randomperson123 wrote:Is the hovercraft written test self-scheduled at nationals?
Based on the tentative schedule in the rulebook, hovercraft is completely self schedule.
randomperson123 wrote:Is the hovercraft written test self-scheduled at nationals?
BasuSiddha23 wrote:How many rolls of pennies can your hovercrafts hold?
I just want to see our performance compared to other teams.
jgrischow1 wrote:Sorry if something like this has already been posted, but we found a really good way to increase the number of pennies we could hold. When my kids were just randomly poking holes in the bottom of their skirt, they could only get 3-4 pennies on. We created a simple device in which graph paper was taped on to a piece of cardboard. We then poked thumbtacks through the intersections of the graph paper. We must have had hundreds of tacks on it. The kids pierced their skirts with the device, and the even holes really helped, enabling them to get 12-13 penny roles and a 15 second time.
IvanGe wrote:Does anyone know if they will give you credit for an answer being 51.50N when the answer key says 51.52N? Does that small difference affect if you get credit or not?
MIScioly1 wrote:IvanGe wrote:Does anyone know if they will give you credit for an answer being 51.50N when the answer key says 51.52N? Does that small difference affect if you get credit or not?
100% dependent on the supervisor. If the supervisor is requesting the calculation to be done in a certain way using sig figs, then I suppose that difference could affect credit. That being said, I would give it full credit if I were the supervisor.
CMS AC wrote:How significant is sig figs though? Like how many points would they take off?(No pun intended)
MIScioly1 wrote:IvanGe wrote:Does anyone know if they will give you credit for an answer being 51.50N when the answer key says 51.52N? Does that small difference affect if you get credit or not?
100% dependent on the supervisor. If the supervisor is requesting the calculation to be done in a certain way using sig figs, then I suppose that difference could affect credit. That being said, I would give it full credit if I were the supervisor.
IvanGe wrote:i didn't feel like posting this in the Question Marathon, so i guess i'll just post it here because it's more like a check-my-work-please.
A bungee jumper of mass 40 kg is attached to a bungee cord with a constant of 80 n/m. The unstretched length of the bungee is 15m. What is the maximum velocity of the jumper? (g = 10 m/s^2)
I did it by using the potential energy of the spring and when it is at maximum velocity, the potential energy should completely change into kinetic energy(even though it doesn't in this case). So :
Hooke's law
f = kx to find x
PE = KE + PEspring
mgh = 1/2kx^2 + 1/2mv^2
assuming h is equal to 15 + x
The answer key said 17.32 m/s and I did not get that by doing this, instead I got 18.71 m/s^.
MattChina wrote:IvanGe wrote:i didn't feel like posting this in the Question Marathon, so i guess i'll just post it here because it's more like a check-my-work-please.
A bungee jumper of mass 40 kg is attached to a bungee cord with a constant of 80 n/m. The unstretched length of the bungee is 15m. What is the maximum velocity of the jumper? (g = 10 m/s^2)
I did it by using the potential energy of the spring and when it is at maximum velocity, the potential energy should completely change into kinetic energy(even though it doesn't in this case). So :
Hooke's law
f = kx to find x
PE = KE + PEspring
mgh = 1/2kx^2 + 1/2mv^2
assuming h is equal to 15 + x
The answer key said 17.32 m/s and I did not get that by doing this, instead I got 18.71 m/s^.
Idk man, Got the same answer.
UTF-8 U+6211 U+662F wrote:MattChina wrote:IvanGe wrote:i didn't feel like posting this in the Question Marathon, so i guess i'll just post it here because it's more like a check-my-work-please.
A bungee jumper of mass 40 kg is attached to a bungee cord with a constant of 80 n/m. The unstretched length of the bungee is 15m. What is the maximum velocity of the jumper? (g = 10 m/s^2)
I did it by using the potential energy of the spring and when it is at maximum velocity, the potential energy should completely change into kinetic energy(even though it doesn't in this case). So :
Hooke's law
f = kx to find x
PE = KE + PEspring
mgh = 1/2kx^2 + 1/2mv^2
assuming h is equal to 15 + x
The answer key said 17.32 m/s and I did not get that by doing this, instead I got 18.71 m/s^.
Idk man, Got the same answer.
I got the same answer.
- For those confused about how the answer was obtained (like I was)
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