Get Help Solving Questions on Octane, Pressure, and Density | Expert Tips"

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The discussion addresses several physics questions related to octane, pressure, and density. For octane, the melting point is misunderstood; the correct approach involves recognizing that the boiling point is the upper limit, leading to a melting point of -57°C. In calculating pressure exerted by a solid cylinder, the area should only consider the circular end, not an additional rectangle. For a submerged metal object, Archimedes' principle explains the difference in apparent weight, emphasizing the relationship between displaced water volume and the object's volume. Lastly, converting units from g/cm³ to kg/m³ involves careful unit analysis, confirming that 1 g/cm³ equals 1000 kg/m³.
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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 I am 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 isn't 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
 
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jai6638 said:
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 I am supposedly supposed ot get -57 and convert that to K... anyone know how to answer this question ? thanks


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??


6638 said:
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 isn't correect for some reason... can anyone please temme why?

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

jai6638 said:
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...

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

jai6638 said:
Q4) How do i convert g/cm3 to kg/m3?

How about appying the definition??

Daniel.
 
\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

how can the length be 183 degrees celcius?? how is it possible? also, to find lower limit then, wouldn't i just subtract ( 183 - 126) same thing what i did... how is it -57?? sorry am confused..


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

stupid me :(.. don't know 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?

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

what is the second law?

How about appying the definition??

well i was doing this

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

is that correct?

thanks much for ur help
 
jai6638 said:
what is the second law?

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}
 
havent learned f=dp/dt... have learned f=ma... though, don't see how i could use that for the above question..
 
jai6638 said:
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 I am supposedly supposed ot get -57 and convert that to K... anyone know how to answer this question ? thanks

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.

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 isn't correect for some reason... can anyone please temme why?

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 ?

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...

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 ?

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

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
 
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