lol.. ah come on, physics and calculus r da bomb..

I'm like you sk8.. When it comes to english, I don't know jack. I don't understand how people can pull out all these symbols and junk from books and poems.. Why can't you just read somethn and not try to link it to some crazy allusion lol.

No better at spelling.. I usually copy everything I write, paste it in a word document, then use the spell checker lol

The past few days in my English class have been stories all about symbolism and I didn't get any of it. So with the most recent one we were supposed to read I tried looking for symbols in it. The next day in school the teacher tells us that the story was written to show a point that too much symbolism is in short stories and that there actually wasn't any in his story. Mathematics is so much more straightfoward to me.

commented: I agree lol -Josh +2

The past few days in my English class have been stories all about symbolism and I didn't get any of it. So with the most recent one we were supposed to read I tried looking for symbols in it. The next day in school the teacher tells us that the story was written to show a point that too much symbolism is in short stories and that there actually wasn't any in his story. Mathematics is so much more straightfoward to me.

haha yea same here.. when it comes to symbolism and all that other bs, I go straight to sparknotes.com lol.. math is very straightforward.

And again, since I have already asked a question, the next person to post (who hasn't already) gets to ask a question.. I'm hoping this will encourage more people to play:)

The average power exerted by a bicyclist ( m = 75.0kg ) is 6.50 W per kilogram of his body mass. (a) How much work does he do during a 135-km race in which his average speed is 12.0m/s? (b) Express the work done in part (a) in terms of nutritional Calories.

hah. Calories and calories

a) 5484375 Joules

b) 1310797.083 calories or 1310.797083 Calories.

I got 5,484.375 J and 1 310.79708 calories ?

commented: hey cutie.. it's me again. yay! i love you mucho grande. +4

I got 5,484.375 J and 1 310.79708 calories ?

I think you forgot to change kilometers into meters.

hm, how did you get 5.48 x 10^6 instead of x 10^3?

so who's go?

We three had the same answer (besides the km to m conversion) but she hasn't said that's right yet.

Oh sorry.. lol
It's right.

trying to trick us up.. that's not nice

;)

Okay, I'll go one more.

When the carnival comes to town, there is a ride where passengers get into a bullet shaped cart. The cart is attached to a stiff metal arm that is 13m long and attached to a rotating motor. When the passengers are fastened into the cart, the ride turns the arm in circles like the hands of a clock. The passengers are turned upside-down as they are spun in circles. If the cart has a weight of 5395.5N and it has 5 passengers with masses of 85kg, 65kg, 100kg, 75kg, and 50kg, what is the potential energy of the cart and passengers when the ride is spun to 20 degrees above the horizontal (ignore the mass of the arm).

I'm going to bed though, I'll verify answers tomorrow. If you think you have it then go ahead to the next question if you want to.

Okay, I'll go one more.

When the carnival comes to town, there is a ride where passengers get into a bullet shaped cart. The cart is attached to a stiff metal arm that is 13m long and attached to a rotating motor. When the passengers are fastened into the cart, the ride turns the arm in circles like the hands of a clock. The passengers are turned upside-down as they are spun in circles. If the cart has a weight of 5395.5N and it has 5 passengers with masses of 85kg, 65kg, 100kg, 75kg, and 50kg, what is the potential energy of the cart and passengers when the ride is spun to 20 degrees above the horizontal (ignore the mass of the arm).

I'm going to bed though, I'll verify answers tomorrow. If you think you have it then go ahead to the next question if you want to.

4.327x10^5 J ?

assuming g=10 m/s^2

There's not enough information to do the problem. We need to know where is the center of mass of the cart-person system, relative to the point at which the metal arm connects to the cart. We also need to know the distance from the metal arm's connection point to the motor and the center of the axis of rotation. We also need to know what you consider to be the level of zero potential, relative to some part of the system you've described.

I just regarded the info. he didn't give us as the "all else equal" principle. I think I worked it correctly considering that everything else was negligible

Okay, so I can prove that there is enough information considering this was on a test in my physics honors class (boo yah). Potential energy is mass times gravity (I like to use 9.81 but 10 works too if you like to round) time height. You have the mass, it is the mass of the cart plus the people inside. You have gravity because it is common information. All you need is the height. Only the vertical height counts so when the 13m (which is the length of the arm which was given) arm is positioned twenty degrees above the horizontal the opposite side of the angle must be found. Well, sin20(degrees)=opp/hyp=x/13. This can be solved for x which gives us a height of 4.446m But you must add this to the height from the horizontal to the absolute minimum the ride can go which is the length of the arm (when the car would be as low as it can go) so that gives us 17.446m for the height, 9.81m for gravity and 925kg. (The F=ma equation is used to find the mass of the cart where acceleration is again 9.81m/s^2). This gives us a potential energy of 158309.3655 Joules

There's not enough information to do the problem. We need to know where is the center of mass of the cart-person system, relative to the point at which the metal arm connects to the cart. We also need to know the distance from the metal arm's connection point to the motor and the center of the axis of rotation. We also need to know what you consider to be the level of zero potential, relative to some part of the system you've described.

The center of mass of the cart-person system relative to the point at which the metal arm connects to the cart isn't needed to do the problem. The distance from the metal arm's connection point to the motor is given and is 13m. The center of the axis of rotation isn't necessary either (it's where the arm connects to the motor, I don't know how else to put it). The level of zero potential is always and is always assumed to be the lowest possibility in the situation. Usually this is the ground, but in this case it would be the lowest altitude the ride can go. This is always assumed in any scenario. Take a swing for example; the person on the swing can go no lower than the lowest altitude of the swing and so zero potential is assumed to be the minimal point of swinging.

Okay, so I can prove that there is enough information considering this was on a test in my physics honors class (boo yah). Potential energy is mass times gravity (I like to use 9.81 but 10 works too if you like to round) time height. You have the mass, it is the mass of the cart plus the people inside. You have gravity because it is common information. All you need is the height. Only the vertical height counts so when the 13m (which is the length of the arm which was given) arm is positioned twenty degrees above the horizontal the opposite side of the angle must be found. Well, sin20(degrees)=opp/hyp=x/13. This can be solved for x which gives us a height of 4.446m But you must add this to the height from the horizontal to the absolute minimum the ride can go which is the length of the arm (when the car would be as low as it can go) so that gives us 17.446m for the height, 9.81m for gravity and 925kg. (The F=ma equation is used to find the mass of the cart where acceleration is again 9.81m/s^2). This gives us a potential energy of 158309.3655 Joules

ah, I used the wrong height.. well damn, I keep missing these..

lol, it's okay. I had to draw a picture myself. I almost forget to add the extra thirteen meters. (You had the one with the fly, I just didn't explain it clearly enough.) Who's next?