# Density Of the Object

1. Nov 28, 2008

### Naldo6

An object with density of 831 kg/m^3 and *** of 1003 kg is thrown into te ocean. Find the volume that sticks out of the water. ( use density of water= 1024 kg/m^3)

I use F_b= m'g
m_water*g=m_object*g
p_water*V_water*g=p_object*V_object*g
v_object=(p_water*V_water)/p_object

but what is the volume of the water?....

are my formulas ok?....

2. Nov 29, 2008

### horatio89

Hey Naldo6,

By the Achimedes' Principle, the volume of water displaced = volume of the block submerged. However, since the object is floating, the volume of water displaced is only a fraction of the volume of the whole object. Your equations are correct, however, your problem may have arisen from your definitions of the variables.

V_object=(p_water*V_water)/p_object

Consider this equation - the V_object here is: the volume of the object itself, or the volume of the object submerged? What should be the subject of the equation? (Hint: look at the statement at the beginning of the post)

Regards,
Horatio

Last edited: Nov 29, 2008
3. Nov 29, 2008

### Naldo6

i really dont understand....

4. Nov 29, 2008

### Naldo6

V_object=(1003)(831)=8.33x10^5

but then that is the wrong answer.... how i do this exercise?..help plz

5. Nov 29, 2008

### Staff: Mentor

So far, so good. Now find V_object and V_water and compare.

That's not "the answer", it's a step toward the answer. Good. Now figure out V_water (the volume of water displaced) using your equations. Then calculate out how much of the object's volume sticks out of the water.

6. Nov 29, 2008

### Naldo6

V_water=(V_object*p_object)/p_water
so V_water = (8.33x10^5)(831)/(1024)= 6.76x10^5 m^3

ok but how i can calcuta how much of the objects's volume sticks out of the water

7. Nov 29, 2008

### Staff: Mentor

Hint: The part that sticks out is the part that doesn't displace water.

8. Nov 29, 2008

### Naldo6

ok, but how i know that?....

9. Nov 29, 2008

### Staff: Mentor

You calculated two volumes. One represents the volume under water, the other represents the total volume.

10. Nov 29, 2008

### Naldo6

so that the answer to the problem is the diference between the total volumne and the volume under the water?....

11. Nov 29, 2008

### Staff: Mentor

Yes.

12. Nov 29, 2008

### Naldo6

i did:

8.33x10^5 6.76x10^5= 1.57x10^5 m^3 and that is not the crrect answer...plz can u cheq me in my calculus aboves...

13. Nov 29, 2008

### Staff: Mentor

Oops... this calculation is incorrect.

Density = Mass/Volume, so:

Volume = Mass/Density.

(Redo your values for V_object and V_water.)

14. Nov 29, 2008

### Naldo6

ok ty.... i get my corrects answer... but can u help me with the other exercise i post about density...
The text passage is a discussion thread from a physics forum where a user is asking for help with a density problem and other users are providing guidance and feedback.
single_question_with_answer

the passage contains a single mathematical question about finding the volume of an object that sticks out of the water given its density and mass, and the density of water. the answer to the question is provided through the discussion, with the final correct approach being to calculate the difference between the total volume of the object and the volume of the object submerged in water.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Density Of the Object

1. Nov 28, 2008

### Naldo6

An object with density of 831 kg/m^3 and *** of 1003 kg is thrown into te ocean. Find the volume that sticks out of the water. ( use density of water= 1024 kg/m^3)

I use F_b= m'g
m_water*g=m_object*g
p_water*V_water*g=p_object*V_object*g
v_object=(p_water*V_water)/p_object

but what is the volume of the water?....

are my formulas ok?....

2. Nov 29, 2008

### horatio89

Hey Naldo6,

By the Achimedes' Principle, the volume of water displaced = volume of the block submerged. However, since the object is floating, the volume of water displaced is only a fraction of the volume of the whole object. Your equations are correct, however, your problem may have arisen from your definitions of the variables.

V_object=(p_water*V_water)/p_object

Consider this equation - the V_object here is: the volume of the object itself, or the volume of the object submerged? What should be the subject of the equation? (Hint: look at the statement at the beginning of the post)

Regards,
Horatio

Last edited: Nov 29, 2008
3. Nov 29, 2008

### Naldo6

i really dont understand....

4. Nov 29, 2008

### Naldo6

V_object=(1003)(831)=8.33x10^5

but then that is the wrong answer.... how i do this exercise?..help plz

5. Nov 29, 2008

### Staff: Mentor

So far, so good. Now find V_object and V_water and compare.

That's not "the answer", it's a step toward the answer. Good. Now figure out V_water (the volume of water displaced) using your equations. Then calculate out how much of the object's volume sticks out of the water.

6. Nov 29, 2008

### Naldo6

V_water=(V_object*p_object)/p_water
so V_water = (8.33x10^5)(831)/(1024)= 6.76x10^5 m^3

ok but how i can calcuta how much of the objects's volume sticks out of the water

7. Nov 29, 2008

### Staff: Mentor

Hint: The part that sticks out is the part that doesn't displace water.

8. Nov 29, 2008

### Naldo6

ok, but how i know that?....

9. Nov 29, 2008

### Staff: Mentor

You calculated two volumes. One represents the volume under water, the other represents the total volume.

10. Nov 29, 2008

### Naldo6

so that the answer to the problem is the diference between the total volumne and the volume under the water?....

11. Nov 29, 2008

### Staff: Mentor

Yes.

12. Nov 29, 2008

### Naldo6

i did:

8.33x10^5 6.76x10^5= 1.57x10^5 m^3 and that is not the crrect answer...plz can u cheq me in my calculus aboves...

13. Nov 29, 2008

### Staff: Mentor

Oops... this calculation is incorrect.

Density = Mass/Volume, so:

Volume = Mass/Density.

(Redo your values for V_object and V_water.)

14. Nov 29, 2008

### Naldo6

ok ty.... i get my corrects answer... but can u help me with the other exercise i post about density...
user
The text passage is a discussion thread from a physics forum where a user is asking for help with a density problem and other users are providing guidance and feedback.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question about finding the volume of an object that sticks out of the water given its density and mass, and the density of water. The answer to the question is provided through the discussion, with the final correct approach being to calculate the difference between the total volume of the object and the volume of the object submerged in water.