Asst 11 Final

course Phy 232

011. `Query 10

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Question: **** Query introductory set six, problems 11-14 **** given the length of a string how do we determine the wavelengths of the first few harmonics?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&&

******

if you have the length of the string and the number of waves, you can get the wavelength by dividing length of string/waves, and that should be the wavelength.

******

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confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc..

So you get

1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be

1 * 1/2 `lambda = L so `lambda = 2 L.

For 2 wavelengths fit into the string you get

2 * 1/2 `lambda = L so `lambda = L.

For 3 wavelengths you get

3 * 1/2 `lambda = L so `lambda = 2/3 L; etc.

Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc..

FOR A STRING FREE AT ONE END:

The wavelengths of the first few harmonics are found by the node - antinode distance between the ends. The first node corresponds to 1/4 wavelength. The second harmonic is from node to antinode to node to antinode, or 4/3. the third and fourth harmonics would therefore be 5/4 and 7/4 respectively. **

Your Self-Critique: OK

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Question: **** Given the wavelengths of the first few harmonics and the velocity of a string, how do we determine the frequencies of the first few harmonics?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&&

******

v/lambda = frequency

******

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confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** The frequency is the number of crests passing per unit of time.

We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second.

So frequency is equal to the wave velocity divided by the wavelength. **

Your Self-Critique: OK

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Question: **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&&

******

v^2 = T/(m/length)

v = sqrt(m/length)

******

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confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** We divide tension by mass per unit length:

v = sqrt ( tension / (mass/length) ). **

Your Self-Critique: OK

Your Self-Critique rating #$&* OK

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Question: query univ phy problem 15.48 (19.32 10th edition) y(x,t) = .75 cm sin[ `pi ( 250 s^-1 t + .4 cm^-1 x) ] What are the amplitude, period, frequency, wavelength and speed of propagation?

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*****

y(x,t) = .75sin[pi ( 250 s^-1 t + .4 cm^-1 x)]

a)

Amplitude = .75cm

Wavelength = 2/(.4rad/cm) = 5cm

Frequency = omega/(2pi) = 250pis^-1/(2pi) = 125 Hz

T = 1/f = 0.008s

v = lambda*f = 125s^-1*5cm = 625cm/s

*****

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confidence rating #$&*

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Given Solution:

** y(x, t) = A sin( omega * t + k * x), where amplitude is A, frequency is omega / (2 pi), wavelength is 2 pi / x and velocity of propagation is frequency * wavelength. Period is the reciprocal of frequency.

For A = .75 cm, omega = 250 pi s^-1, k = .4 pi cm^-1 we have

A=.750 cm

frequency is f = 250 pi s^-1 / (2 pi) = 125 s^-1 = 125 Hz.

period is T = 1/f = 1 / (125 s^-1) = .008 s

wavelength is lambda = (2 pi / (.4 * pi cm^-1)) = 5 cm

speed of propagation is v = frequency * wavelength = 125 Hz * 5 cm = 625 cm/s.

Note that v = freq * wavelength = omega / (2 pi) * ( 2 pi ) / k = omega / k. **

STUDENT COMMENT:

2*pi is one full cycle, but since the function is cos, everything is multiplied by pi. So does this mean that the cos function only represents a half a cycle?

It's late and I might well be missing something, but I don't think the cosine is mentioned in this problem. Let me know if I'm wrong.

I believe that only the sine function is mentioned. However there isn't much difference between the two (only a phase difference of pi/2) and everything below would apply to either:

pi is there because of the periodicity of the sine and cosine functions, as described below.

A sine or cosine function completes a full cycle as its argument changes by 2 pi.

The argument might be a function of clock time, or of position.

If the argument is of the form omega * t, then a period is completed every time omega * t changes by 2 pi. This occurs when t changes by 2 pi / omega. So the period of the function is 2 pi / omega.

If the argument is of the form k x, then it changes by 2 pi every time x changes by 2 pi / k, so the wavelength of the function is 2 pi / k.

STUDENT COMMENT

the book has this as position = Acos(k*x+ -omega*t)

and velocity omega*A*sin(k*x-omega*t)

this is no big deal, they mean the same as the student sort of mentions, that the sine is 2pi shifted which is in the wave number.

INSTRUCTOR RESPONSE

It doesn't make a lot of difference. The sine or cosine is a valid function to use, and whether you use k x - omega t, kx + omega t, -kx + omega t or -kx - omega t you get the equation of a traveling harmonic wave that satisfied the wave equation y_tt = 1/c * y_xx, where for example _tt represents second derivative with respect to t, and c represents the propagation velocity of the wave.

The general form of a traveling harmonic wave can be taken as

y(x, t) = A cos(k x - omega t + phi)

where phi is the initial phase. Any initial phase is possible. If phi = -pi/2 then since sin(theta) = cos(theta - pi/2) the equation could be

y(x, t) = A sin(kx - omega t).

If phi = 3 pi / 2 then since sin(theta) = - sin(theta + pi) = sin(-theta) we have

y(x, t) = A sin(omega t - k x).

Recall that the graph of y = f( x - h) is identical in shape to the graph of y = f(x), but translated h units in the horizontal direction.

Now kx - omega t can be written as k ( x - omega / k * t), and f(k ( x - omega / k * t) ) can therefore be seen as a horizontal translation of the graph of y = f(kx), the amount of the translation being omega / k * t . This translation is positive, so as t increases the horizontal translation increases, and the graph progresses to the right. The rate of progression with respect to t is omega / k, which is therefore the speed of propagation.

Thus the argument k x - omega t guarantees that the function y = f( k x - omega t) represents an unchanging shape that moves to the right with speed omega / k.

If f is a sine or cosine function, or any superposition of sine and cosine functions, we have a harmonic wave.

Another interpretation is that k x - omega t = -omega ( t - k / omega * x). In this case k / omega * x is regarded as the time required for the wave to propagate from position 0 to position x. The function f(-omega * t) that describes the motion of the point x = 0 in time is translated in space, so that the same motion occurs at position x after a 'time lag' of k / omega * x.

Similar analysis shows that a function of the form y = f( omega t + k x) represents a wave traveling in the negative x direction.

You'll probably have more questions, which I'll welcome.

Your Self-Critique: In the first line of the solution, it says that 2 pi / x is the wavelength, but I thought it would be 2pi/k? Is this a typo or do I have my k and x mixed up?

That's a typo; x is a variable to 2 pi / x doesn't make sense.

Thanks for pointing out the error.

Your Self-Critique rating #$&* OK

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Question: **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&

*****

For my t=0s, the crest is right over the t=0, for my t= .0005, and t =.0010s, the crest has moved to the left (moreso for .0010s than .0005s).

&&&&&

*****

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** Basic precalculus: For any function f(x) the graph of f(x-h) is translated `dx = h units in the x direction from the graph of y = f(x).

The graph of y = sin(k * x - omega * t) = sin(k * ( x - omega / k * t) ) is translated thru displacement `dx = omega / k * t relative to the graph of sin(k x).

At t=0, omega * t is zero and we have the original graph of y = .75 cm * sin( k x). The graph of y vs. x forms a sine curve with period 2 pi / k, in this case 2 pi / (pi * .4 cm^-1) = 5 cm which is the wavelength. A complete cycle occurs between x = 0 and x = 5 cm, with zeros at x = 0 cm, 2.5 cm and 5 cm, peak at x = 1.25 cm and 'valley' at x = 3.75 cm.

At t=.0005, we are graphing y = .75 cm * sin( k x + .0005 omega), shifted -.0005 * omega / k = -.313 cm in the x direction. The sine wave of the t=0 function y = .75 cm * sin(kx) is shifted -.313 cm, or .313 cm left so now the zeros are at -.313 cm and every 2.5 cm to the right of that, with the peak shifted by -.313 cm to x = .937 cm.

At t=.0010, we are graphing y = .75 cm * sin( k x + .0010 omega), shifted -.0010 * omega / k = -.625 cm in the x direction. The sine wave of the t = 0 function y = .75 cm * sin(kx) is shifted -.625 cm, or .625 cm left so now the zeros are at -.625 cm and every 2.5 cm to the right of that, with the peak shifted by -.625 cm to x = +.625 cm.

The sequence of graphs clearly shows the motion of the wave to the left at 625 cm / s. **

Your Self-Critique: OK

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Question: ** Velocity of propagation is

v = sqrt(T/ (m/L) ). Solving for T:

v^2 = T/ (m/L)

v^2*m/L = T

T = (6.25 m/s)^2 * 0.5 kg/m so

T = 19.5 kg m/s^2 = 19.5 N approx. **

Your Self-Critique: I really like questions like this. :)

Your Self-Critique rating #$&* OK

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Question: **** What is the average power?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&

*****

According to the book, P_average = .5sqrt(m/L*F)*omega^2*A^2

*****

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confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** The text gives the equation Pav = 1/2 sqrt( m / L * F) * omega^2 * A^2 for the average power transferred by a traveling wave.

Substituting m/L, tension F, angular frequency omeage and amplitude A into this equation we obtain

Pav = 1/2 sqrt ( .500 kg/m * 19.5 N) * (250 pi s^-1)^2 * (.0075 m)^2 =

.5 sqrt(9.8 kg^2 m / (s^2 m) ) * 62500 pi^2 s^-2 * .000054 m^2 =

.5 * 3.2 kg/s * 6.25 * 10^4 pi^2 s^-2 * 5.4 * 10^-5 m^2 =

5.5 kg m^2 s^-3 = 5.5 watts, approx..

The arithmetic here was done mentally so double-check it. The procedure itself is correct. **

Your Self-Critique: I see.

Your Self-Critique rating #$&* OK" "

011. `Query 10

*********************************************

Question: **** Query introductory set six, problems 11-14 **** given the length of a string how do we determine the wavelengths of the first few harmonics?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&&

******

if you have the length of the string and the number of waves, you can get the wavelength by dividing length of string/waves, and that should be the wavelength.

******

&&&&&&

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc..

So you get

1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be

1 * 1/2 `lambda = L so `lambda = 2 L.

For 2 wavelengths fit into the string you get

2 * 1/2 `lambda = L so `lambda = L.

For 3 wavelengths you get

3 * 1/2 `lambda = L so `lambda = 2/3 L; etc.

Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc..

FOR A STRING FREE AT ONE END:

The wavelengths of the first few harmonics are found by the node - antinode distance between the ends. The first node corresponds to 1/4 wavelength. The second harmonic is from node to antinode to node to antinode, or 4/3. the third and fourth harmonics would therefore be 5/4 and 7/4 respectively. **

Your Self-Critique: OK

Your Self-Critique rating #$&* OK

*********************************************

Question: **** Given the wavelengths of the first few harmonics and the velocity of a string, how do we determine the frequencies of the first few harmonics?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&&

******

v/lambda = frequency

******

&&&&&&

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The frequency is the number of crests passing per unit of time.

We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second.

So frequency is equal to the wave velocity divided by the wavelength. **

Your Self-Critique: OK

Your Self-Critique rating #$&* OK

*********************************************

Question: **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&&

******

v^2 = T/(m/length)

v = sqrt(m/length)

******

&&&&&&

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** We divide tension by mass per unit length:

v = sqrt ( tension / (mass/length) ). **

Your Self-Critique: OK

Your Self-Critique rating #$&* OK

*********************************************

Question: query univ phy problem 15.48 (19.32 10th edition) y(x,t) = .75 cm sin[ `pi ( 250 s^-1 t + .4 cm^-1 x) ] What are the amplitude, period, frequency, wavelength and speed of propagation?

&&&&&

*****

y(x,t) = .75sin[pi ( 250 s^-1 t + .4 cm^-1 x)]

a)

Amplitude = .75cm

Wavelength = 2/(.4rad/cm) = 5cm

Frequency = omega/(2pi) = 250pis^-1/(2pi) = 125 Hz

T = 1/f = 0.008s

v = lambda*f = 125s^-1*5cm = 625cm/s

*****

&&&&&

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** y(x, t) = A sin( omega * t + k * x), where amplitude is A, frequency is omega / (2 pi), wavelength is 2 pi / x and velocity of propagation is frequency * wavelength. Period is the reciprocal of frequency.

For A = .75 cm, omega = 250 pi s^-1, k = .4 pi cm^-1 we have

A=.750 cm

frequency is f = 250 pi s^-1 / (2 pi) = 125 s^-1 = 125 Hz.

period is T = 1/f = 1 / (125 s^-1) = .008 s

wavelength is lambda = (2 pi / (.4 * pi cm^-1)) = 5 cm

speed of propagation is v = frequency * wavelength = 125 Hz * 5 cm = 625 cm/s.

Note that v = freq * wavelength = omega / (2 pi) * ( 2 pi ) / k = omega / k. **

STUDENT COMMENT:

2*pi is one full cycle, but since the function is cos, everything is multiplied by pi. So does this mean that the cos function only represents a half a cycle?

It's late and I might well be missing something, but I don't think the cosine is mentioned in this problem. Let me know if I'm wrong.

I believe that only the sine function is mentioned. However there isn't much difference between the two (only a phase difference of pi/2) and everything below would apply to either:

pi is there because of the periodicity of the sine and cosine functions, as described below.

A sine or cosine function completes a full cycle as its argument changes by 2 pi.

The argument might be a function of clock time, or of position.

If the argument is of the form omega * t, then a period is completed every time omega * t changes by 2 pi. This occurs when t changes by 2 pi / omega. So the period of the function is 2 pi / omega.

If the argument is of the form k x, then it changes by 2 pi every time x changes by 2 pi / k, so the wavelength of the function is 2 pi / k.

STUDENT COMMENT

the book has this as position = Acos(k*x+ -omega*t)

and velocity omega*A*sin(k*x-omega*t)

this is no big deal, they mean the same as the student sort of mentions, that the sine is 2pi shifted which is in the wave number.

INSTRUCTOR RESPONSE

It doesn't make a lot of difference. The sine or cosine is a valid function to use, and whether you use k x - omega t, kx + omega t, -kx + omega t or -kx - omega t you get the equation of a traveling harmonic wave that satisfied the wave equation y_tt = 1/c * y_xx, where for example _tt represents second derivative with respect to t, and c represents the propagation velocity of the wave.

The general form of a traveling harmonic wave can be taken as

y(x, t) = A cos(k x - omega t + phi)

where phi is the initial phase. Any initial phase is possible. If phi = -pi/2 then since sin(theta) = cos(theta - pi/2) the equation could be

y(x, t) = A sin(kx - omega t).

If phi = 3 pi / 2 then since sin(theta) = - sin(theta + pi) = sin(-theta) we have

y(x, t) = A sin(omega t - k x).

Recall that the graph of y = f( x - h) is identical in shape to the graph of y = f(x), but translated h units in the horizontal direction.

Now kx - omega t can be written as k ( x - omega / k * t), and f(k ( x - omega / k * t) ) can therefore be seen as a horizontal translation of the graph of y = f(kx), the amount of the translation being omega / k * t . This translation is positive, so as t increases the horizontal translation increases, and the graph progresses to the right. The rate of progression with respect to t is omega / k, which is therefore the speed of propagation.

Thus the argument k x - omega t guarantees that the function y = f( k x - omega t) represents an unchanging shape that moves to the right with speed omega / k.

If f is a sine or cosine function, or any superposition of sine and cosine functions, we have a harmonic wave.

Another interpretation is that k x - omega t = -omega ( t - k / omega * x). In this case k / omega * x is regarded as the time required for the wave to propagate from position 0 to position x. The function f(-omega * t) that describes the motion of the point x = 0 in time is translated in space, so that the same motion occurs at position x after a 'time lag' of k / omega * x.

Similar analysis shows that a function of the form y = f( omega t + k x) represents a wave traveling in the negative x direction.

You'll probably have more questions, which I'll welcome.

Your Self-Critique: In the first line of the solution, it says that 2 pi / x is the wavelength, but I thought it would be 2pi/k? Is this a typo or do I have my k and x mixed up?

Your Self-Critique rating #$&* OK

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Question: **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&

*****

For my t=0s, the crest is right over the t=0, for my t= .0005, and t =.0010s, the crest has moved to the left (moreso for .0010s than .0005s).

&&&&&

*****

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** Basic precalculus: For any function f(x) the graph of f(x-h) is translated `dx = h units in the x direction from the graph of y = f(x).

The graph of y = sin(k * x - omega * t) = sin(k * ( x - omega / k * t) ) is translated thru displacement `dx = omega / k * t relative to the graph of sin(k x).

At t=0, omega * t is zero and we have the original graph of y = .75 cm * sin( k x). The graph of y vs. x forms a sine curve with period 2 pi / k, in this case 2 pi / (pi * .4 cm^-1) = 5 cm which is the wavelength. A complete cycle occurs between x = 0 and x = 5 cm, with zeros at x = 0 cm, 2.5 cm and 5 cm, peak at x = 1.25 cm and 'valley' at x = 3.75 cm.

At t=.0005, we are graphing y = .75 cm * sin( k x + .0005 omega), shifted -.0005 * omega / k = -.313 cm in the x direction. The sine wave of the t=0 function y = .75 cm * sin(kx) is shifted -.313 cm, or .313 cm left so now the zeros are at -.313 cm and every 2.5 cm to the right of that, with the peak shifted by -.313 cm to x = .937 cm.

At t=.0010, we are graphing y = .75 cm * sin( k x + .0010 omega), shifted -.0010 * omega / k = -.625 cm in the x direction. The sine wave of the t = 0 function y = .75 cm * sin(kx) is shifted -.625 cm, or .625 cm left so now the zeros are at -.625 cm and every 2.5 cm to the right of that, with the peak shifted by -.625 cm to x = +.625 cm.

The sequence of graphs clearly shows the motion of the wave to the left at 625 cm / s. **

Your Self-Critique: OK

Your Self-Critique rating #$&* OK

*********************************************

Question: ** Velocity of propagation is

v = sqrt(T/ (m/L) ). Solving for T:

v^2 = T/ (m/L)

v^2*m/L = T

T = (6.25 m/s)^2 * 0.5 kg/m so

T = 19.5 kg m/s^2 = 19.5 N approx. **

Your Self-Critique: I really like questions like this. :)

Your Self-Critique rating #$&* OK

*********************************************

Question: **** What is the average power?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

&&&&&

*****

According to the book, P_average = .5sqrt(m/L*F)*omega^2*A^2

*****

&&&&&

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The text gives the equation Pav = 1/2 sqrt( m / L * F) * omega^2 * A^2 for the average power transferred by a traveling wave.

Substituting m/L, tension F, angular frequency omeage and amplitude A into this equation we obtain

Pav = 1/2 sqrt ( .500 kg/m * 19.5 N) * (250 pi s^-1)^2 * (.0075 m)^2 =

.5 sqrt(9.8 kg^2 m / (s^2 m) ) * 62500 pi^2 s^-2 * .000054 m^2 =

.5 * 3.2 kg/s * 6.25 * 10^4 pi^2 s^-2 * 5.4 * 10^-5 m^2 =

5.5 kg m^2 s^-3 = 5.5 watts, approx..

The arithmetic here was done mentally so double-check it. The procedure itself is correct. **

Your Self-Critique: I see.

Your Self-Critique rating #$&* OK"

&#Good work. See my notes and let me know if you have questions. &#