Jump to content

School help - Physics

Wano97
By Wano97
in Off Topic
 Share
Posted

I was wondering if anyone can help me with a few things for school.

Some things possibly don't make sense, I used google translate for certain words.

 

1. Calculate the power(/output) of a machine which gets 15.000 kilo (=33069.3393 pounds) coal out of a mine which is 150 meters deep.

I'm pretty sure you'll have to use gravity or acceleration due to gravity here which is 9.81meter/second² (m/s²)

 

 

2. A block of steel (bulk density of steel = 7800kg/m3) with a volume of 80 centimeters3 (cm3) gets immersed in petroleum (bulk density of petroleum = 800kg/m3)

 

A) How large is the weight of the block of steel

B ) How large is apparent weight of this immersed steel?

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

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

-snip-

I'm just going to give some notes:

Power = Force * Distance / Time : It's the rate of doing work, where work is force x distance. 

So 1 really confuses me since a "time" isn't given, so I assume it's really asking for work. Which means you have to use gravity * weight to get the required force to move the coal out of the mine, then multiply that by the distance (150m).

2a is just the bulk density multiplied by the volume considering decimal places (boldened because this is where I make errors 90% of the time).

2b is confusing since the total volume of the block of steel + the petroleum isn't given (or the volume of the petroleum itself), unless immersed doesn't mean what I think it means. It didn't mean what I thought it meant. See this post.

2c isn't something I can help on since I don't know anything about buoyancy in a "liquid" (quotes because petroleum is usually a viscous liquid AFAIK).

† 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.

 

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

W=m*g*h should help you out with the first one if I'm not mistaken.

It is said, that civilised man seeks out good and intelligent company, so that through learned discourse he may rise above the savage and closer to God. Personally, however, I like to start the day with a total dickhead to remind me I'm best! 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
  • Author

I'm just going to give some notes:

Power = Force * Distance / Time : It's the rate of doing work, where work is force x distance. 

So 1 really confuses me since a "time" isn't given, meaning I assume it's really asking for work, in which you have to use gravity * weight to get the required force to move the coal out of the mine, then multiply that by the distance (150m).

Yeah I had the second part for 50%. I calculated the Force (F) by using this formula: F=m*g Or in words Force equals mass times gravity acceleration. 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

c is the same answer as b, i think

Intel 3570k 3,4@4,5 1,12v Scythe Mugen 3 gigabyte 770     MSi z77a GD55    corsair vengeance 8 gb  corsair CX600M Bitfenix Outlaw 4 casefans

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

Yeap I got that now :P Thanks anyways

Also, for your second question, part c. You might want to look/think about the "Law of Archimedes in liquids" ;)

It is said, that civilised man seeks out good and intelligent company, so that through learned discourse he may rise above the savage and closer to God. Personally, however, I like to start the day with a total dickhead to remind me I'm best! 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
  • Author

Also, for your second question, part c. You might want to look/think about the "Law of Archimedes in liquids" ;)

Yeah I thought about that, but I don't know those formulas

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

Yeah I thought about that, but I don't know those formulas

 

F=density*g*V

 

Density of the liquid of course.

It is said, that civilised man seeks out good and intelligent company, so that through learned discourse he may rise above the savage and closer to God. Personally, however, I like to start the day with a total dickhead to remind me I'm best! 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

Yeah I thought about that, but I don't know those formulas

 

Yo, from the wiki (to see the 2nd & 3rd formulas better, don't be using Night Theme on LTT):

apparent immersed weight = weight of object - weight of displaced fluid

237bee86559c5c0d935068fd88c93d5d.png

91afaf73f49521c0376339d4071094f4.png

F=density*g*V

I assume V = Vsteel block? Does it matter that we don't know how much petroleum there is? I can't imagine it would. 

† 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.

 

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

W=m*g*h should help you out with the first one if I'm not mistaken.

That's an equation for pressure, not work.

Spoiler

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

I was wondering if anyone can help me with a few things for school.

Some things possibly don't make sense, I used google translate for certain words.

 

1. Calculate the power(/output) of a machine which gets 15.000 kilo (=33069.3393 pounds) coal out of a mine which is 150 meters deep.

I'm pretty sure you'll have to use gravity or acceleration due to gravity here which is 9.81meter/second² (m/s²)

 

 

2. A block of steel (bulk density of steel = 7800kg/m3) with a volume of 80 centimeters3 (cm3) gets immersed in petroleum (bulk density of petroleum = 800kg/m3)

 

A) How large is the weight of the block of steel

B ) How large is apparent weight of this immersed steel?

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

As stated before, #1 is weird as there is a unit mismatch.

 

2a. mass = density * volume

2b. Not sure what its is asking here.

2c. You are looking for the buoyancy force. Fb = density of fluid * gravity * volume of solid. (although it should be noted that the steel is just going to sink to the bottom like a piece of...well steel...until it hits the bottom of whatever container its sitting in or the density of the fluid increased to the point where it equals gravity, wherein it would begin to float at that level.

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

2b. Not sure what its is asking here.

It's basically asking how much the object effectively weighs after considering the force of buoyancy pushing up on the steel block. That's what apparent immersed weight is. The weight of an object in a liquid after considering the force of buoyancy, since the weight is apparently less than it is (due to buoyancy fighting gravity) while the object isn't in a liquid.

From Google:

Archimedes principle says that when an object is immersed in a liquid the apparent loss of weight of an object is equal to the upthrust and this is also equal to the weight of the liquid displaced.

† 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.

 

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

1. Calculate the power(/output) of a machine which gets 15.000 kilo (=33069.3393 pounds) coal out of a mine which is 150 meters deep.

 

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)

Spoiler

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
  • Author

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)

I meant work in that case, since I have to get a outcome with wattage

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

It's basically asking how much the object effectively weighs after considering the force of buoyancy pushing up on the steel block. That's what apparent immersed weight is. The weight of an object in a liquid after considering the force of buoyancy, since the weight is apparently less than it is (due to buoyancy fighting gravity) while the object isn't in a liquid.

From Google:

Ahhh that thing...I remember having to derive that bloody formula in my fluid dynamics midterm because I forgot it. Funny thing was that my answer was more accurate than everyone elses because the accepted version makes certain estimates that I didn't...I was like 0.01% more correct YAY!

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

I meant work in that case, since I have to get a outcome with wattage

Wattage is power which is energy over a specific time. Joules which is the result of the calculations is energy. 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

 

2. A block of steel (bulk density of steel = 7800kg/m3) with a volume of 80 centimeters3 (cm3) gets immersed in petroleum (bulk density of petroleum = 800kg/m3)

 

A) How large is the weight of the block of steel

 

rho (kg/m^3)  =  M (kg) / V (m^3)

 

Hence : 

 

M (kg) = rho (kg/m^3) * V (m^3)

           = 7,800 * 0.00008

           = 0.624 kg

Spoiler

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
  • Author

Wattage is power which is energy over a specific time. Joules which is the result of the calculations is energy. 

Wasn't joules the one with a specific time?

 

ee79aac2f7e6202503a843f449500315.png

Link to comment
Share on other sites
Link to post
Share on other sites
Posted
  • Author

rho (kg/m^3)  =  M (kg) / V (m^3)

 

Hence : 

 

M (kg) = rho (kg/m^3) * V (m^3)

           = 7,800 * 0.00008

           = 0.624 kg

Shouldn't that be like 6240 kg? Since it has a large volume. 0.624 seems a bit light

I've done this: m = 7800kg/m^3 * 0.8m^3 = 6240 Kg

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

2. A block of steel (bulk density of steel = 7800kg/m3) with a volume of 80 centimeters3 (cm3) gets immersed in petroleum (bulk density of petroleum = 800kg/m3)

 

 

B ) How large is apparent weight of this immersed steel?

 

Mass (kg)

Weight (N)

 

 

Apparent weight : weight of object minus weight of displaced volume of surroundings

 

Weight_steel (N) = rho_steel (kg/m^3) * V_steel (m^3) * a_earth (m/s^2)

                           = 7,800 * 0.00008 * 9.81

                           = 6.12144 N

 

Weight_displaced_petroleum (N) = rho_petroleum (kg/m^3) * V_steel (m^3) * a_earth (m/s^2)

                                                    = 800 * 0.00008 * 9.81

                                                     = 0.62784 N

 

Apparent weight (N) = weight_steel (N) - weight_petroleum (N)

                                = 6.12144 - 0.62784

                                 = 5.4936 N

Spoiler

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

Shouldn't that be like 6240 kg? Since it has a large volume. 0.624 seems a bit light

I've done this: m = 7800kg/m^3 * 0.8m^3 = 6240 Kg

0.8 m^3  DOES NOT EQUAL 80 cm^3

Spoiler

 

Link to comment
Share on other sites
Link to post
Share on other sites
Posted

Wasn't joules the one with a specific time?

 

ee79aac2f7e6202503a843f449500315.png

Joules has the units of kg*m^2 per s^2. So both units technically have a time component.

Link to comment
Share on other sites
Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share
Go to topic listing Off Topic

×