Introduction to Physics. I need help with some homework questions.?

Question

Please do not answer "I don't know." I do report those who try to get points when they don't know an answer. Question #1. When a 4kg object is suspended by a scale in water, the scale reads 3.2kg. What is the density of the object? Question #2. Two charged particles are separated by 5m. The attractive force between them is 54N. Find the force between them when they are separated by 15cm.
Please tell me as much as you can (i.e. formulas)to find the answers.

Answer 1

1. the object is pushed up by the weight of its equivalent volume of water, 4-3.2=0.8l

So the object's density is equal to its mass divided by its volume:

4kg/0.8l=4kg/0.0008m3=40,000kg/8m3=5000kg/m3



2. the force decreases with the square of distance, so it will be 3^2=9 times smaller, which is 6N.

Answer 2

I tell you what, you start with what you know and I'll try to help. But I'm not doing your homework for you.



Hint the density issue is an a/b = c/d type equation.



Secondly Intensity is a function of r^2. If the attractive force is gravity, then use a form of (k m1m2/r^2)



As far as reporting I don't know as abuse, It is still a valid response. What a putz!

Answer 3

#1 Compute for the mass of water displaced



mass of water displaced = 4 kg - 3.2 kg



The volume of water is equal to the volume of the object.



volume of water = mass of water / density of water



Compute for the density of the object.



density object = mass of object / volume of water





#2 Coulombs law:



F = G Q q / d^2



G is constant or:



F d^2 / Q q = constant or:



F1 d1^2 / Q1 q1 = F2 d2^2 / Q2 q2 = constant



In the problem, the charges did not change so they will cancel:



F1 d1^2 = F2 d2^2



54 (5^2) = F2 (0.15^2)



Solve for F2.



F1 = initial force

d1 = initial distance

etc.

Answer 4

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Answer 5

I know the precise answer to both of your questions, and I have a nice detailed explanation.



However, I don't like that you threatened to "report" everyone, whatever that means. No answer from me! Nyaaah!

Answer 6

LOSS OF WT OF THE OBJECT= 0.8KG = WT. OF H20 DISPLACED

NOW THE DENSITY OF H2O IS 1000KG/M^3

SO VOL. OF H2O DISPLACED WOULD BE = MASS/DENSITY

0.8/1000= VOL. OF OBJECT

MASS OF OBJECT= 4KG

DENSITY =

5000 KG/M^3

Answer 7

To find the density you need the mass and volume{d=m/v}.you have the mass but not the volume.(I think)

Answer 8

the attraction force between 2 charged particles=q1q2/r^2

so:

54=q1q2/25>>>>q1q2=25*54

f=q1q2/(0.15)^2>>>f=25*54/(0.15)^2