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

Answers

Answer 1

Answer:

His angular velocity will increase.

Explanation:

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

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

where

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

ω' is the initial angular velocity

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

where m is the mass of the system

and r' is the initial radius of rotation

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

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

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

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

therefore, the angular velocity increases.


Related Questions

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.
a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?

Answers

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

A) Applying the energy equation

The positive terms is :   ΔUg The negative terms is :  ΔEth The zero term are :  ΔK  and ΔUs

B) The energy output by the athlete is ; 800 Joules

C) The metabolic power is : 2000 w

D) When he performs the task in 1.2 s

The metabolic energy he expends is : the same His metabolic power is :  more

Given data :

Weight of barbell = 400 N

Height = 2.0 m

Time = 1.6 secs

efficiency of the human body = 25%

Speed = constant

A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its  potential energy increases.  ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement

B)  Determine the energy output of the athlete

weight of barbell * Height  = 400 * 2 = 800 J

C) Determine the metabolic power

Metabolic power = energy output / Time

where ; energy output = 4 * 800 = 3200

∴ Metabolic power = 3200 / 1.6

                                = 2000 w

D) When performs same task at 1.2 s

The metabolic energy he expends is  the same  and His metabolic power is  more

Hence we can conclude that the answers to your questions are as listed above

Learn more : https://brainly.com/question/17807740

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

Answers

Answer:

Explanation:

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

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

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

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

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

Answers

Answer:

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

Explanation:

From the question we are told that

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

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

     

Generally the apparent depth of the coin is mathematically represented as

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

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

So

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

        [tex]D' = 0.00376 \ m[/tex]

The apparent depth will be 0.00376 m.

What is an index of refraction?

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

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

n is the refractive index of water =1.33

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

The apparent depth of the coin is given as;

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

Hence the apparent depth will be 0.00376 m.

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

https://brainly.com/question/23750645

A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is so large that the ball flies up at an angle of 30 degrees above the ground. The player however claims (s)he aimed her/his foot at a 40 degree angle above the ground. Calculate the average kicking force magnitude and the final speed of the ball, if you are given that the foot was in contact with the ball for one hundredth of a second.

Answers

Answer:

v_{f} = 74 m/s, F = 230 N

Explanation:

We can work on this exercise using the relationship between momentum and moment

        I = ∫ F dt = Δp

bold indicates vectors

we can write this equations in its components

X axis

       Fₓ t = m ( -v_{xo})

Y axis  

        t = m (v_{yf} - v_{yo})

in this case with the ball it travels horizontally v_{yo} = 0

Let's use trigonometry to write the final velocities and the force

        sin 30 = v_{yf} / vf

        cos 30 = v_{xf} / vf

        v_{yf} = vf sin 30

        v_{xf} = vf cos 30

         sin40 = F_{y} / F

         F_{y} = F sin 40

         cos 40 = Fₓ / F

         Fₓ = F cos 40

let's substitute

      F cos 40 t = m ( cos 30 - vₓ₀)

      F sin 40 t = m (v_{f} sin 30-0)

we have two equations and two unknowns, so the system can be solved

        F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)

        F sin 40 0.1 = 0.4 v_{f} sin 30

we clear fen the second equation and subtitles in the first

         F = 4 sin30 /sin40     v_{f}

         F = 3.111 v_{f}

        (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80

        v_{f} (3,111 cos 40 -4 cos30) = - 80

        v_{f} (- 1.0812) = - 80

        v_{f} = 73.99

        v_{f} = 74 m/s

now we can calculate the force

          F = 3.111 73.99

          F = 230 N

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

Answers

Answer:

a) The value of g at such location is:

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

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

[tex]T=1.9767 sec[/tex]

Explanation:

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

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

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

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

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

The location of a particle is measured with an uncertainty of 0.15 nm. One tries to simultaneously measure the velocity of this particle. What is the minimum uncertainty in the velocity measurement. The mass of the particle is 1.770×10-27 kg

Answers

Answer:

198 ms-1

Explanation:

According to the Heisenberg uncertainty principle; it is not possible to simultaneously measure the momentum and position of a particle with precision.

The uncertainty associated with each measurement is given by;

∆x∆p≥h/4π

Where;

∆x = uncertainty in the measurement of position

∆p = uncertainty in the measurement of momentum

h= Plank's constant

But ∆p= mΔv

And;

m= 1.770×10^-27 kg

∆x = 0.15 nm

Making ∆v the subject of the formula;

∆v≥h/m∆x4π

∆v≥ 6.6 ×10^-34/1.770×10^-27 × 1.5×10^-10 ×4×3.142

∆v≥198 ms-1

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

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

Answers

Answer:

a

  [tex]Q = 5.34 *10^{19} \ J[/tex]

b

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

Explanation:

From the question we are told that

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

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

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

       

Generally the volume of the ice is mathematically represented is

         [tex]V = a_i * t[/tex]

substituting value

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

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

Generally the mass of the ice is

       [tex]m_i = \rho_i * V[/tex]

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

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

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

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

        [tex]Q = m _i * H_f[/tex]

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

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

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

Considering part b

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

So  the time taken to melt the ice is

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

        [tex]T = 0.445 \ years[/tex]

converting to days

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

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answers

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is [tex]E^o_{cell} = 2.35 V[/tex]

Explanation:

From the question we are told that

   The ionic equation is  

               [tex]3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}[/tex]

Now under standard conditions the reduction  half reaction  is

      [tex]Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V[/tex]

And the oxidation half reaction is

      [tex]Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V[/tex]

The emf of this cell under standard conditions  is mathematically represented as

     [tex]E^o_{cell} = E^o _r - E^o _o[/tex]

substituting values

     [tex]E^o_{cell} = 1.61 - (- 0.74)[/tex]

    [tex]E^o_{cell} = 2.35 V[/tex]

     

Rank these electromagnetic waves on the basis of their speed (in vacuum). Rank from fastest to slowest.

a. Yellow light
b. FM radio wave
c. Green light
d. X-ray
e. AM radio wave
f. Infrared wave

Answers

Answer:

From fastest speed to slowest speed, the electromagnetic waves are ranked as(up to down):

d. X-ray

c. Green light

a. Yellow light

f. Infrared wave

b. FM radio wave

e. AM radio wave

Explanation:

Electromagnetic waves are waves produced as a result of vibrations between an electric field and a magnetic field. The waves have three properties and these properties are frequency, speed and wavelength, which are related by the relationship below

V = Fλ

where:\

V = speed (velocity)

F = frequency

λ = wavelength.

From the relationship above, it is seen that the speed of a wave is directly proportional to its frequency. The higher the frequency, the higher the speed. Therefore, from the list given, the waves with  the highest to lowest frequencies/ from left to right are:

X-ray (3×10¹⁹ Hz to 3×10¹⁶Hz), Green light (5.66×10¹⁴Hz), Yellow light (5.17×10¹⁴Hz), Infrared wave (3×10¹¹Hz), FM radio wave (10.8×10⁸Hz to 8.8×10⁷Hz), AM radio wave (1.72 × 10⁶Hz to 5.5×10⁵Hz).

This corresponds to the speed from highest to lowest from left to right.

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

Answers

Answer:

1 volt and 0.5 volt

Explanation:

Given data

voltage supplied Vs= 1.5 volts

resistance R1= 1000 ohms

resistance R2= 500 ohms

The total resistance is

Rt= 1000+ 500

Rt= 1500 ohms

The current I is given as

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

Voltage across R1

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

Voltage across R2

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

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

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

Answers

Answer:

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

Explanation:

Net force on the hoop is given by;

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

where;

a is linear acceleration

m is the mass

Net torque on the hoop is given by;

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

where;

I is moment of inertia

α is the angular acceleration

But, τ = Fr

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

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

let the angular acceleration of the smaller hoop = α₁

let the radius of the smaller hoop = r₁

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

let the angular acceleration of the larger hoop = α₂

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

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

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

Answers

Explanation:

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

Then from the figure we can observe that

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

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

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

There is a river in front of you that flows due South at 3.0m/s. You launch a toy boat across the river with the front of the boat pointed due East. When you tested the boat on a still pond, the boat moved at 4.0m/s. Now as it moves to the opposite bank, it travels at some speed relative to you, sitting in your chair. What is this speed

Answers

Answer:

5.0 m/s

Explanation:

If the river moves towards the south at 3m/s and the both moves towards the east at 4.0m/s, the speed of the boat relative to me will be the resulting displacement of both velocities of the river and that of the boat. This can be gotten using pythagoras theorem.

Let Vr be the relative speed. According to the theorem;

[tex]V_r^2 = V_s^2 + V_e^2\\\\V_r^2 = 3.0^2 + 4.0^2\\\\V_r^2 = 9+16\\\\V_r^2 = \sqrt{25}\\ \\V_r = 5.0m/s[/tex]

Hence this relative speed is 5.0 m/s

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

Answers

Answer:

Kidney

Explanation:

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

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

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

The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps and dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 1/2 in).

If we treat the spring assembly as a single spring, what is the approximate spring constant?

k= ____________

Answers

Answer:

The approximate  spring constant is  [tex]k = 55533.33 \ N/m[/tex]

Explanation:

From the question we are told that

   The  mass of the person is  [tex]m = 68 \ kg[/tex]

     The  dip of the car is  [tex]x = 1.2 \ cm = 0.012 \ m[/tex]

Generally according to hooks law  

        [tex]F = k * x[/tex]

here the force F is the weight of the person which is mathematically represented as

         [tex]F = m * g[/tex]

=>    [tex]m * g = k * x[/tex]

=>     [tex]k = \frac{m * g }{x }[/tex]

=>    [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]

=>   [tex]k = 55533.33 \ N/m[/tex]

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

Answers

Answer:

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

Explanation:

The velocity of the ball v = 30 m/s

The angle it makes with the horizontal ∅ = 40°

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

[tex]V_{x}[/tex] = v cos ∅

[tex]V_{x}[/tex]  = 30 cos 40°

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

A sinusoidal voltage Δv = (100 V) sin (170t) is applied to a series RLC circuit with L = 40 mH, C = 130 μF, and R = 50 Ω.

Required:
a. What is the impedance of the circuit?
b. What is the maximum current in the circuit?

Answers

Answer:

See attached file

Explanation:

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

Answers

Answer:

m = 22,877 kg / s

Explanation:

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

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

           Q = 300 60% / 40%

           Q = 450 MW

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

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

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

let's use the maximum temperature change allowed

           (T_{f} - T_{o}) = 5

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

     

power and energy are related

              W = Q / t

               Q = W t

             

let's calculate

             m = 450 10⁶ / (3934 5)

             m = 22,877 kg / s

A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)

Answers

Answer:I don’t know

Explanation:

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

Answers

Answer:

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

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

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

#FreeMelvin

A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement. How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.

Answers

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

Answer:

14,400 J

Explanation:

Its the answer

At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field to have a maximum value of the induced emf equal to 8.0 V

Answers

Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turns of the coil, 200 turn

cross sectional area of the coil, A = 300 cm² = 0.03 m²

magnitude of the magnetic field, B = 30 mT = 0.03 T

Maximum value of the induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

Where;

ω is angular frequency = 2πf

E = NBA(2πf)

f = E / 2πNBA

f = (8) / (2π x 200 x 0.03 x 0.03)

f = 7.07 Hz

Therefore, the frequency of the coil is 7.07 Hz

A 120-V rms voltage at 60.0 Hz is applied across an inductor, a capacitor, and a resistor in series. If the peak current in this circuit is 0.8484 A, what is the impedance of this circuit?
A) 200 Ω
B) 141 Ω
C) 20.4 Ω
D) 120 Ω
E) 100 Ω

Answers

Answer:A  200

Explanation:

Vp=1.41*Vrms

Vp=169.7 v

Z=Vp/Ip

Z=169.7/.8484

Z=200.03 ohm

Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minimum film thickness will result in minimum reflection of this light?

Answers

Answer:

tmin= lambda/2

Explanation:

See attached file pls

g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube

Answers

Answer:

The length of the tube is 85 cm

Explanation:

Given;

speed of sound, v = 340 m/s

first harmonic of open-closed tube is given by;

N----->A , L= λ/₄

λ₁ = 4L

v = Fλ

F = v / λ

F₁ = v/4L

Second harmonic of open-closed tube is given by;

L = N-----N + N-----A, L = (³/₄)λ

[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]

Third harmonic of open-closed tube is given by;

L = N------N + N-----N + N-----A, L = (⁵/₄)λ

[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]

The difference between second harmonic and first harmonic;

[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]

The difference between third harmonic and second harmonic;

[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]

Thus, the difference between successive harmonic of open-closed tube is

v / 2L.

[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]

Therefore, the length of the tube is 85 cm

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

Answers

Answer:

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

Explanation:

From the question we are told that

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

    The distance of the screen is   x  =  4.80 m

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

    The order of the fringe is  n  =  4

   

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

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

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

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

substituting values

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

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

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?

Answers

Answer:

The current is  [tex]I = 8.9 *10^{-5} \ A[/tex]

Explanation:

From the question we are told that

     The  radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]

      The current density is  [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]

      The distance we are considering is  [tex]r = 0.5 R = 0.001585[/tex]

Generally current density is mathematically represented as

          [tex]J = \frac{I}{A }[/tex]

Where A is the cross-sectional area represented as

         [tex]A = \pi r^2[/tex]

=>      [tex]J = \frac{I}{\pi r^2 }[/tex]

=>    [tex]I = J * (\pi r^2 )[/tex]

Now the change in current per unit length is mathematically evaluated as

        [tex]dI = 2 J * \pi r dr[/tex]

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         [tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]

         [tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]

substituting values

        [tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]

        [tex]I = 8.9 *10^{-5} \ A[/tex]

Helium-neon laser light (λ = 6.33 × 10−7 m) is sent through a 0.30 mm-wide single slit. What is the width of the central maximum on a screen 1.0 m from the slit?

Answers

Answer:

The width is [tex]w_c = 0.00422 \ m[/tex]

Explanation:

From the question we are told that

   The  wavelength is  [tex]\lambda = 6.33*10^{-7} \ m[/tex]

    The  width of the slit is  [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]

    The distance of the screen is  [tex]D = 1.0 \ m[/tex]

     

Generally the central maximum is mathematically represented as

      [tex]w_c = 2 * y[/tex]

Here  y is the width of the first order maxima which is mathematically represented as

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

substituting values

      [tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]

       [tex]y = 0.00211 \ m[/tex]

So  

    [tex]w_c = 2 *0.00211[/tex]

     [tex]w_c = 0.00422 \ m[/tex]

A rope, under a tension of 153 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by . where at one end of the rope, is in meters, and is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

Answers

Complete question is;

A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by

y = (0.15 m) sin[πx/3] sin[12π t].

where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?

Answer:

A) Length of rope = 4 m

B) v = 24 m/s

C) m = 1.0625 kg

D) T = 0.11 s

Explanation:

We are given;

T = 153 N

y = (0.15 m) sin[πx/3] sin[12πt]

Comparing this displacement equation with general waveform equation, we have;

k = 2π/λ = π/2 rad/m

ω = 2πf = 12π rad/s

Since, 2π/λ = π/2

Thus,wavelength; λ = 4 m

Since, 2πf = 12π

Frequency;f = 6 Hz

A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;

λ = 2L/n

Since second harmonic, n = 2 and λ = L = 4 m

Length of rope = 4 m

B) speed is given by the equation;

v = fλ = 6 × 4

v = 24 m/s

C) To calculate the mass, we will use;

v = √T/μ

Where μ = mass(m)/4

Thus;

v = √(T/(m/4))

Making m the subject;

m = 4T/v²

m = (4 × 153)/24²

m = 1.0625 kg

D) Now, the rope oscillates in a third harmonic.

So n = 3.

Using the formula f = 1/T = nv/2L

T = 2L/nv

T = (2 × 4)/(3 × 24)

T = 0.11 s

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

Answers

Answer: 56°

Explanation:

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

The formula is;

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

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

= 56.30993247

= 56°

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