calculusxy
  • calculusxy
MEDAL!!!!!!!!!!!!! Problem stated below.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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calculusxy
  • calculusxy
I did an experiment today relating to torques. The materials that I had were a meter stick, masking tape, and weights. We had to use the ruler and place the two weights on each side, to try to make it balance to each other. Also, we had to note how far the tape (holding the weight) was from the center (or 50cm). Here is my first data: Mass 1: 200g Distance from 50cm: 19cm Mass 2: 100g Distance from 50cm: 36 cm Then we had to multiply the mass times the distance for each and were asked to look for a pattern. Here are my products: Mass 1: 200g x 19 cm = 3800 Mass 2: 100g x 36 cm = 3600 I don't notice a pattern and need help in finding one.
calculusxy
  • calculusxy
@Nnesha
calculusxy
  • calculusxy
@ikram002p

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DanJS
  • DanJS
looks pretty close
anonymous
  • anonymous
okay so you say your lab involved torque right
calculusxy
  • calculusxy
Yes
DanJS
  • DanJS
The real world will have small errors, not ideal... you did it correct
calculusxy
  • calculusxy
@DanJS So can my conclusion be like since they are both near of each other, it means that they are almost balanced to each other.
DanJS
  • DanJS
Are you discovering torque moments for the first time here, or do you know already
DanJS
  • DanJS
torque = (force perpendicular to the line of action) *(distance)
calculusxy
  • calculusxy
Well it is the second to last day of class, and my teacher didn't want to explain torques to us fully and just let us do the experiment. So we needed to find out pattern with the products.
DanJS
  • DanJS
The rotational moment about a point (here 50cm mark) = distance from that point, times the perpendicular force
anonymous
  • anonymous
here this is what i used to study torque last year https://www.youtube.com/watch?v=sAMNoRBBjzg i hope it helps :)
DanJS
  • DanJS
so if you have a force on an angle, you need to break into components first..
calculusxy
  • calculusxy
I just need to know if the statement that I wrote earlier would be fine for my conclusion or not.
DanJS
  • DanJS
hah, if your teacher is cramming it in this late, probably wont be much of it on the final
calculusxy
  • calculusxy
I already took the final. It's just an extra thing.
DanJS
  • DanJS
oh, is that the only single trial point you did? did you try different distances?
calculusxy
  • calculusxy
yes until the weights were balanced (almost)
DanJS
  • DanJS
I would say something like, The distance multiplied by the force seems almost the same... Need to try further experiments with different masses and distances to test if this is true
DanJS
  • DanJS
one test isn't enough
calculusxy
  • calculusxy
I did another trial, but with three different weights ( two on one side and one on the other). Mass 1 : 100g Distance from 50cm: 28cm Mass 2: 200g Distance from 50cm : 20cm Mass 3: 300g( this weight was kept on the other side) Distance from 50cm: 20cm
calculusxy
  • calculusxy
@DanJS
DanJS
  • DanJS
ok, apply the same reasoning as before... If a torque is clockwise it is negative, and if it is counter clockwise about a point, it is positive... The sum of all the torques should add to zero if the thing is balanced and not moving
DanJS
  • DanJS
so 100*28 + 200*20 ideally would be 300*20
calculusxy
  • calculusxy
Okay I get it thanks. I have some other questions, will you be able to help me on them?
calculusxy
  • calculusxy
This is an example of a velocity a. 45 miles per hour b. 34 meters per second c. 23 centimeters per year northeast d. 12 days per mile north e. 0.1 milliseconds per kilometer f. all of the above are examples of velocity
calculusxy
  • calculusxy
It's C right?
DanJS
  • DanJS
back
DanJS
  • DanJS
yes, a velocity is a vector quantity
DanJS
  • DanJS
has a magnitude and a direction for distance/time
calculusxy
  • calculusxy
second one austin is out for a run. he starts at 98th street and 5th ave and runs 1km north to 110th street in 10 minutes. then, he turns left and runs 0.8km in 9 minutes. then, he turns left and runs down central park west. this takes him 11 min to run down the remaining 1.0km down to 98th street. what is the average velocity?
calculusxy
  • calculusxy
@DanJS
DanJS
  • DanJS
sorry, here
DanJS
  • DanJS
First need to draw all those vectors
DanJS
  • DanJS
|dw:1438729327608:dw|
DanJS
  • DanJS
The average velocity will be the resultant of those 3 vectors, the sum of them
DanJS
  • DanJS
|dw:1438729501422:dw|
DanJS
  • DanJS
The average velocity is the total displacement/ total time Displacement is the start point to the end point, independent of the path
calculusxy
  • calculusxy
I got the velocity of 33.33m/s
calculusxy
  • calculusxy
I am wrong, right?
DanJS
  • DanJS
since they made a square, the magnitude or R is the same as the top piece
calculusxy
  • calculusxy
0.8km/9min = V_avg
DanJS
  • DanJS
|dw:1438729828843:dw|
DanJS
  • DanJS
His net displacement is 0.8 km straight west, the total time is (10 + 9 + 11) min
DanJS
  • DanJS
Average Velocity = Total Disp[lacement / Total Time
calculusxy
  • calculusxy
0.8 / 1800 sec
calculusxy
  • calculusxy
0.0004
DanJS
  • DanJS
I would just leave it in the units the prob already is in , km/min, since it didn't ask to change it
calculusxy
  • calculusxy
Our teacher likes seconds to be used that's why.
DanJS
  • DanJS
0.8km / 30 min
DanJS
  • DanJS
oh, then convert to whatever units you need, but that is the prob
DanJS
  • DanJS
about 0.027 km/min
DanJS
  • DanJS
west
calculusxy
  • calculusxy
THANK YOU!!!

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