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School help - Physics

Wano97

0.8 m^3  DOES NOT EQUAL 80 cm^3

As I said, decimals. They are important @Wano1997 :P

0.8 m3 = 800 cm3 (unless I just don't know what I'm talking about regarding volume)

80 cm3 = 0.08 m3

† Christian Member †

For my pertinent links to guides, reviews, and anything similar, go here, and look under the spoiler labeled such. A brief history of Unix and it's relation to OS X by Builder.

 

 

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2. A block of steel (bulk density of steel = 7800kg/m3) with a volume of 80 centimeters(cm3) gets immersed in petroleum (bulk density of petroleum = 800kg/m3)

 

C) How big is the upward force/uplift that the block of steel encounters from the petroleum?

The answer is weight_petroleum calculated in 2b)

Spoiler

 

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As I said, decimals. They are important @Wano1997 :P

0.8 m3 = 800 cm3 (unless I just don't know what I'm talking about regarding volume)

80 cm3 = 0.08 m3

 

 

The answer is weight_petroleum calculated in 2b)

Oh right I see! Great, thanks guys  :lol:

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Before helping someone, please be sure you are absolutely sure about what you're talking about. Same goes for tech.

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Before helping someone, please be sure you are absolutely sure about what you're talking about. Same goes for tech.

Alright thanks a lot for the help, all of you  :lol:

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Before helping someone, please be sure you are absolutely sure about what you're talking about. Same goes for tech.

(unless I just don't know what I'm talking about regarding volume)

Disclaimers exist for a reason. 

Edit: Just realized there are 100 cm in a meter. I wish the US ran on Metric too. *sigh*

† Christian Member †

For my pertinent links to guides, reviews, and anything similar, go here, and look under the spoiler labeled such. A brief history of Unix and it's relation to OS X by Builder.

 

 

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Disclaimers exist for a reason.

Edit: Just realized there are 100 cm in a meter. I wish the US ran on Metric too. *sigh*

1m = 100cm

1m2 = 10 000cm2

1m3 = 1 000 000cm3

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F (N) = m (kg) a (m/s^2)

 

F = 15,000 * 9.81

   = 147,150 N

 

W (J) = F (N) s (m)   where s is the displacement parallel to the force vector

 

W = 147,150 * 150

    = 22,072,500 J = 22,072.5 kJ

 

This is the WORK done by the machine. it's not it's POWER OUTPUT. To determine it's POWER OUTPUT, you are missing data. you'll need how much time it take for the machine to lift the mass. That is because :

 

P (W) = W (J) / t (s)

^^^Yes, yes, and a little more yes

1.

 

Yep, no time mentioned in the question makes it impossible to give a power output of the machine. A 1W motor could theoretically lift the load, it would just take a long time. All you can do is state the energy/work needed to lift it.

 

Also, using the basic GPE equation is much easier to calculate this value unless this course requires calculus based workflow (it is derived by the same equations anyway). GPE=mgh=(15,000)(150)(9.8)=22,050kj

 

2a.

 

80cm3=8E-5m3

8E-5m3(7800kg/m3)=.624kg (mass of steel block)

 

2b.

 

8E-5m3(800kg/m3)=.064kg (mass of displaced oil)

apparent mass=acutal-displaced=(.624kg)-(.064kg)=.56kg

 

2c.

 

From 2b, uplift=displaced mass=.064kg

 

@Wano1997, sorry if this is too late to be of any use. It looks like everyone has you covered pretty much already. I just wanted to wrap up all the questions in one post. BTW, E(x)=10x just in case you use different notation over there (8E-5=8*10-5)

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^^^Yes, yes, and a little more yes

1.

 

Yep, no time mentioned in the question makes it impossible to give a power output of the machine. A 1W motor could theoretically lift the load, it would just take a long time. All you can do is state the energy/work needed to lift it.

 

Also, using the basic GPE equation is much easier to calculate this value unless this course requires calculus based workflow (it is derived by the same equations anyway). GPE=mgh=(15,000)(150)(9.8)=22,050kj

 

2a.

 

80cm3=8E-5m3

8E-5m3(7800kg/m3)=.624kg (mass of steel block)

 

2b.

 

8E-5m3(800kg/m3)=.064kg (mass of displaced oil)

apparent mass=acutal-displaced=(.624kg)-(.064kg)=.56kg

 

2c.

 

From 2b, uplift=displaced mass=.064kg

 

@Wano1997, sorry if this is too late to be of any use. It looks like everyone has you covered pretty much already. I just wanted to wrap up all the questions in one post. BTW, E(x)=10x just in case you use different notation over there (8E-5=8*10-5)

Yeah I got everything done already, but thanks anyways!

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