Lenny Conundrum - Answers only

I'll explain the previous LC since there wasn't a decent explanation before.

I'm pretty sure everyone got the first equation easily enough:
P = pgh

Do some elementary substitution (50 kPa = 50,000 Pa):
50,000 = .998*9.81*h
h = 5.107 m

Great, now for the slightly harder part: finding the volume of the water taken up by 5.107 m above the bottom of a sphere. This requires some basic integral calculus and a little bit of simple geometry.

The radius of a cross-section of a sphere is sqrt( 8.6h - h^2 ), found using some simple pythagorean geometry.
Therefore the area of a cross-section is:
A = pi*r^2
A = pi*( 8.6h - h^2 )

Almost done now. All we have to do is integrate the area on the interval 0 to 5.107 with respect to h:
V = int_(0 --> 5.107) (A dh)
V = int_(0 --> 5.107) ( pi*( 8.6h - h^2 ) dh )

You could use your graphing calculator to find this integral for you, but it would be slighly inaccurate due to the approximation method. Thus we would need to find the indefinite integral first:

int ( pi*( 8.6h - h^2 ) dh ) = pi*( 4.3h^2 - 1/3 *h^3 )+C
C = 0, for obvious reasons if you've taken calculus and understand what we are doing

Then just solve the definite integral found 2 steps ago, convert from m^3 to litres, round, and you get:
V = 212,280 litres

Now that wasn't that hard now was it?

Hope that helped

 

first: sorry i don't want to spam, but i have to say this!

There was a descent explanation, if you know some basic geometrical formulas...
the way you did it looks to me very complicated... all i did was: using google!

1 Pa = 0,0001020 m water column (water head, static head, pressure head... don't know, which is the right english word)

so 50000 PA = 5,1 m water above the drain... that'll be "h" to put in the following formula

that's the one you find using google with partial sphere volumen... you got all you need already...

H= 5.1 (or 5.0... depends on the temperature)
pi= ~3.1415...
r = 4,3 (because 8.6 is the diameter)

now: fill in and find out

 

New lenny is out !
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm.

If the density of the steel is exactly 7.8 g/cm3, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s2, what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.

 

Using the stress equation [1]:
rho = F/A

where rho = 400 MPa = 400,000,000 kg/m·s2 is the tensile stress that steel can withstand, F is the force that the steel has to withstand and A is the crossed area of the steel bar.

The crossed area is a regular pentagon can be calculated by the following equation [2]:
A = 5/4 t2 tan(54°)

where t = 2 cm = 0.02 m is the side of the pentagon.

Now, we can calculate the force that the steel bar can withstand:
F = rho * A = rho * 5/4 t2 tan(54°)

Using Newton's gravity law [3], we have:
F = m * g

Therefore, the mass that the steel bar can widthstand is:
m = F /g = rho * 5/4 t2 tan(54°) / g

where g = 9.81 m/s2 is the gravitational acceleration.

To make it a little bit easier, you can put in the numbers and calculate the mass m now. Don't round m yet. The mass is contributed by 3 components: the big sphere m1, small spheres m2, and the steel bar m3. Since all of them are steel, they have the same density. We have:
m1 + int(h/3) * m2 + m3 = m
<=>[V1 + int(h/3) * V2 + A * h] * d = m

where h is the length of the steel bar, int is the floor function (every 3 meter, we have one small sphere), d = 7.8 g/cm3 = 7,800 kg/m3 is the density of steel, V1, V2 are the volumes of big and small spheres. The volume of the steel bar is the product of the crossed area with the length of the bar.

We can calculate V1, V2 by the sphere volume formula [4]:
V1 = 4/3 * pi * (r1)3
V2 = 4/3 * pi * (r2)3

where the radius r1 = d1/2 = 25 cm = 0.25 m, r2 = d2/2 = 30 cm = 0.15 m. We already know A, so the equation now should be:
V1 + int(h/3) * V2 + A * h = m/d
<=>int(h/3) * V2 + A * h = m/d - V1

We know the right hand side of the equation but the left hand side is a little bit tricky. So, we can do a method of trial-and-error to test. Solve h by the following equation:
h/3 * V2 + A * h = m/d - V1
<=>(V2/3 + A) * h = m/d - V1
<=>h = (m/d - V1)/(V2/3 + A)

Now round h down to an integer. Then, test h into the equation int(h/3)* V2 + A * h = m/d - V1.
We then compare the left hand side with the right hand side. If the left hand side is smaller than the right hand side, add 1 to h and test again. If the left hand side is greater than the right hand side, subtract 1 from h and test again.

The largest h that is satisfied that the left hand side is smaller than the right hand side will be the answer. Remember do not round during calculate all the above equations (except for the last equation to calculate h). Make sure you use SI units when calculating also (kg, m... not cm, g...)

 

Here's what I found from another forum, not the same as fastbullet..

THE ANSWER:
rho = F /A

rho = 400,000,000 kg/m*s^2

rho is stress, F = force, A = cross area of bar

A = 5/4*t^2*tan(54) (54 is degrees)

t = 0.02 m

F = rho * 5/4*t^2*tan(54)

F = m*g

m = F /g

g = 9.81 obviously

m = (mass of big sphere) + (mass of little spheres) + (mass of bar)

mass of little spheres = l/3*(mass of 1 little sphere, which is volume of little sphere * d) (l = length of bar rounded down)
mass of bar = A*l*d (d = density which is 7800 kg/m^3)
mass of big sphere = d*(volume of big sphere)

volume of big sphere = 4/3 * pi * (.25m)^3 (the .25m is the radius of the big sphere)
volume of bar = 4/3 * pi * (.15m)^3

if we multiply out by the densities we get

(volume of big sphere) + (l/3)*(volume of little sphere) + A*l = m/d
(l/3)*(volume of little sphere) + A*l = m/d - (volume of big sphere)

now we solve for l (the length it can be)

l = (m/d - (volume of the big sphere))/((volume of little sphere)/3 + A)

(volume of big sphere) = 0.0654497917
(volume of little sphere) = 0.014137155

F = rho * A
F = 275276.384
m=F/g
m =

l = (28060.7935/7800 - (0.0654497917)) / (0.014137155/3 + 0.00068819096)
l = 654.020583


Then you round l and plug it into the previous forumla

l/3* (volume of the small sphere) + A*l = m/d - (volume of the big sphere)
l/3 * 0.014137155 + 0.00068819096 * l = 28060.7935/7800 - 0.0654497917
l/3 * 0.014137155 + 0.00068819096 * l = 3.53208784


If the left < the right add 1 to l, if it's >, take away one from l, the largest l you get without going over is the answer

(but you don't have to go any higher then 654 which you got for l)


They got 654 and submitted that as the right answer.

 

What's the next word in this sequence?

Cream
Radish
And
______

Submit only a one-word answer.

 

its from the Kelp
Menu 1 (Appetizers), Item 1, 2nd WORD IS "Cream"
Menu 2 (Main Courses), Item 2, 2nd WORD IS "Radish"
Menu 3 (Desserts), Item 3, 2nd WORD IS "and"
Menu 4 (Cocktails), Item 4, 2nd WORD IS "Juice"
So im thinking its Juice

 

A new LC is out now

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

How many people did not properly follow the instructions on last week's Lenny Conundrum, and submitted more than a single word when the Conundrum specifically required a one-word answer?

Please round to the nearest 50, and please submit only a number, otherwise you will be disqualified like all those people from last week.

 

Haha, this time's lenny conundrum is complete random guessing,
I think I would guess around the 10k range?

 

Since it says to round to the nearest 50, I think it is under 10k.

~M

 

its not random guess its a math problem

 

It was a random guess, and the answer was 750.

The current competition... no idea.

 

I love these types of problems >

1 = 1!
721 = 1!+6!
727 = 1!+6!+3!
5767 = 1!+6!+3!+7!
5791 = 1!+6!+3!+7!+4!
5792 = 1!+6!+3!+7!+4!+1!

Look that the numbers 1, 6, 3, 7, 4, 1 and then consider the phrase "A Chomby was walking down a path one day." The numbers of letters in each word of that phrase are 1, 6, 3, 7, 4, 1. Coincidence? I think not. After seeing that the, the rest of the problem is really quite simple. The next number would be 1!+6!+3!+7!+4!+1!+(number of letters in the next word)! which is equal to 1!+6!+3!+7!+4!+1!+4! which is also equal to 5816

 

I forgot my answer but I just got this. I remember getting it from here though.

Dear ~-~-~,

Congratulations! You have guessed correctly in the Lenny Conundrum game (round 263). You have won 1714 NP!

Yours Sincerely,
The Neopets Team!

Web : http://www.neopets.com
Dont forget to tell your friends about us - send them a greeting at http://www.neopets.com/sendgreeting.phtml.

 

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

What is the missing word in this sequence?
allowed
here
entry
________
you
here
you

Please enter only a single word, with no other information. Otherwise, your entry will be disqualified.

 

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

You are presented with the following gibberish:

WLTT ZXPYGWO WR IRKBPP PK BOX OPHBVUALPN XR OOX YFHOMXPTL BVCRVDH?

"Here's a tool to help you," Jhudora said as she handed this to you:

[big wheel right here]

What is the answer to the coded question? Please submit only a single-word answer.

 

thankyou SOC seldom see ppls help out LC ans nowaday!!!