| New Reply |
Ping-Pong balls in a room? |
Share Thread | Thread Tools |
| Feb17-10, 10:45 PM | #1 |
|
|
Ping-Pong balls in a room?
1. The problem statement, all variables and given/known data
Estimate the number of Ping - Pong balls that would fit into a typical-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. (Assume 25% of the space cannot be filled due to spherical packing.) (a) Find the volume of the room (in m^3) (b) Find the volume of a ball (in cm^3) (c) Find the number of balls (order of magnitude only) (balls) 2. Relevant equations 75% of 1000m^3 = 750m^3 750m^3(1 000 000 cm^3/1m^3) 3. The attempt at a solution (a)I said that an average room had 1000m^3 It said "Your response differs from the correct answer by orders of magnitude." (b) I said the volume of a ball had 100cm^3 It said "Your response differs from the correct answer by more than 100%." (c) I said the number of balls equaled 7.5e6 It said "Your response differs from the correct answer by more than 100%." Any help would be greatly appreciated! |
| Feb17-10, 11:43 PM | #2 |
|
|
Anyone?
|
| Feb18-10, 01:03 AM | #3 |
|
|
a) Do you really think a typical room is 1000m^3? that would be 20 x 20 x 2.5 m^3
b) google for "size of pingpong ball". Use formula for volume of sphere. |
| Feb18-10, 03:46 AM | #4 |
|
|
Ping-Pong balls in a room?
volume of pingpong ball = 4/3[tex]\pi[/tex]r3
Volume of room = Length*Width*Height let's say that a standard room has a length of 7m width of 4m and a height of 2.5m. and the diameter of a pingpong ball is 4cm or 0.04m. Volume of room = 7*4*2.5 = 70m^2 Volume of room @ 75% = 70*0.75 = 52.5m^3 volume of a pingpong ball = 4/3([tex]\pi[/tex]0.02^3) = 0.0000335m^3 Fitting pingpong balls into room = volume of room / volume of pingpong ball. = 52.5/0.0000335 = 1,567,164 ping pong balls. |
| Feb18-10, 11:57 AM | #5 |
|
|
All of this was very helpful. I did the math on my own and my answer matched yours. I went to enter it in and this is what I got. Attempt 1: (c) Find the number of balls (order of magnitude only) 1.56e6 Balls Your response differs from the correct answer by 10% to 100%. Attempt 2: (I tried rounding) (c) Find the number of balls (order of magnitude only) 1.57e6 Balls Your response differs from the correct answer by 10% to 100%. I only have one more shot at the answer and I know I am correct with the math. What am I doing wrong? |
| Feb18-10, 05:21 PM | #6 |
|
|
Do they not state the size of the room? and specific size of the ball? that's probably why the answers are wrong, because the values we're using are just random.
|
| Feb18-10, 07:30 PM | #8 |
|
|
|
| Feb18-10, 07:39 PM | #9 |
|
|
|
| Feb18-10, 07:46 PM | #10 |
|
Recognitions:
|
|
| Aug28-12, 05:22 PM | #11 |
|
|
I think the issue here is answer format. A correct answer would be written as " ~ 10x #/balls in a room of x*y*z size ", where x clearly denotes the order of magnitude on base-10. Still seems like a sensitive answer to write in online submission though. Note that x #/ping pong balls is clearly separate from volume.
|
| New Reply |
| Thread Tools | |
Similar Threads for: Ping-Pong balls in a room?
|
||||
| Thread | Forum | Replies | ||
| Reliable Method for Launching Ping Pong Balls at Varying Speeds? | Classical Physics | 1 | ||
| How to hallucinate with ping-pong balls and a radio | General Discussion | 3 | ||
| New uses for ping-pong balls | General Discussion | 4 | ||
| Video object tracking: ping-pong balls | Computing & Technology | 1 | ||
| Quantum Ping Pong Balls | Quantum Physics | 0 | ||