Post Your Maths Problems Here

[Mujteba]

69

Not accounting for air resistance though.

[ZacDaMan72]

Guys search your questions on Wolfram Alpha first, then ask if your questions didn't get answered.

[LukaP]

Guys search your questions on Wolfram Alpha first, then ask if your questions didn't get answered.

This. Wolfram usually does step by step for most of these

 

I have a question for fellow physicists, not because i wouldnt know it, but because i love how unrelated it will all seem.

Sally falls from a height of 20m and takes 2 seconds to reach the ground. Provided you know the orbit of the earth around the sun(149.59787 million kilometers), calculate both the masses and gravitational force between both objects.

Dont worry, it can be done, even though it seems like one of those sally has 4 apples, calculate the mass of the sun thing

[Mujteba]

This. Wolfram usually does step by step for most of these

 

I have a question for fellow physicists, not because i wouldnt know it, but because i love how unrelated it will all seem.

Sally falls from a height of 20m and takes 2 seconds to reach the ground. Provided you know the orbit of the earth around the sun(149.59787 million kilometers), calculate both the masses and gravitational force between both objects.

Dont worry, it can be done, even though it seems like one of those sally has 4 apples, calculate the mass of the sun thing

This is tricky. I presume it would somehow involve constant acceleration formula as well as Newton's law of gravity (or even Kepler's). Pretty sure Kepler's though since we are given the radius.

 

EDIT: To clarify, we want to find mass of Sally and the Sun or the Sun and the Earth??

[LukaP]

This is tricky. I presume it would somehow involve constant acceleration formula as well as Newton's law of gravity (or even Kepler's). Pretty sure Kepler's though since we are given the radius.

 

EDIT: To clarify, we want to find mass of Sally and the Sun or the Sun and the Earth??

Sun and Earth. You cant find the mass of Sally from this dataset

[Mujteba]

Sun and Earth. You cant find the mass of Sally from this dataset

Haha. But I already know the mass of the Sun and Earth 

 

EDIT: Master of page 5

[LukaP]

Haha. But I already know the mass of the Sun and Earth 

 

EDIT: Master of page 5

I know you do, but thats not the point the point is to figure out how to get it from this dataset and its not really that hard ill give you a hint. get g

[Mujteba]

so from g=GM/R^2 I'll have one equation, and I guess work from there

[LukaP]

so from g=GM/R^2 I'll have one equation, and I guess work from there

oh yeah, i guess you need to know the radius of the earth yeah... two circles, for two masses

 

and dont forget Sally falling

[Mujteba]

Something's up here. I managed to get g=10ms^-2 using d=ut+0.5at^2 and then after subbing into g=GM/R^2 found Msun=3.36E33 <---WAAAY off

[LukaP]

Something's up here. I managed to get g=10ms^-2 using d=ut+0.5at^2 and then after subbing into g=GM/R^2 found Msun=3.36E33 <---WAAAY off

where is g measured at? i know what youre doing wrong, but try finding it yourself.

[Mujteba]

where is g measured at? i know what youre doing wrong, but try finding it yourself.

I realise that g is varying so I guess constant acceleration is not the appropriate way to go about it. Although 20m above Earth surface shouldn't be that significant of a difference. Hmmmm this needs some time to ferment..

[LukaP]

I realise that g is varying so I guess constant acceleration is not the appropriate way to go about it. Although 20m above Earth surface shouldn't be that significant of a difference. Hmmmm this needs some time to ferment..

nonono, im not thinking height, im thinking body you were measuring g on earth, then using it to calculate the mass of the sun cant do that

[Mujteba]

nonono, im not thinking height, im thinking body you were measuring g on earth, then using it to calculate the mass of the sun cant do that

Oh.... Ooops didn't even realise that. derp

I'm so use to satellite questions with Kepler's law where the mass of the orbiting body is independent of it's radius

[LukaP]

Oh.... Ooops didn't even realise that. derp

when you get the mass of the earth, and you have its orbit, its trivial

[Mujteba]

when you get the mass of the earth, and you have its orbit, its trivial

Ok so managed to get Msun=1.99E30kg (yay!) Mearth=6.09E24kg (close enough lol) and grav acceleration between two bodies = 5.93E-3ms^-2

[LukaP]

Ok so managed to get Msun=1.99E30kg (yay!) Mearth=6.09E24kg (close enough lol) and grav acceleration between two bodies = 5.93E-3ms^-2

these look rather correct (i dont know them haha) 

 

so congrats. hope someone else does it as well

[Nineshadow]

u Asian? 

Romanian.

[A.D.A.M]

Uhh i need help again.

Not giving me the answer would be nice since i would like to be able to do it myself. For a) i got 30-4x/2pi

[Mujteba]

Uhh i need help again.

Not giving me the answer would be nice since i would like to be able to do it myself. For a) i got 30-4x/2pi PARENTHESES PLEASE!!

IMG_20141201_192917.jpg

Part a

The perimetre is found by taking the initial length of the wire and subtracting the length used for the square; take that expression and equate it to the circumference of a circle using C=2pi*r and then rearrange in terms of x [And please USE BRACKETS when writing out fractions like above]

 

Part b

Use area formulae Acircle=pi*r^2 and Asquare=length^2. For the circle you can replace the radius in the expression with the relationship you found in part a. Then just factorise it to get it in the required form.

 

Part c

Differentiate the area expression in terms of x and equate it to zero (for a minimum value [or could even be maximum which you can verify later on]). Find the value of x and done

[A.D.A.M]

Part a

The perimetre is found by taking the initial length of the wire and subtracting the length used for the square; taking that expression equate that to the circumference of a circle using C=2pi*r and then rearrange in terms of x

so x=2pi*r-30/-4

[Mujteba]

so x=2pi*r-30/-4

Correct but you haven't exactly done what the question has asked you to do; first it asks for perimetre in terms of r (which is uber simple and a bit of a silly question really) just Perimetre=2pi*r and then you have to work from there to find r=blaha blah x-something (because it asked r in terms of x) so essentially your answer just rearranged.

 

Don't forget about the word 'HENCE' there, meaning using this result go on to show that. So start with P=2pi*r and also P=length original wire - Length used for square and then equate the two and solve for r in terms of x.

 

EDIT: PARANTHESES FOR THE FRACTIONS PLEASE or else it looks like two separate terms.

[A.D.A.M]

Correct but you haven't exactly done what the question has asked you to do; first it asks for perimetre in terms of r (which is uber simple and a bit of a silly question really) just Perimetre=2pi*r and then you have to work from there to find r=blaha blah x-something (because it asked r in terms of x) so essentially your answer just rearranged.

 

Don't forget about the word 'HENCE' there, meaning using this result go on to show that. So start with P=2pi*r and also P=length original wire - Length used for square and then equate the two and solve for r in terms of x.

 

EDIT: PARANTHESES FOR THE FRACTIONS PLEASE or else it looks like two separate terms.

So for b i got for the area of the circle: ((225+4x^2-60x)/(pi)) does that seem correct

[Guest]

GOD DAMN IT!

WHY DOES YOUR STUFF HAVE TO BE SOOO SIMPLE.We were doing stuff like that in the 5th grade or something.

I'm in the 9th grade too and ...well...have a look at my last test :

1.Demonstrate that is divisible by 19 whatever value of n (natural)

2.Demonstrate that  (n natural and bigger or equal to 2)

3. [] = ?

where [x] is the smallest integer smaller or equal to x.

4. Calculate 

5. a + b + c + d = 1

Demonstrate that < 8

6.Calculate 

7.Demonstrate that 

8.Demonstrate that >= 3/2  ( Nesbit's inequality)

Also that is basic algebra 1/2

[Mujteba]

So for b i got for the area of the circle: ((225+4x^2-60x)/(pi)) does that seem correct

Yep that's right.

 

Also part c can be done without even doing b since they gave you the expression. It's a handy little thing; whenever you get a 'show that' question just remember that you can use that equation for the remainder of the question even if you were unable to derive it (just some exam advise  )