Pls can u help solve this question

  • Thread starter abuhotlove
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In summary, the planet Jupiter has a mass density that is 317.4 times that of the Earth, as its mass is also 317.4 times that of the Earth. The expression x = vt + ½ at is dimensionally correct, with each term having the same dimensions of length (L). The density of the Earth can be calculated using its mass and radius, while the density of Jupiter can be calculated using its mass and radius. A completely solved example may be helpful in understanding the process.
  • #1
abuhotlove
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1. The planet Jupiter has an average radius 10.95 times that of the average radius of the Earth and a mass 317.4 times that of the Earth. Calculate the ratio of Jupiter’s mass density to the mass density of the Earth.

2. Show that the expression x = vt + ½ at is dimensionally correct , where x is a
coordinate and has units of length, v is speed, a is acceleration, and t is time.
 
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  • #2
Welcome to PF :smile:

First show us what you know, and your thoughts about how to solve the problems. (That's how things work here.)
 
  • #3
for the (2) I did something like this x=L, V=LT^-1, t=T, a=LT^-2 and t^2 = T^2
there4 the eqn is L= LT^-1*T +1/2 LT^-2*T^2
L= L+1/2 L
So I don't know if it is correct so I need help and the first 1 I'm thinking of solving 4 mass/volume bt i don't knw if it correct and that of the Earth I don't know
Pls can u help me on this.
thnx
 
  • #4
abuhotlove said:
for the (2) I did something like this x=L, V=LT^-1, t=T, a=LT^-2 and t^2 = T^2
there4 the eqn is L= LT^-1*T +1/2 LT^-2*T^2
L= L+1/2 L
So I don't know if it is correct so I need help
That's good. Then you'd want to remark that each term in the equation has the same dimensions (L). The "1/2" doesn't matter.

If instead you had something like
L = T + 1/2 L T-1
that would be a problem, because one can't say a length is equal to a time, also one can't add a time to a velocity (L T-1)


... and the first 1 I'm thinking of solving 4 mass/volume bt i don't knw if it correct and that of the Earth I don't know
Pls can u help me on this.
thnx
Yes, we'll use
density = mass/volume​
And you can start by writing an expression for the Earth's density. Let it have mass "Me" and radius "Re", how would you express the density of Earth?
 
  • #5
that I don't know may be u can help me. but I know of desity of the Earth = M/(4pair^3/3)
thnx just help on that I find it difficult that is why I need ur help.
thnx
 
  • #6
the for the eqn of motion the answer in the text said I should solve so that L=L but I got L =3L bcos L+1/2L = 3L , so can u help me too so that I can arrive @ d answer .
thnx
 
  • #7
abuhotlove said:
that I don't know may be u can help me. but I know of desity of the Earth = M/(4pair^3/3)
thnx just help on that I find it difficult that is why I need ur help.
thnx

Okay, yes, that's the density of Earth. But I'm just going to rewrite it:
density of Earth = M/(4 pi r^3/3)​
Next step is to write out the density of Jupiter.
Last step will be to take the ratio of the two densities and work through some algebra.
 
  • #8
abuhotlove said:
the for the eqn of motion the answer in the text said I should solve so that L=L but I got L =3L bcos L+1/2L = 3L , so can u help me too so that I can arrive @ d answer .
thnx
I would look at an example problem from your book or teacher's lecture. That will (hopefully) show how you did indeed get L=L, the correct answer.
 
  • #9
thnx but 1 thing am confuse about is that d mass and radius of the Earth is not given but that of Jupiter is given so how can I calculate 4 earth?
 
  • #10
ok the questn is an exercise in a text, which I supplied d question to u. then at d back of the text the answer is L=L ie x=L = right hand side = L too, so it bcomes L=L so I don't know how they got that.
 
  • #11
abuhotlove said:
thnx but 1 thing am confuse about is that d mass and radius of the Earth is not given ...
That's okay. Just leave the expression as it is,
density of Earth = M/(4 pi r^3/3)​
where M is the mass of the Earth and r is the radius of the Earth.
... but that of Jupiter is given so how can I calculate 4 earth?
Since they don't actually ask for the density of the Earth, it is not necessary to calculate it.

Can you write an expression for the density of Jupiter now?
 
  • #12
abuhotlove said:
ok the questn is an exercise in a text, which I supplied d question to u. then at d back of the text the answer is L=L ie x=L = right hand side = L too, so it bcomes L=L so I don't know how they got that.
What I meant was, you'll need a completely solved example to look at.

Also, in the future please ask different questions in different separate posts. It gets confusing talking about different problems together.
 
  • #13
it still the same formula
 
  • #14
abuhotlove said:
it still the same formula

Well, not quite. M is the density of the Earth, and r is the radius of the Earth. You would need to use the mass and radius of Jupiter instead.

Hint: the mass of Jupiter is 317.4M, according to the problem statement.
 
  • #15
p.s. I'll be logging off the internet for a while. Good luck!
 

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