setting the magnetic force equal to the gravitational force, we get:

[tex]\[ \frac{\mu_0 I^2 L}{2 \pi d} = m g \][/tex]

Solving for I, we have:

[tex]\[ I^2 = \frac{m g 2 \pi d}{\mu_0 L} \][/tex]

[tex]\[ I = \sqrt{\frac{m g 2 \pi d}{\mu_0 L}} \][/tex]

[tex]\[ I = \sqrt{\frac{0.11 \text{ kg} \times 9.81 \text{ m/s}^2 \times 2 \pi \times 8.3 \times 10^{-3} \text{ m}}{4\pi \times 10^{-7} \text{ T·m/A} \times 1.1 \text{ m}}} \][/tex]

[tex]\[ I = \sqrt{\frac{0.11 \times 9.81 \times 2 \times 8.3 \times 10^{-3}}{4\pi \times 10^{-7} \times 1.1}} \][/tex]

[tex]\[ I = \sqrt{\frac{0.11 \times 9.81 \times 16.6 \times 10^{-3}}{4\pi \times 10^{-7} \times 1.1}} \][/tex]

[tex]\[ I = \sqrt{\frac{1.77558 \times 10^{-2}}{4\pi \times 10^{-7}}} \][/tex]

[tex]\[ I = \sqrt{\frac{1.77558 \times 10^{-2}}{1.256637 \times 10^{-6}}} \][/tex]

[tex]\[ I = \sqrt{1413.55} \][/tex]

[tex]\[ I \approx 132.75 \text{ A} \][/tex]

Answer:

Inductance of first is one-4th that of the second inductor

Explanation:

See attached file

Answer:

the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]

the angle through which the apparatus turns in radians or degrees is :   [tex]\theta = 68.54^0[/tex]

Explanation:

Given that :

mass of uniform disk M = 1.2 kg

Radius R = 0.11 m

lower radius (b) = 0.14 m

small mass (m) = 0.4 kg

Force (F) = 21 N

time (∆t) =0.2 s

Moment of Inertia of [tex]I_{disk, CM}[/tex] = [tex]\frac{1}{2}MR^2[/tex]

= [tex]\frac{1}{2}*1.2*0.11^2[/tex]

= 0.00726 kgm²

Point mass  [tex]I_{point \ mass}[/tex] = mb²

But since four low rods are attached ; we have :

[tex]I_{point \ mass}[/tex] = 4 × mb²

= 4  × 0.4 (0.14)²

= 0.03136 kgm²

Total moment of Inertia =  [tex]I_{disk, CM}[/tex] + [tex]I_{point \ mass}[/tex]

= (0.00726 + 0.03136) kgm²

= 0.03862 kgm²

Assuming ∝ = angular acceleration = constant;

Then; we can use the following kinematic equations

T = FR

T = 2.1 × 0.11 N

T = 2.31 N

T = I × ∝

2.31 = 0.03862 × ∝

∝ = [tex]\frac{2.31}{0.03862}[/tex]

∝ = 59.81 rad/s²

Using the formula [tex]\omega_f = \omega_i + \alpha \delta T[/tex] to determine the angular speed of the apparatus; we have:

[tex]\omega_f =0 + 59.81*0.2[/tex]         since  [tex]( w_i \ is \ at \ rest ; the n\ w_i = 0 )[/tex]

[tex]\omega_f =11.962 \ rad/s[/tex]

∴ the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]

b) Using the formula :

[tex]\theta = \omega_i t + \frac{1}{2}*\alpha*(t)^2\\\\\theta = 0 *0.2 + \frac{1}{2}*59.81*(0.2)^2 \\ \\ \theta = 29.905 *(0.2)^2 \\ \\ \theta = 1.1962 rads \ \ \ ( to \ degree; \ we \ have) \\ \\ \theta = 68.54^0[/tex]

Thus, the angle through which the apparatus turns in radians or degrees is :   [tex]\theta = 68.54^0[/tex]

To solve the problem, compute the angular acceleration using the torque and moment of inertia. Given the angular acceleration and the time, calculate the angular velocity. Finally, use the angular acceleration and time to compute the angle turned through.

The first part of this problem involves calculating angular acceleration, which is the rate of change of angular velocity. Using the formula, angular acceleration (α) = Torque (τ) / Moment of Inertia (I), the torque can be calculated using the formula Torque = Force * Radius, and the moment of inertia is calculated using the given formulas for disk and point masses.

For the disk, I_(disk,CM) = 1/2 M R^2 and for the point masses, I_(point mass) = 4 * m * b^2. Summing these gives the total moment of inertia. The angular acceleration is then obtained by dividing the torque by the total moment of inertia.

Having the angular acceleration and the time, we can calculate the angular velocity (ω) using the formula ω = α * Δt.

For the second part, the angle through which the apparatus turns can be calculated using the formula θ = 0.5 * α * (Δt)^2, as the initial angular velocity was zero.

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Final answer:

To find the magnitude of the emf induced in the loop after 8.00 seconds has passed since the circumference started to decrease, we can use Faraday's law of electromagnetic induction. We calculate the rate of change of magnetic flux through the loop based on the changing area of the loop, and then determine the magnitude of the emf induced in the loop. The emf induced is -253.30 V, indicating that the induced current flows in a direction that opposes the change in magnetic flux.

Explanation:

To find the magnitude of the emf induced in the loop after 8.00 seconds has passed since the circumference started to decrease, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the emf induced in a loop is equal to the rate of change of magnetic flux through the loop. In this case, as the loop shrinks, its area decreases, resulting in a decrease in magnetic flux.

We know that the circumference of the loop is decreasing at a constant rate of 15.0 cm/s. Using the formula for the circumference of a circle, we can determine the radius of the circle at the given time: r = C / (2*pi), where C is the circumference and pi is a mathematical constant approximately equal to 3.14159. Substituting the given values, we get r = 168 cm / (2*3.14159) = 26.79 cm.

Next, we can calculate the area of the loop as a function of time using the equation A = pi*r^2. Substituting the value of the radius at 8.00 seconds, we get A = 3.14159 * (26.79 cm)^2 = 2252.68 cm^2.

Since the magnetic field is perpendicular to the loop and uniform in magnitude, we can calculate the rate of change of magnetic flux as: dPhi/dt = B*dA/dt, where B is the magnitude of the magnetic field and dA/dt is the rate of change of the area.

Finally, we can calculate the magnitude of the emf induced in the loop as: EMF = -dPhi/dt. The negative sign indicates that the induced current flows in a direction that opposes the change in magnetic flux. Substituting the given values, we get EMF = -0.900 T * (2252.68 cm^2) / 8.00 s = -253.30 V.

Answer:

[tex]2.429783984\times 10^{-14}\ A[/tex]

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]

Number of electrons passing per second

[tex]n_e=\dfrac{6020}{0.0476}\\\Rightarrow n_e=126470.588[/tex]

Number of protons passing per second

[tex]n_p=\dfrac{1681}{0.0662}\\\Rightarrow n_p=25392.749[/tex]

Current due to electrons

[tex]I_e=n_ee\\\Rightarrow I_e=126470.588\times 1.6\times 10^{-19}\\\Rightarrow I_e=2.0235\times 10^{-14}\ A[/tex]

Current due to protons

[tex]I_p=n_pe\\\Rightarrow I_p=25392.749\times 1.6\times 10^{-19}\\\Rightarrow I_p=4.06283984\times 10^{-15}\ A[/tex]

Total current

[tex]I=2.0235\times 10^{-14}+4.06283984\times 10^{-15}\\\Rightarrow I=2.429783984\times 10^{-14}\ A[/tex]

The average current is [tex]2.429783984\times 10^{-14}\ A[/tex]

The problem is likely an excessive combined power draw from the three heating elements, exceeding the circuit breaker's capacity; the recommendation is to upgrade the circuit breaker or adjust the furnace's operation to stay within the breaker's limit.

We have,

The problem seems to be related to the load on the circuit exceeding the capacity of the circuit breaker.

When the furnace is turned on and all three 5000-W heating elements turn on in stages, the combined power consumption becomes 3 * 5000 W = 15000 W.

This is a substantial load that exceeds the circuit breaker's capacity, which is likely 240 V * 60 A = 14400 W (due to the product of voltage and current rating).

As a result, the circuit breaker is tripping to protect the circuit from overloading, as the total power drawn from the furnace exceeds its rated capacity.

To correct this issue and prevent the circuit breaker from tripping, you could consider the following recommendations:

- Check the Circuit Breaker Rating:

Confirm the rating of the circuit breaker connected to the furnace. If it's indeed 60 A, you might need to upgrade the circuit breaker to a higher rating that can handle the combined power consumption of all three heating elements.

- Reduce Load:

Alternatively, you could reconfigure the furnace to operate only one or two heating elements at a time to reduce the load and stay within the circuit breaker's capacity.

This may involve adjusting the furnace's internal settings or installing additional controls to manage the heating elements' activation.

- Consider Energy Management:

Implement an energy management system that staggers the activation of the heating elements over time.

This would ensure that the power demand doesn't exceed the circuit breaker's capacity during startup.

- Professional Electrician:

Thus,

The problem is likely an excessive combined power draw from the three heating elements, exceeding the circuit breaker's capacity; the recommendation is to upgrade the circuit breaker or adjust the furnace's operation to stay within the breaker's limit.

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The issue is likely that the electric furnace surpasses the 60-A limit of the circuit breaker when all heating elements are active, causing it to trip. A potential solution could be to upgrade the circuit breaker to a higher amperage rating or adjust the furnace so that it never exceeds 60 A.

The problem the homeowner is encountering with their electric furnace is likely due to the furnace exceeding the capacity of the 240-V, 60-A circuit breaker when all three heating elements activate. Each 5000-W heating element at 240 V consumes about 20.8 A of current. If all three elements are powered simultaneously, the total current draw is 62.4 A, which surpasses the 60-A circuit breaker limit and causes it to trip.

One potential solution would be to upgrade the circuit breaker to one with a higher amperage rating. However, any modifications should comply with local electricity standards and should be carried out by a qualified electrician. Alternatively, the furnace could be rewired or adjusted to ensure that the cumulative draw never surpasses 60 A at any given time.

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Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = [tex]5.676*10^{-3} \ mm[/tex]

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  [tex]\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}][/tex]

where;

[tex]\epsilon = 0[/tex]

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

= [tex]\sqrt { \frac{100*32.2}{200}}[/tex]

= 4.0124

replacing them into the above equation and making X the subject of the formula:

[tex]X =[/tex] [tex]Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}[/tex]

[tex]X =[/tex] [tex]0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}[/tex]

[tex]X =[/tex] [tex]5.676*10^{-3} \ mm[/tex]

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = [tex]5.676*10^{-3} \ mm[/tex]

Answer:

Hi Carter,

The complete answer along with the explanation is shown below.

I hope it will clear your query

Pls rate me brainliest bro

Explanation:

The magnitude of the magnetic field on the axis of a circular loop, a distance z  from the loop center, is given by Eq.:

B = NμοiR² / 2(R²+Z²)³÷²

where

R is the radius of the loop

N is the number of turns

i is the current.

Both of the loops in the problem have the same radius, the same number of turns,  and carry the same current. The currents are in the same sense, and the fields they  produce are in the same direction in the region between them. We place the origin  at the center of the left-hand loop and let x be the coordinate of a point on the axis  between the loops. To calculate the field of the left-hand loop, we set z = x in the  equation above. The chosen point on the axis is a distance s – x from the center of  the right-hand loop. To calculate the field it produces, we put z = s – x in the  equation above. The total field at the point is therefore

B = NμοiR²/2 [1/ 2(R²+x²)³÷²   + 1/ 2(R²+x²-2sx+s²)³÷²]

Its derivative with respect to x is

dB /dx=  - NμοiR²/2 [3x/ (R²+x²)⁵÷²   + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]

When this is evaluated for x = s/2 (the midpoint between the loops) the result is

dB /dx=  - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷²   - 3(s/2)/(R²+s²/4)⁵÷² ] =0

independent of the value of s.

Answer:

Explanation:

find the solution below

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g=[tex]426\times 10^{-3} kg[/tex]

1 kg=1000 g

Radius,r=[tex]\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m[/tex]

1m=100 cm

Height,h=5m

[tex]I=\frac{2}{2}mr^2[/tex]

a.By law of conservation of energy

[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh[/tex]

[tex]\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh[/tex]

[tex]v=\omega r[/tex]

[tex]gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2[/tex]

[tex]\omega^2=\frac{6}{5r^2}gh[/tex]

[tex]\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s[/tex]

Where [tex]g=9.8m/s^2[/tex]

b.Rotational kinetic energy=[tex]\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J[/tex]

Rotational kinetic energy=8.35 J