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iwanttogotostanford
 one year ago
I
iwanttogotostanford
 one year ago
I

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iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix @triciaal

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0I need to know the acceleration due to gravity and the average G value for each of them given what I have

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0E Earth Moon Mars Distance (m) 2.25 3.08 4.04 4.57 4.68 3.55 Time (s) 0.680 0.793 2.20 2.26 1.58 1.38 Acceleration due to Gravity (m/s^2) 9.732 9.795 1.669 1.789 3.749 3.728 Average G value for the location (m/s^2)

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0I NEED TO KNOW THE AVERAGE G VALUE FOR THE LOCATION

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0I need the equation or formula please I am confused

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0a picture of the chart/table will work better methinks but the acceleration could be obtained from the slope I'd think

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0and average... is more or less rate of change, which brings us back to slope again

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0Earth Moon Mars Distance (m) 2.25 3.08 4.04 4.57 4.68 3.55 Time (s) 0.680 0.793 2.20 2.26 1.58 1.38 Acceleration due to Gravity (m/s^2) 9.732 9.795 1.669 1.789 3.749 3.728 Average G value for the location (m/s^2)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0then again, a picture of the table woiuld help :)

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0This is the chart but it looks a little messed up but it is in line

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0well, surely doesn't look like that on your screen... so just screenshot it methink

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0I'm on a macbook

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0but I will try

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0Here it is

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0it is a download did you get it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://prntscr.com/8q0dt3 There's a screenshot if you can't download the document for some strange reason. For future reference, press command + shift + 4 to screenshot something on a Mac.

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0yes.... well you have two values per each location so the average "G"ravity will be the sum of both, divided by 2 say for earth will be \(\bf \cfrac{9.732+9.795}{2}\)

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0So would they be in order= 14.6295 2.5635 5.613

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0@jdoe0001 Please check

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal Did i do the table right?? :)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0no idea haven't look yet

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal could you check please

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0not sure of final for each location same process velocity = distance /time acceleration = velocity/time for Earth (3.08  2.25)/ (0.7930.680) = velocity velocity/time(1) = 9.732 velocity/time(2) = 9.795 average G = velocity/(t2  t1)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444528438601:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate can you help with this?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1@iwanttogotostanford You would need the kinematics equation for an object in free fall, \(\Delta x= (1/2)gt^2\) where g=acceleration due gravity, t=time it takes, (initial velocity = 0 for free fall) Solve for g to get \(g=2\Delta x /t^2\) That's how @jdoe001 got 9.732 and 9.795 for the first pair of data. Continue with the two other pairs for moon and mars, and you'll have a nice table.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Average value would be, as @jdoe0001 stated, the sum divided by 2, for each pair.

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate I know but what about for the last row of the table i am confused

iwanttogotostanford
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 how do you figure out the last row i know the third row but did i do the last row correctly?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Location Earth Moon Mars Acceleration due to Gravity (m/s^2) \(\color{red}{(}\)9.732 9.795\(\color{red}{)}\) \(\color{red}{(}\)1.669 1.789\(\color{red}{)}\) \(\color{red}{(}\)3.749 3.728\(\color{red}{) }\) Average acceleration 9.764 1.729 3.739 For earth, the average is (9.732+9.795)/2=9.764 For moon, the average is (1.669+1.789)/2=1.729 For mars, the average is (3.749+3.728)/2=3.739
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