Michael A. Bayne - Computer Systems Engineer, Department of Computer Science, University of Virginia - has accumulated 700 points.
Michael A. Bayne email: bayne@cs.virginia.edu
2*pi*r = 25000 mi, the circumference of the earth and what he was expecting
2*pi*(r + x) = 25000 mi + 1 yd, The circumference of his string, what he got
Massaging the second equation:
2*pi*(r + x) - 1 yd = 25000 mi
Setting the two equal to each other:
2*pi*r = 2*pi*(r + x) - 1 yd
r = r + x - (1 / (2 *pi)) yd
x = 1/ (2 * pi) yd
So the string is 1 / (2 * pi) yd above the surface, or about 5.73
inches.
The fun thing about this, though, is that it doesn't matter that the
string was wrapped about the earth. Since the circumference of the
earth did not factor in the final equation, adding the yard to the
circumference of the sun, Jupiter, or a tennis ball would have resulted
in the extra 5.73" above the surface.
War VI - battle 18
Michael A. Bayne - email: bayne@cs.virginia.edu
Drew Smith - email: drewsmith@aol.com
Bernard Frazer - email: b.frazer@sympatico.ca
War VI - battle 17
Drew Smith - email: drewsmith@aol.com
GLFlispart@aol.com - email: GLFlispart@aol.com
...The zero flag (ZF) is set to 1 if the buffer is empty,
or to 0 if a character is waiting. In the latter case, a
copy of the character at the head of the buffer is placed
in AX without removing it from the buffer. AL returns the
character code for one-byte ASCII characters, or it returns
ASCII 0 for extended codes, in which case the code number
appears in AH.
Extended codes exist as a mechanism for handling special keys
which exceed the original 128-character ASCII design, such as
the Function keys (F1-F12), Insert, Home, etc. For example,
the capital 'A' would show up as 41H in AL, while PgUP would
show up as 0 in AL, and 49H in AH.
Michael A. Bayne - email: bayne@cs.virginia.edu
War VI - battle 16
People had trouble with this forestry question. Some said because the picture was not clear enough. As a result I will let two players who together got five of the answers share the twenty points available for 10 points each. I will also add a clearer picture to be seen by clicking on the picture above. The two winners are:
(a) My yard right now is covered with black walnuts from the walnut tree, Juglans nigra. Normally by this time of year, I've had requests to come into my yard and pick them up from people who are out collecting along the roadside, picking up the nuts to deliver to processors such as the one in Stamping Ground, Kentucky. For a real treat, add walnut meats to turkey dressing at Thanksgiving. (b) Horse chestnut or Buckeye (genus Aesculus). The nuts have all kinds of great uses: carrying one in the pocket will not only bring good luck, but ease the pain of a toothache. (c) On my 212-acre farm, I have only two hickory trees, genus Carya. I was amazed to discover them, and have no idea how they got there, so far from any mates. The wood is a favorite for tool handles. (e) is the fruit of the Common Persimmon, genus Diospyros. They're not as tasty as the best cultivated types, but all the sweeter for being wild. But wait till they're ripe!! (g) pear (genus pyrus) The two trees these guys didn't get are: (d) Dogwood, genus Cornus. This fall specimen has the distinctive reddish purple leaves and, though a little hard to see, red berries. (f) Carolina buckthorn, genus Rhamnus.
Drew Smith - email: drewsmith@aol.com
Anne Lurie email: ALurie6171@aol.com
The bottom line:
Chambers hid the film.
Hiss was the alleged spy.
Richard M. Nixon was the figure who became President.
1948 was the year when the details began to surface.
See:
http://detnews.com/1996/menu/stories/75183.htmWar VI - battle 15
Tricholoma magnivelare
Cantharellus formosus
Boletus edulis
Tuber gibbosum
Leucangium carthusiana
Hydnum repandum
In what two ways are these significant in the forest?
FOR Z=1 TO 200
FL$="A:\"+RIGHT$(STR$(Z),LEN(STR$(Z))-1)
OPEN FL$ FOR OUTPUT AS 1
PRINT #1,FL$
CLOSE
NEXT Z
Drew Smith - email: drewsmith@aol.com
The term mycorrhiza refers to an association or symbiosis between plants and fungi that colonize the cortical tissue of roots during periods of active plant growth. The association is characterized by the movement of plant- produced carbon to the fungus and fungal-acquired nutrients to the plant.
Question 2 was unanswered in War VI.
Zaheer Jhetam has 20 points for his answer to the question below:
Michael A. Bayne - email: bayne@cs.virginia.edu has 20 points for his answer to the question below:
Given the muzzle velocity (400 ft/s) and the acceleration due to
gravity (16 * t * t), the position of the bullet at time t is:
s(t) = -16 * t * t + 400 * t
The bullet will strike the soldier when s = 0, so setting s(t) = 0 (and
ignoring the trivial answer of t=0s):
0 = -16 * t * t + 400 * t
16 * t * t = 400 * t
16 * t = 400
t = 25 s
The bullet will reach maximum height at half this time, or 12.5 s. The
maximum height will be:
s(12.5) = -16 * 12.5 * 12.5 + 400 * 12.5
s(12.5) = -2500 + 5000
s(12.5) = 2500 ft
So his life expectancy is 25 seconds, and the bullet reaches a maximum
height of 2500 ft (plus however tall he is since I placed my x-axis at a
point from where the bullet was fired).
I can see how they won the war, though. If he is accurate enough to
shoot straight up and hit himself, he would be absolutely deadly on the
battlefield. Even being off by a tenth of a degree would have resulted
in missing himself by 17 ft.
s_x = 400 * sin(0.1) * t = 400 * 0.001745 * 25 = 17.45
War VI - battle 14
Michael A. Bayne - email: bayne@cs.virginia.edu has 80 points for his answers to the questions below:
Drew Smith - email: drewsmith@aol.com
War VI - battle 13
DIM MONTH$(20),DAYS(20)
DATA JANUARY,31,FEBRUARY,28,MARCH,31,APRIL,30,MAY,31,JUNE,30
DATA JULY,31,AUGUST,31,SEPTEMBER,30,OCTOBER,31,NOVEMBER,30,DECEMBER,31,END,0
MO=0
WHILE Z$<>"END"
READ Z$,Z
MONTH$(MO)=Z$:DAYS(MO)=Z
MO=MO+1
WEND
IF YEAR/4=INT(YEAR/4) THEN DAYS(2)=29
JD=JULIANDATE
MONTH=1
WHILE JD>DAYS(MONTH)
JD=JD-DAYS(MONTH):MONTH=MONTH+1
WEND
PRINT "Julian date"+STR$(JULIANDATE)+" of year"+STR$(YEAR)+" is "+MONTH$(MONTH)+STR$(JD)+","+STR$(YEAR)+"."
Zaheer Jhetam has 20 points for his answer to the question below:
Selling now and investing at 10% per annum would give you (after 10 years): * 1,000 x $150 x (1.10^10) = $389,061 Should you not sell, you would, after 10 years, have some grade 2 timber... * Size of tract, given growth in volume: 1,000,000 x(1.05^10)=1,628,895 * Proportion of total tract that would be grade 2: (0.95^10)=0.5987 * Inflated price: $150 x (1.03^10)= $201.59 * Total value of grade 2 timber = 1,628.895 x $201.59 x 0.5987 =$196,592 ...and some grade 1 timber... * Size of tract, given growth in volume: 1,000,000 x (1.05^10)=1,628,895 * Proportion of total tract that would be grade 1: 1-0.5987 = 0.4013 * Inflated price: $250 x (1.03^10)= $335.98 * Total value of grade 1 timber = 1,628.895 x $335.98 x 0.4013 =$219,621 The total value of your tract would thus be $416,213 You would thus be $27 thousand better off if you kept your tract.From Duane:
VG2=1000000!:VG1=0:PG2=150:PG1=250:C=PG1*VG1/1000+PG2*VG2/1000
VR=1.05:PR=1.03:GR=.05:SR=1.1
CLS
PRINT " YEAR STOCKS VOL GR1 VOL GR2 PRICE 1 PRICE 2 TIMBER VALUE"
Y=C
PRINT USING"########";Z;C;VG1;VG2;PG1;PG2;Y
FOR Z=1 TO 10
VG2=VG2*VR:VG1=VG1*VR:X=GR*VG2:VG2=VG2-X:VG1=VG1+X
PG1=PG1*PR:PG2=PG2*PR
C=C*SR
Y=PG1*VG1/1000+PG2*VG2/1000
PRINT USING"########";Z;C;VG1;VG2;PG1;PG2;Y
NEXT
END
YEAR STOCKS VOL GR1 VOL GR2 PRICE 1 PRICE 2 TIMBER VALUE
0 150000 0 1000000 250 150 150000
1 165000 52500 997500 258 155 167633
2 181500 107494 995006 265 159 186850
3 199650 165106 992519 273 164 207787
4 219615 225469 990037 281 169 230586
5 241577 288719 987562 290 174 255404
6 265734 355002 985093 299 179 282411
7 292308 424470 982630 307 184 311788
8 321538 497281 980174 317 190 343733
9 353692 573604 977723 326 196 378462
10 389061 653615 975279 336 202 416205
Michael A. Bayne - email: bayne@cs.virginia.edu has 20 points for his answer to the question below:
Anne Lurie email: ALurie6171@aol.com
War VI - battle 12
Michael A. Bayne email: bayne@cs.virginia.edu
Complete directions can be found at http://www.nucleus.com/~kmcmurdo/win95logo.html.
War VI - battle 11
What family is this and what is the common name of this plant?
see Answer
ra
bmp
wmp
wav
jpg
mid
You know, we haven't got any business in those faraway wars.
Seven thousand miles is a long way to go to shoot somebody,
especially if you are not right sure they need shooting, and
you are not sure whether you are shooting the right side or
not. You see, it's their war and they have a right to fight
it as they see fit, without any advice from us.
Who was this American icon?
Michael A. Bayne - email: bayne@cs.virginia.edu has 20 points for his answer to the question below:
Question 2 was unanswered in War VI
sound picture
wav bmp - straight data mapping
ra jpg - compressed data formats
mid wmp - representing data by formulas
Anne Lurie email: ALurie6171@aol.com
War VI - battle 10
PRINT CHR$(218)+CHR$(196)+CHR$(191)
PRINT CHR$(192)+CHR$(196)+CHR$(217)
"Stop," says the second man. "I know their ages."
What are they?
see Answer
Zaheer Jhetam gains 80 points for the correct answers to the questions below:
9, 2, 2
6, 6, 1
The last bit of information excluded the oldest being a twin, so the only
acceptable set of ages is 9,2,2.
Michael A. Bayne - email: bayne@cs.virginia.edu
answered correctly both questions 2 and 4 also but about 17 minutes after Zaheer.
Michael gave the entire set of 3 numbers adding up to 13 and their product below:
1 1 11 11
1 2 10 20
1 3 9 27
1 4 8 32
1 5 7 35
1 6 6 36
2 2 9 36
2 3 8 48
2 4 7 56
2 5 6 60
3 3 7 63
3 4 6 72
3 5 5 75
4 4 5 80
So the age of the man who answered was 36.
Anne Lurie email: ALurie6171@aol.com
Introduction:
"The Book of the Generations of Adam was the Book of Remembrance commenced by
Adam subsequent to the expulsion from the Garden. In it he recorded his
experiences in the fallen world for the benefit of his posterity. This book
is referred to by Moses in his recounting of the history of the first ages of
the world. Sometime subsequent to the compilation of Moses' history, the Book
of Generations was lost from among men, but it was preserved in the mind of
God, the Holy Spirit, until the last dispensation when it is being brought
forth for the edification of those who are attempting to recreate at the end
of time that Holy Order which existed among our First Parents."
See: http://www.eagle-net.org/livingscripture/scripture/adam.html
War VI - battle 9
IF A>=256^3 THEN X$=CHR$(INT(A/256^3)):A=A-INT(A/256^3)*256^3 ELSE X$=CHR$(0)
IF A>=256^2 THEN X$=X$+CHR$(INT(A/256^2)):A=A-INT(A/256^2)*256^2 ELSE X$=X$+CHR$(0)
IF A>=256 THEN X$=X$+CHR$(INT(A/256)):A=A-INT(A/256)*256 ELSE X$=X$+CHR$(0)
IF A>255 THEN IF A<255.5 THEN A=255 ELSE A=0:MID$(X$,3,1)=CHR$(ASC(RIGHT$(X$,1))+1)
X$=X$+CHR$(A)
What is the largest integer that can be compressed by this algorithm? Design
an algorithm to uncompress this string and return the original integer.
Marino Jaen - email: marino@sinfo.net has 20 points for his answer to the question below:
Michael A. Bayne - email: bayne@cs.virginia.edu has 20 points for his answer to the question below:
A=ASC(MID$(X$,1,1))*256^3
A=A+ASC(MID$(X$,2,1))*256^2
A=A+ASC(MID$(X$,3,1))*256
A=A+ASC(MID$(X$,4,1))
Zaheer Jhetam gains 40 points for the correct answers to questions 3 and 4. His answers were dated a few hours before the answers given below as winners but the email did not reach me for five days for reasons unknown.
Krish Kannan - email: kannank@mindspring.com
Steven Skelton - email: Steven.Skelton@hhsc.state.tx.us
War VI - battle 8
Drew Smith - email: drewsmith@aol.com
War VI - battle 7

Who's this guy?
And what is his philosophical significance?
see Answer
Material Analysis -------- ------- Muriate of potash 0-0-60 DAP 18-46-0 Urea 46-0-0 Ammonia nitrate 34-0-0 Sulfate of potash 0-0-50If they mix my bulk fertilizer from DAP, Urea, and Sulfate of potash, how much actual material will be applied to my 2 acre tobacco patch?
Drew Smith - email: drewsmith@aol.com
Zaheer Jhetam has 20 points for his answer to the question below:
Michael A. Bayne - email: bayne@cs.virginia.edu
You'll need to apply this to your two acres:
Sulfate of potash: 1200 lbs DAP: 870 lbs Urea: 94 lbs Total 2164 lbs
War VI - battle 6
Michael A. Bayne - email: bayne@cs.virginia.edu
unknown left(with unknown) right 1 1 2 1 3 3 3 4 3 + 1 5 3 + 1 9 6 3 9 7 3 9 + 1 8 1 9 9 9 10 9 + 1 11 1 9 + 3 12 9 + 3 13 9 + 3 + 1 14 9 + 3 + 1 27 15 9 + 3 27 16 9 + 3 27 + 1 17 9 + 1 27 18 9 27 19 9 27 + 1 20 9 + 1 27 + 3 21 9 27 + 3 22 9 27 + 3 + 1 23 3 + 1 27 24 3 27 25 3 27 + 1 26 1 27 27 27 28 27 + 1 29 1 27 + 3 30 27 + 3 31 27 + 3 + 1 32 3 + 1 27 + 9 33 3 27 + 9 34 3 27 + 9 + 1 35 1 27 + 9 36 27 + 9 37 27 + 9 + 1 38 1 27 + 9 + 3 39 27 + 9 + 3 40 27 + 9 + 3 + 1 Of course, all this assumes that all the unknown weights will only be measures to the nearest pound.
Zaheer Jhetam has 20 points for his answer to the question below:
War VI - battle 5
He has achieved success who has
lived well,
laughed often
and loved much;
who has enjoyed the trust of pure women,
the respect of intelligent men
and the love of little children;
who has filled his niche
and accomplished his task;
who has left the world better than he found it,
whether by an improved poppy,
a perfect poem,
or a rescued soul;
who has never lacked appreciation of earth's beauty
nor failed to express it;
who has always looked for the best in others
and given them the best he had;
whose life was an inspiration;
whose memory is a benediction.
Anne Lurie - email: ALurie6171@aol.com
Russell Flowers - email: flowershome@accessky.net
War VI - battle 4
Drew Smith - email: drewsmith@aol.com
Michael A. Bayne - email: bayne@cs.virginia.edu
Ryan Bowling - email: rbowling@calclab.math.tamu.edu
10 * t = 2/8 * b t = time, b = length of bridge Tsp * t = b Tsp = train speedTwo equations, two unknowns, simple algebra to know that the train is travelling 4 times faster than the man, or 40 mph.
The game of life is hard to play
I'm going to lose it anyway
The losing card I'll someday lay
And this is all I have to say,
......................
The only way to win is cheat
And lay it down before I'm beat
And to another give a seat
For that's the only painless feat,
......................
The sword of time will pierce our skins
It doesn't hurt when it begins
but as it works its way on in
The pain grows stronger - watch it grin
Click here for an audio clue.Question 1 was unanswered in War VI.
Michael A. Bayne - email: bayne@cs.virginia.edu
Steven Skelton - email: Steven.Skelton@hhsc.state.tx.us
Drew Smith - email: drewsmith@aol.com
says that an excellent discussion of the relationship between Schrodinger's
work and Heisenberg's work can be found at
http://www.aip.org/history/heisenberg/
Drew Smith - email: drewsmith@aol.com
Question 1 was unanswered in War VI.
Drew Smith - email: drewsmith@aol.com
The years 1700, 1800, and 1900 were not leap years.
Zaheer Jhetam
answered later than Drew but gave a more comprehensive meaning for "yin" and "yang" as follows:
Tao, "the way", theorises that everything in the universe is made of two
conflicting forces. The Yin is the negative, passive power, depicted in
black. The yang is the positive, active power, depicted in white. Harmony
can only be achieved when the two are perfectly balanced, as in the circle.
The small circle of the opposite colour is contained in each, signifying
their interdependence. The yin may represent the soul, night, darkness, the
Earth, and sustenance, whereas the yang may represent the spirit, light,
day, heaven creation, and dominance.
Zaheer Jhetam has 20 points for his answer to the question below:
The total time taken will be
* T=distance/speed
* ={(25 + x^2)^(1/2)} / 4 + (5-x) / 6
And the minimum time will be obtained by differentiating this equation, and
setting the derivative equal to zero:
* dT/dx = 0
* x / {4(x^2 + 25)^(1/2)} -1/6 = 0
* which gives 6x = 4 {(x^2 + 25)}^(1/2)
* 36x^2 = 16 (x^2 + 25)
* 36x^2 = 16x^2 + 400
* 20x^2 = 400
* x^2 = 20
* x=20^(1/2)
* Therefore, he should land 4.472 miles away from the closest point
(or 0.528 miles away from his target)
Final point: this calculation assumes that no currents exist, either aiding
or impeding his swim!
Duane got the same answer with use of the following computer program:
CLS:DEFDBL A,D,T,S
B=0:E=5:S=.5
WHILE S>.0001
T=999:A=0
FOR L=B TO E STEP S
D1=SQR(25+L^2):T1=D1/4
D2=5-L:T2=D2/6
IF T1+T2<T THEN T=T1+T2:A=L
NEXT
PRINT A,T
B=A-S
S=S/2
WEND
END
The output of this program is:
4.5 1.76503630479
4.46875 1.76502839724
4.46875 1.76502839724
4.46875 1.76502839724
4.47265625 1.76502835750
4.470703125 1.76502831776
4.4716796875 1.76502827803
4.4716796875 1.76502827803
4.4719238281 1.76502823829
The man will have taken 1.765 hours to travel a total distance of
7.236 miles at an average speed of 4.10 mph.
Drew Smith - email: drewsmith@aol.com
Last revised December 2, 1999.
URL: http://www.webcom.com/duane/warmind6.html
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