Kelp me Please!!!! :")

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Kelp me Please!!!! :")

Physics
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can someone please check the Net force between the person (65kg) and the earth with a mass of 5.98X10^24Kg and radius of 6.38X10^6m. And how do i find the net force between them the second way...i'm not getting the right answer!!!
@PlasmaFuzer Please help me!!!! :")
is it.....|dw:1442282566185:dw|

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Hello whats going on :D
nothing much, just another day stuck on physics as usual :D
i think i got it.... but i don't quite know how!!!! :"D
i was suppose to get the net force between the earth and the person....i got it the first way by applying the conventional formula.... F = G(m1)(m2)/d^2 and got 635N
sorry i forgot to tell you...it's question "b"
and the other way i got the net force was by multiplying the mass off the person with the gravitational force of wearth...9.80N and i got 637N
but what do you call that in terms of an actual formula...?!?!?!?
Ok so the way I would do it is as follows: Since the force is the same (the force of gravity I can cheat and use the ratio of the two.... On Earth F=weight=mg=(65kg)g On Jupiter F'=jupiter weight=mj = (65kg)j Now divide the equations which give you F/F' = g/j
So clearly j=g*(F'\F)
My dear what is F' and m/j, i'm not familiar to those variables.....:"(
i'm sorry if that makes things difficult....but i don't know what they mean...!!! :"(
Ok sorry I am hoping around to much
did i do part b right..?!?!?!? please check ")
Oh that I defined F to be the force of gravity on earth (the weight on earth, F' to be the weight on jupiter and g is the earth grav accel and j is the jupiter grav accel (just made that letter up)
OMG.....i'm sorry, i'm really dumb....please don't mind that :")
Ahhh no your'e part be isnt TOTALLY correct. If you put jupter's mass in a ball with the radius the size of the galaxy it's gravitational acceleration on the surface wouldn't be close to what it is on earth's surface.... what matters is that its Mass/Radius^2 which is honestly why I chose the approach I did...
Oh wait I read the question a bit backwards but my reasoning is still sound... The smaller planet here (earth) definitely DOES NOT have more mass than jupiter... google it ~10^24kg for earth and ~10^27 for jupiter
so your saying it's partly correct?!?!! ?O_oo_O
No now that I reread the question it is totally incorrect sorry :/
Aaawwnnn!!!! :"(
I just glanced at it and thought it was saying the weight on jupiters "surface" was greater than that on earth
but anways i digress
so what have i done wrong dear?!??! Are you sure you read the right part....question A is on the left side of the paper and B is on the right side....i'm sure B is correct because i've double checked it....
Ok so we had j=g*(F'/F)
tow different ways and i', getting the right answer@@@
Ok you will see why I do it this way in a second... it saves having to calculate a lot of numbers
please proceed your highness!!! :")
\[j=g \frac{F_{G_{Jupiter}}}{F_{G_{Earth}}}=g\frac{\frac{GmM_{Jup}}{R_{Jup}^2}}{\frac{GmM_{Earth}}{R_{Earth}^2}} = g \frac{\frac{M_{Jup}}{R_{Jup}^2}}{\frac{M_{Earth}}{R_{Earth}^2}}= g \frac{M_{Jup}R_{Earth}^2}{{R_{Jup}^2M_{Earth}}} \]
So now we have removed all those pesky extra factors... and have expressed it in terms of what we know
Awwww darn I just realized they dont give you the earth mass so you have to go through all the lame calculations..... (thinking) hold on
but it looks scary tho :(
Actually I can fix it hold on:
please don't fix it, let it be..i have a habit following those lame and long calculations....
Oh wait... were you allowed to just look up the values of the radius and mass of the earth?
because if so... all you need to do is take that equation plug in your knowns and evaluate it.... then of course you have found the acceleration on the surface of jupiter (I called it j)... then multiply by the mass to find the weight on the "surface" of jupiter
they were given to me in the question.
Nope I only see Jupiter's mass and radius as givens
But u were definitely using earth's mass and radius in one of the formulas
that's what i did honey.... the second way to get the net force between the person and the earth is by simply multiplying the gravitational force of gravity of earth(9.8m/sec^2) times the mass of the person...65kg...and i get the same answer as in the initial way i got the answer!!!
please ignore the last line on the paper...my second way there is wrong...........i made a mistake there....
Ummm yes part a) has you find the earth weight which is just mg but you did an F_grav calculation with earth's mass and radius which weren't givens... look if your'e teacher is ok with that I am I just wasn't sure because sometimes you have to stick to the knowns in the problem.
Its alright
please ignore the last line on the paper...my second way there is wrong...........i made a mistake there....
So your answer looks correct using the numbers you used int he problem 24.585 which you should express as 24.6 since the largest number of sig figs used in the givens is 3
I think thats how it works... it might be the minimum number of sig figs, but I think I have it right
sorry 24.585 N and 24.6 N shame on me for forgetting units
btw were you referring to art B, because the mass of the earth and the radius were given in the previouse page o i used those to solve for the net force....for part A i on'y used what was given to me...
OOOOO ok then yes then my way totally works... and yes I calculated the answer to part b) using my method... Note how it looks nasty, but I got rid of the extraneous factors before evaluating it
but the question is still there.... do you see the 635N on the paper...????? how do i get the same net force using another way.....
Given the level of accuracy required... it probably doesn't make any difference, but in the real world a scientific calculation should be first expressed as simply as possible with as few variables as possible and then evaluated. Since each term bring with it uncertainties which stack up (though some known values can have absurdly high accuracy like charge/mass ratio of electron)
^^.agree
oh... ummm I thought it was calculate earth's weight and then calculate jupiter's weight two different ways
no just the earth's.....
So the first way is definitely using F_gravity in its full form... then just use mg as the second one
there is another way to find the net force between two objects.....
heck all g represents is this: \[ g= \frac{GM_{Earth}}{R_{Earth}^2} \]
i think it's 9.80N X mass of the person (65kg)
yah...that's it....i get 637N which is pretty close to 635N....
So if you evaluate it long hand by using all those terms... and then evaluate it using just g technically they are the same but I dont know.... No that is incorrect the units of g are acceleration not force ( m/s/s)
i know but i get the right number tho :"(
Correct... but your teacher would also point that out. Units are important
F = mXa a = F/m i know :"(
I know its ok... just a simple error, but I can't let it go
Anyways I have to go and get some food so I will see you next time
so is it right or wrong?!?!?!
You calculated it two different ways and came to pretty much the same answer what do you think? XD
Im not trying to be rude, but since your answers agreed that would suggest they are probably correct.
ok Thank you and love you infinite....GOD bless you :")
Please take care of yourself we need more people like you here on this planet :") Live long.
btw it was on the spot guess kinda thing.....no joke.!! :D
:D your welcome... remember as you progress in your studies you will find it harder and harder for people to be able to help you and then maybe, eventually you will be out there on your own solving problems that don't have known answer. This is a really good time to start practicing your judgement and reasoning skills on problems that are easy, and when you do make an error to learn from it by finding out where you made the error and always being cautious of it down the road.
Again tHank you shoooooooooooooooooooo much!!! :")
:D cya later
Abolutely agree with you....:") Anyways Cya amigo :)

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