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Lionel james

Lionel james

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Lionel james

Lionel james

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Naombeni ushauri kuna mdogo wangu kamaliza kidato cha nne na ana ufaulu wa kiswahil D, English C, Na History D. Je anaweza kujiunga na chuo kwa alama hizi alizopata.
 
Borndi

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Borndi

Borndi

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Naombeni ushauri kuna mdogo wangu kamaliza kidato cha nne na ana ufaulu wa kiswahil D, English C, Na History D. Je anaweza kujiunga na chuo kwa alama hizi alizopata.
Kiongozi jarbu kufuatilia kwenye OPEN UNIVERSITY...Naic anaweza jiunga ..lkn kwanza nenda kwenye ofisi zao wakueleweshe ..wako Tanzania nzima.karbu
 
Samwel Gwiggy

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Samwel Gwiggy

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Naomba kuuliza, stashahada maalumu ya ualimu wa masomo ya sayansi na hisabati ambayo hutolewa muda wa miaka 3 ajira zake ni kufundisha ngazi zipi kati ya sekondari au msingi coz mdogo wang kachaguliwa kuisoma iyo program
 
G-Wu

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G-Wu

G-Wu

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Naombeni msaada wa swali hili

Toa mifano ya misimu iliyozuka katika vipindi vifuatavyo; njaa ya mwaka 1975-1979, pia njaa ya 1998-1999, muda mfupi baada ya kupata uhuru, miaka ya vita vya kagera, kipindi cha hali ngumu ya maisha baada ya vita vya kagera
 
binyampo

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binyampo

binyampo

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msaada kwa anayejua jinsi ya kutuma maombi ya ualimu.
nasikia kuna fomu mtandaoni ya kujaza.
mwenye kujua ikoje tafadhari nijuze
 
selewatatu

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selewatatu

selewatatu

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wakuu nisaidieni nina diploma ya records management, je naweza kujiunga na kozi ya bachelor of law
 
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P

Pencil

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Geography.
Kwa nini hifadhi nyingi za wanyama Tz kuna Tembo? Mfano, Serengeti, Tarangire, Ngorongoro, mikumi n.k.
 
Makanyaga

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Makanyaga

Makanyaga

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Wakuu naomba msaada wa swali la pili hapo. Density of mixture..
Thanks. View attachment 858891
ANSWERS:

QN1:
From the law of floatation: A floating object displaces its own weight in the fluid in which it floats. So the amount of water displaced by the boat is likewise 10000 tonnes


QN2:
The important thing to note here is that, in order for the piece made of both iron and wood (attached together) to float in water, the THE COMBINATION OF THEIR DENSITY must be less than that of water, i.e. assuming their combined total volume is V, then their combined density will be


combined density,(CD)=(mass of iron (MI)+mass of wood)/V................(i)
Also,


The total volume, V=Volume of iron+Volume of wood

V=(Mass of iron,(MI)/density of iron)+(Mass of wood/density of wood)..........................................................(ii)


So the above two equations will provide us with the required solution, bearing in mind that the density of water is 1.0g/cc
With the following information is provided,


Density of iron=0.8g/cc
Density of wood=0.3g/cc
Mass of wood=100g
Volume of iron=?


then in order to get the required volume (of iron), we need first to know the mass of iron (MI), after which we will get its volume by taking the ratio of its mass to its density (given above as 0.8g/cc)

Therefore from (i) above:
CD=(MI+100)/V
and at the moment let's consider this quantity to be equal to the density of water, but bearing in mind that it is supposed to be less that that of water for floating conditions for the pieces of woood and iron to be satisfied.
Therefore:


(MI+100)/V=1
from which we get
V=MI+100


Again from (ii) above

(MI/0.8)+(100/0.3)=V


From these two equations therefore, we finally find that:

(MI/0.8)+(100/0.3)=MI+100 or simply

30MI+8000=24MI+2400

from which upon solving, we get a NEGATIVE VALUE OF MI. Now since the mass of iron cannot be negative, it implies that IT IS THEREFORE IMPOSSIBLE for such a combination of masses of wood and iron to be attached together so that it floats in water!

On the other hand however, assuming the MI were a positive value, we could simply divide MI to the density of iron to get the THRESHOLD volume i.e. the value we were to get would be the maximum volume above which there would be no possibility for the two attached pieces to flow. This is true bearing in mind the assumption we made when we equated to 1.0g/cc, the combined density(CD) of the two pieces (i.e to the density of water, 1.0g/cc)

QN3:

(a) UPTHRUST=weight in air-weight in water=40N-30N=10N

(b) From the law of flotation, Upthrust=weight of water displaced

UPTHRUST=10N=1kgf=1000gf, assuming the acceleration due to gravity is 10 meters per second per second!

(c)Mass of displaced water is 1000g

(d)Volume of displaced water is 1000g/(1.0g/cc)=1000cc

(e) Mass and volume of the object
Weight of object in air is 40N. The mass of the object is therefore 4kg or 4000g
Volume of object=volume of water displaced=mass of water displaced/density of water=1000cc


(f)Relative density (RD) =(Mass/volume)/(density of water)=(4000g/1000cc)/1.0g/cc)=4.0

(g)Density=RDx density of water=4.0x1g/cc=4.0g/cc

QN4:
Given, weight in air 60N=6kgf=6000gf
weight in water 40N


Therefore:
Upthrust=20N=Weight of water displaced=2kgf=2000gf
mass of water displaced=2000g


Volume of the object=2000gxdensity of water=2000gx1.0g/cc=2000cc

Hence:
(a) RD=(6000g/2000cc)/1.0g/cc)=3.0


(b) Density=3.0x1.0g/cc=3.0g/cc

Kama kuna sehemu nimekosea, naomba mnirekebishe, ni siku nyingi. Natumia copyrigt ya ubongo wa mwaka 1987 (More than 30 years!)
 
Caine Chuhira

Caine Chuhira

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Caine Chuhira

Caine Chuhira

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Naombeni Msaada was swali hili, Wadau wa Jukwaa hili LA Elimu :

By connecting dots of historical events, Explain why some regions in the Great lakes are ungovernable by using Burundi, Kenya ,Uganda, Tanzania and Congo Dr.

Msaada please Wadau.!!
 
dronedrake

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dronedrake

dronedrake

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ANSWERS:

QN1:
From the law of floatation: A floating object displaces its own weight in the fluid in which it floats. So the amount of water displaced by the boat is likewise 10000 tonnes


QN2:
The important thing to note here is that, in order for the piece made of both iron and wood (attached together) to float in water, the THE COMBINATION OF THEIR DENSITY must be less than that of water, i.e. assuming their combined total volume is V, then their combined density will be


combined density,(CD)=(mass of iron (MI)+mass of wood)/V................(i)
Also,


The total volume, V=Volume of iron+Volume of wood

V=(Mass of iron,(MI)/density of iron)+(Mass of wood/density of wood)..........................................................(ii)


So the above two equations will provide us with the required solution, bearing in mind that the density of water is 1.0g/cc
With the following information is provided,


Density of iron=0.8g/cc
Density of wood=0.3g/cc
Mass of wood=100g
Volume of iron=?


then in order to get the required volume (of iron), we need first to know the mass of iron (MI), after which we will get its volume by taking the ratio of its mass to its density (given above as 0.8g/cc)

Therefore from (i) above:
CD=(MI+100)/V
and at the moment let's consider this quantity to be equal to the density of water, but bearing in mind that it is supposed to be less that that of water for floating conditions for the pieces of woood and iron to be satisfied.
Therefore:


(MI+100)/V=1
from which we get
V=MI+100


Again from (ii) above

(MI/0.8)+(100/0.3)=V


From these two equations therefore, we finally find that:

(MI/0.8)+(100/0.3)=MI+100 or simply

30MI+8000=24MI+2400

from which upon solving, we get a NEGATIVE VALUE OF MI. Now since the mass of iron cannot be negative, it implies that IT IS THEREFORE IMPOSSIBLE for such a combination of masses of wood and iron to be attached together so that it floats in water!

On the other hand however, assuming the MI were a positive value, we could simply divide MI to the density of iron to get the THRESHOLD volume i.e. the value we were to get would be the maximum volume above which there would be no possibility for the two attached pieces to flow. This is true bearing in mind the assumption we made when we equated to 1.0g/cc, the combined density(CD) of the two pieces (i.e to the density of water, 1.0g/cc)

QN3:

(a) UPTHRUST=weight in air-weight in water=40N-30N=10N

(b) From the law of flotation, Upthrust=weight of water displaced

UPTHRUST=10N=1kgf=1000gf, assuming the acceleration due to gravity is 10 meters per second per second!

(c)Mass of displaced water is 1000g

(d)Volume of displaced water is 1000g/(1.0g/cc)=1000cc

(e) Mass and volume of the object
Weight of object in air is 40N. The mass of the object is therefore 4kg or 4000g
Volume of object=volume of water displaced=mass of water displaced/density of water=1000cc


(f)Relative density (RD) =(Mass/volume)/(density of water)=(4000g/1000cc)/1.0g/cc)=4.0

(g)Density=RDx density of water=4.0x1g/cc=4.0g/cc

QN4:
Given, weight in air 60N=6kgf=6000gf
weight in water 40N


Therefore:
Upthrust=20N=Weight of water displaced=2kgf=2000gf
mass of water displaced=2000g


Volume of the object=2000gxdensity of water=2000gx1.0g/cc=2000cc

Hence:
(a) RD=(6000g/2000cc)/1.0g/cc)=3.0


(b) Density=3.0x1.0g/cc=3.0g/cc

Kama kuna sehemu nimekosea, naomba mnirekebishe, ni siku nyingi. Natumia copyrigt ya ubongo wa mwaka 1987 (More than 30 years!)
aisee
 
Michael joshua

Michael joshua

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Michael joshua

Michael joshua

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Msaada jinsi ya kujibu swali hili
" Account for the impacts of various anthropogenic activities on global climate "
 

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