# Help needed to solve this

1. Jan 15, 2005

### jai6638

The following are questions from my test which i got wrong.. need help solving them :

q1) Octane, a substance found in petroleum, boils at 126 Celcius and is a liquid over a range of 183 celcius degrees. What is the melting point of octane in kelvins?

I subtracted ( 183 - 126 ) and got 57 degrees celcius as my answer.. and converted that into K.. however, checked the answer and im supposedly supposed ot get -57 and convert that to K... anyone know how to answer this question ? thanks

q2) A 75 kg solid cylinder, 2.5 m long and with an end radius of 5 cm, stands on one end. How much pressure does it exert?

I first found the area of the cylinder by adding AREA OF THE CIRCLE + AREA OF RECTANGLE and got 2.57x10^-1 m^2 ...

I used eq = P = f/a ..
p = 75x9.8/ 2.57x10^-1 = 2.85x10^3 PA..

however, it turns out that this isnt correect for some reason... can anyone please temme why?

q3) A metal object is suspended from a spring scale. The scale reads 920 N when the object is suspended in air and 750 N when the object is completely submerged in water ( density = 1000 kg/m^3)

A) Find the volume of the object
B) Find the density of the metal

for A, i used the equation Fb= pVg but turns out that the answer i got was wrong
For B, i used p=fB/VG...

Q4) How do i convert g/cm3 to kg/m3?

thanks much

2. Jan 15, 2005

### dextercioby

Okay.Think of it this way:imagine a temperatura scale in degrees Celsius.The liquid state is marked by two points on this scale:the melting point (the lower one) and the boiling point (the upper one).Between these temps.the substance is in liquid state.
In your problem,u're given the boiling point (the upper limit of the interval) and the length of the interval.If the length is 183 and the upper limit is 126,what is the lower limit??

What rectangle are u talking about??The circle has no area...It has only length...

How about applying the principles of dynamics??Make a vector diagram,write the second law for the 2 situations and draw conclusions...

Daniel.

3. Jan 16, 2005

### jai6638

how can the length be 183 degrees celcius?? how is it possible??? also, to find lower limit then, wouldnt i just subtract ( 183 - 126) same thing what i did... how is it -57?? sorry am confused..

stupid me :(.. dunno where i got the rectangle stuck in my head.... so basically i should have just used the formula for area of a circle , i.e. , pi (r)2 right?

what is the second law?

well i was doin this

1gm/cm3 x (1/1000 / 1/10000) = which would equal 100 kg/m3...

is that correct?

thanks much for ur help

4. Jan 16, 2005

### da_willem

Newtons second law, aka 'the' second law:

$$\vec{F}=\frac{d\vec{p}}{dt}$$

or for constant masses the derivative to mass vanishes and what remains is:

$$\vec{F}=m\frac{\vec{v}}{dt}=m\vec{a}$$

5. Jan 16, 2005

### jai6638

havent learnt f=dp/dt... have learnt f=ma... though, dont see how i could use that for the above question..

6. Jan 16, 2005

### Curious3141

Octane undergoes the transition from gas -> liquid -> solid. You're given the boiling point value (the temperature for the 1st transition). You're know that the second value is 183 degrees lower. So what is the second transition point (melting point) ? (hint : you're subtracting the wrong way around).

I assume you know how to convert to Kelvin after that.

Bad question. Pressure exerted on what ?

Assuming the cylinder is standing on its end on a flat surface, and they're asking for the pressure on the surface, why are you adding the area of the 'rectangle' ? You got the force right (the weight) but shouldn't the area be only the area of the end of the cylinder ?

Why is the apparent weight of a solid object lower when it is immersed in liquid ? What is Archimedes' principle ? What can you say about the volume of water displaced vs the volume of the metal object ?

Just work through the units carefully.

$$1 g/{cm}^3 = 10^{-3}kg/(10^{-2}m)^3 = 10^{-3}kg/(10^{-6}m^3) = 1000 kg/m^3$$