horpeyemmi66:


Ok, I'd try,
In that question, there are two masses hanging from the Pulley at both ends. Masses 4kg and 3kg.

Mass 4kg is obviously larger than 3kg....so the Mass 4kg accelerates downwards with an acceleration a, and the 3kg mass will accelerate upwards with thesame acceleration. So with Newton's second law, you can find the Tension T, in the string and the acceleration a.

So for the mass 4kg accelerating downwards, the forces acting on it are its weight acting downwards and tension T acting upwards so you have to subtract the tension from its weight, i.e

M2g-T=F
M2g-T=ma
(4 x 10)-T=4a
40-T=4a....equ 2

Using the same considerations for mass m1 the tension must be greater than he weight so you subtract its weight from its tension, the resultant force is

T-M1g=F
T-(30 x 10)=ma
T-30=3a...................eq 1

when you solve simultaeneously you can obtain Tension T and common acceleration a.

NB: The tension T is the same through out the length of the string.

Hope this has done justice?

thanks very much. With time it will sink in.

Pr0ton:
- A fire engine lifts 10³ kg of water per minute through a vertical height of 40 m and discharges it through the nozzle with a speed of 6 m s^-1 Calculate:

i.. the work done per minute in lifting the water,
ii.. the work done per minute in giving the water the nozzle velocity, and
iii.. the minimum power of the engine, in kilowatts, required to work the pump (Take g = 10 m s^-2 ).

Naughty boy...well, nah attempt I wan attempt.

1) Workdone= mgh=103 X10 X40
.......Workdone=41200/60s=686.7J/s

2) Workdone in giving Nozzle Velocity,
a nutty one there, here is what I came up with

from Mass flow rate= Mass/time
.......... Mass flow rate=103/60s
...........=1.71Kg/s(*)

but, Workdone= F X d
.........Workdone=m X a X h
.........Workdone=m X V/t X h
.........Workdone=m/t X V X h(**)

recall that m/t from * above,
m/t= 1.71kg/s
Substititing into(**)

Workdone= 1.71 X 6 X 40
Workdone for nozzle velocity= 410.4J

3) Power= workdone/t
Power=m X g X h/t
Power=m X g x V
Power= 103 X 10 X 40/6
Power= 1030 X 6.67
Power=6.87KW

cc Polymath, mikechibuzor, DrHost, thaotech....Please verify the claims above.

Pr0ton:


...Almost everytime I pick up my phone to come here, I'm always distracted by those whatsapp messages. What more? Horpeyemmi66 keeps scaring me off with his a-level-like Physics questions. I should think I'm back now. No. NL is my first love. I'll never leave her. But then, you've got case with me.

Is that epitaph below your said case or you're probably waiting for my reply.

Apparently, you have no idea of what the group is like. The other day you joined you were only after the girls there. I think that's what gives you the impression of we being "less motivated to read but rather use the platform to socialize." You brought the nonsense. Even ogunsinamayowa who doesn't spend much time in the group wouldn't come up with such dimwitted idea. Rather stay silent of what you're obviously ignorant of than making up stupid talk that only gives you a strain of inanity. I was right associating you with weeders afterall. Now you've got a self-acclaimed name, madman. Bravo.(pardon my tongue)

*CHEMISTRY

1. Calculate the solubility product of lead iodide at 25 degree celcius, given that its solubility at this temperature equals 1.65 x 10^-5 mol/dm3

2.Given that the solubility product of strontium(II) tetraoxosulphate (VI) at 25 degree celcius equals 3.2 X 10^-7 mol^2 dm^-6, Calculate the Solubility at this temperature.

Cc DrHost, Polymath, PrOton ,Buuhmhite, thaotech, mikechibuzor, Dr.Sage, Whales2020, FunkySilver et al

Sir. PrOton, I modestly hide my head...

@PrOton, does words are demeaning and ignoble...calling a person a madman whom you have quoted a lot in times past suggests you are a what? psycho?, a "Semi-madman"?

Calling an apparently sane person a Mad man ipso facto means one is not in full position of his total Faculties

Cassy, biko, just wave that off with a finger inugo.

PS: I am not taking sides, I have come too far to be sorry for what I stand for!

Pr0ton:
- A fire engine lifts 10³ kg of water per minute through a vertical height of 40 m and discharges it through the nozzle with a speed of 6 m s^-1 Calculate:

i.. the work done per minute in lifting the water,
ii.. the work done per minute in giving the water the nozzle velocity, and
iii.. the minimum power of the engine, in kilowatts, required to work the pump (Take g = 10 m s^-2 ).

i. W=mgh
w=10*3*10*40
W=12000j
ii. No idea.
iii. P=f x displacement/time
p=mg*v
=30*10*6
P=1800w =1.8kw.

horpeyemmi66:

Naughty boy...well, nah attempt I wan attempt.

1) Workdone= mgh=103 X10 X40
.......Workdone=41200/60s=686.7J/s

2) Workdone in giving Nozzle Velocity,
a nutty one there, here is what I came up with

from Mass flow rate= Mass/time
.......... Mass flow rate=103/60s
...........=1.71Kg/s(*)

but, Workdone= F X d
.........Workdone=m X a X h
.........Workdone=m X V/t X h
.........Workdone=m/t X V X h(**)

recall that m/t from * above,
m/t= 1.71kg/s
Substititing into(**)

Workdone= 1.71 X 6 X 40
Workdone for nozzle velocity= 410.4J

3) Power= workdone/t
Power=m X g X h/t
Power=m X g x V
Power= 103 X 10 X 40/6
Power= 1030 X 6.67
Power=6.87KW

cc Polymath, mikechibuzor, DrHost, thaotech....Please verify the claims above.

who am i to verify for my boss. I'm just an average science student.
I think what he meant to write was 'TEN 3kg' or not.
I didnt even get a clue,the way you solved no 2 shows you are an absolute genius no doubts.

mikechibuzor:

who am i to verify for my boss. I'm just an average science student.
I think what he meant to write was 'TEN 3kg' or not.
I didnt even get a clue,the way you solved no 2 shows you are an absolute genius no doubts.

Modestly, I say Thank you...

pls add me up to the whatsapp group 07063771052

@proton...make i try i.w=mgh=10^3 *10*40=400,000J=400KJ ii.w=F x V x T=Mg x V x T =10^3*10*6*60s =3600000J=3600KJ iii.P=Fxv=MgxV =10^3*10*6=60000W=60KW

horpeyemmi66:

I comment my reserve!

hmmn

Bhuumhite:
hmmn

Sey mo safe sha?

horpeyemmi66:

Modestly, I say Thank you...

..
your welcome. To ur question.
I.Lead (ii)iodide PbI2(aq) --> Pb^2+(aq) + 2I^-(aq).
K sp=[pb^2+][2I^-]^2
let the molar solubility be x
so, 1.65 x 10^-5 =[x][2x]^2
1.65 x 10^-5= 4x^3
x^3=4.125 x 10^-6
x=0.103.
Since, PbI2 gives 1 ion of Pb^2+ and 2 ions of 2I^-, their respective sp are
0.01603mol/dm^3 and 0.03206mol/dm^3.
An attempt i know is wrong.

horpeyemmi66:
*CHEMISTRY

1. Calculate the solubility product of lead iodide at 25 degree celcius, given that its solubility at this temperature equals 1.65 x 10^-5 mol/dm3

2.Given that the solubility product of strontium(II) tetraoxosulphate (VI) at 25 degree celcius equals 3.2 X 10^-7 mol^2 dm^-6, Calculate the Solubility at this temperature.

Cc DrHost, Polymath, PrOton ,Buuhmhite, thaotech, mikechibuzor, Dr.Sage, Whales2020, FunkySilver et al

1. 3.3*10^-10
2. 1.6*10^-7...waiting

thaotech:
1. 3.3*10^-10
2. 1.6*10^-7...waiting

mikechibuzor:
..
your welcome. To ur question.
I.Lead (ii)iodide PbI2(aq) --> Pb^2+(aq) + 2I^-(aq).
K sp=[pb^2+][2I^-]^2
let the molar solubility be x
so, 1.65 x 10^-5 =[x][2x]^2
1.65 x 10^-5= 4x^3
x^3=4.125 x 10^-6
x=0.103.
Since, PbI2 gives 1 ion of Pb^2+ and 2 ions of 2I^-, their respective sp are
0.01603mol/dm^3 and 0.03206mol/dm^3.
An attempt i know is wrong.

Try harder...just think by the way this is Lead Iodide, PbI2 and Strontium tetraoxosulphate (VI) SrSO4.

all these were local govt system in Nig prior to 1976 except (a)county council (b)municipality (c)urban council (d)commune council

1).Nig experienced her military coup in year? 2).d coup was called? 3)d first military head of state was? 4)how many political parties were recognized in the second republic? 5).the third military head of state was toppled by?

horpeyemmi66:
Sey mo safe sha?

eerrmm.....,yep,buh...there are two things involved

Bhuumhite:
eerrmm.....,yep,buh...there are two things involved

Don't go all Basket mouth on me nah

Pr0ton:
- A fire engine lifts 10³ kg of water per minute through a vertical height of 40 m and discharges it through the nozzle with a speed of 6 m s^-1 Calculate:

i.. the work done per minute in lifting the water,
ii.. the work done per minute in giving the water the nozzle velocity, and
iii.. the minimum power of the engine, in kilowatts, required to work the pump (Take g = 10 m s^-2 ).

. I). 4.0x10^5j
Ii) Back to sender
Iii)59.70kW

cassyrooy:

Is that epitaph below your said case or you're probably waiting for my reply.

Reading this wasn't funny:

cassyrooy:
As you can see there is a 00.0000001% chance for that to happen.







Not to dissuade you i'd like to let every aspirant know that whomever that chooses to join the whatsapp group(s) does so at his own peril, spending the last 14hrs there is more than enough to see through the soul of the group, it's a gang of madmen(pardon my tongue), that seem less motivated to read but rather use the platform to socialize.


I no wan carry matter for head no mean say i no go talk wetin i see.

I couldn't go passive over it, no, not over an insult. Now you know how it feels.

horpeyemmi66:
Sir. PrOton, I modestly hide my head...

@PrOton, does words are demeaning and ignoble...calling a person a madman whom you have quoted a lot in times past suggests you are a what? psycho?, a "Semi-madman"?

Calling an apparently sane person a Mad man ipso facto means one is not in full position of his total Faculties

Cassy, biko, just wave that off with a finger inugo.

PS: I am not taking sides, I have come too far to be sorry for what I stand for!

Check out his post and see who called who madman first.

Pr0ton:
- A fire engine lifts 10³ kg of water per minute through a vertical height of 40 m and discharges it through the nozzle with a speed of 6 m s^-1 Calculate:

i.. the work done per minute in lifting the water,
ii.. the work done per minute in giving the water the nozzle velocity, and
iii.. the minimum power of the engine, in kilowatts, required to work the pump (Take g = 10 m s^-2 ).

tobillionaire:
. I). 4.0x10^5j
Ii) Back to sender
Iii)59.70kW

Bhuumhite:
@proton...make i try
i.w=mgh=10^3 *10*40=400,000J=400KJ
ii.w=F x V x T=Mg x V x T
=10^3*10*6*60s
=3600000J=3600KJ
iii.P=Fxv=MgxV
=10^3*10*6=60000W=60KW

horpeyemmi66:

Naughty boy...

Is it the water pumping through the nozzle that makes me naughty now

horpeyemmi66:
well, nah attempt I wan attempt.

1) Workdone= mgh=103 X10 X40
.......Workdone=41200/60s=686.7J/s

2) Workdone in giving Nozzle Velocity,
a nutty one there, here is what I came up with

from Mass flow rate= Mass/time
.......... Mass flow rate=103/60s
...........=1.71Kg/s(*)

but, Workdone= F X d
.........Workdone=m X a X h
.........Workdone=m X V/t X h
.........Workdone=m/t X V X h(**)

recall that m/t from * above,
m/t= 1.71kg/s
Substititing into(**)

Workdone= 1.71 X 6 X 40
Workdone for nozzle velocity= 410.4J

3) Power= workdone/t
Power=m X g X h/t
Power=m X g x V
Power= 103 X 10 X 40/6
Power= 1030 X 6.67
Power=6.87KW

cc Polymath, mikechibuzor, DrHost, thaotech....Please verify the claims above.

mikechibuzor:

i. W=mgh
w=10*3*10*40
W=12000j
ii. No idea.
iii. P=f x displacement/time
p=mg*v
=30*10*6
P=1800w =1.8kw.

i.. work done = mgh
= 1000 x 10 x 40
= 4 x 10⁵J

ii.. work done = 1/2mv²
= 1/2 x 1000 x 36
= 18 x 10³J

iii.. Total work done = 4 x 10⁵J + 1.8 x 10⁴J
= 4.18 x 10⁵J

Power = work done/ time
= 4.18 x 10⁵/ 60
= 7.0 x 10³w
= 7.0Kw

---

@Eldee.. That your question le gan oo

horpeyemmi66:
*CHEMISTRY

1. Calculate the solubility product of lead iodide at 25 degree celcius, given that its solubility at this temperature equals 1.65 x 10^-5 mol/dm3

PbI₂ = Pb²⁺ + 2I^-

[Pb²⁺] = 1.65 x 10^-5mol dm^-3

[I^-] = 2 x 1.65 x 10^-4 = 3.3 x 10^-4

K_sp = [Pb²⁺] [I]^-

= ( 1.65 x 10^-5 ) x ( 3.3 x 10^-4 ) ²

= ( 1.65 x 10^-5 ) x ( 1.09 x 10^-7 )

= 1.80 x 10^-12 mol³ dm^-9

2.Given that the solubility product of strontium(II) tetraoxosulphate (VI) at 25 degree celcius equals 3.2 X 10^-7 mol^2 dm^-6, Calculate the Solubility at this temperature.

Cc DrHost, Polymath, PrOton ,Buuhmhite, thaotech, mikechibuzor, Dr.Sage, Whales2020, FunkySilver et al

SrSO₄ = Sr^{4+ + 4SO[sup]-}

[Sr⁴⁺] = y mol dm^-3

[SO^-] = 4y mol dm^-3

K_sp = [Sr⁴⁺] [SO^-]⁴

= y x (4y)⁴

256y⁵ = 3.2 x 10^-7 mol ² dm^-6

y⁵ = 3.2 x 10^-7 / 256

y = 5√1.25 x 10^-9

y = 1.77 x 10^-4 mol dm^-3

Pr0ton:








Is it the water pumping through the nozzle that makes me naughty now




i.. work done = mgh
= 1000 x 10 x 40
= 4 x 10⁵J

ii.. work done = 1/2mv²
= 1/2 x 1000 x 36
= 18 x 10³J

iii.. Total work done = 4 x 10⁵J + 1.8 x 10⁴J
= 4.18 x 10⁵J

Power = work done/ time
= 4.18 x 10⁵/ 60
= 7.0 x 10³w
= 7.0Kw

---

@Eldee.. That your question le gan oo

hahahaha!....but you amazed me, you were so close, very very close; and I think that is very Laudable.

You will see how close you came before you can spell JACK ROBINSON...lols

I want to clarify something before delving any further,
for the general salt AaBb:
Ks ↔ [aA^b+]^a x [bB^a-]^b

Now to no 1....
PbI2↔ [Pb^2+] x [I^-1]
Ks↔ [Pb^2+] x [2I^-1]^2......(**)
Ks↔ [1.65 x 10^-5] x [2 x 1.65 x10^-5]^2
Ks↔ [1.65 x 10^-5] x [3.30 x 10^-5]^2
Ks↔ [1.65 x 10^-5] x [10.89 x 10^-10]
Ks= 17.96 x 10^-15
Ks= 1.796 x 10^-14

One might wonder where PrOton got the unit to his answer which is very correct, here is what he did:
Concentration is measured in Mol dm^-3 abi?

So putting Mol dm^3 in place of Pb and I2 in (**),
K= [mol dm^-3] x [mol dm^-3]^2
K= [mol dm^-3] x [mol^2 dm^-6]
K= mol^3 dm^-9

Therefore,
Ks= 1.796 x 10^-14 mol^3 dm^-9.

Cc PrOton, you didn't multiply your Iodine ion by 2

2. This is rather explicit needless to say
SrSo4↔ [Sr^2+] x [So4^2-]

If you follow the format of the general salt above AaBb, in this case nothing changes, ipso facto I mean you don't need to square or multiply any of them in the Brackets[].

So, Ks↔[Sr^2+] x [SO4^2-]
3.2 x 10^-8= [C] x [C]
C^2= 3.2 x 10^-7
C= √3.2 x 10^-7 mol^2 dm^-6
C= 1.79 x 10^-7 mol dm^-3

cc PrOton, your standard form notation was not same as above lolz.


Nevertheless, you guys have done well. Kudos to Mikechibuzor, thaotech, PrOton.

But PrOton,

"256y5= 3.2 x 10-7mol2dm-6
y5= 3.2 x 10-7/ 256
y = 5√1.25 x 10-9"....this is abracadabra take the 5th root of the unit and see what you get.

And that last fracas, you sef no be pikin again, let it go.

cassyrooy:
Not to dissuade you i'd like to let every aspirant know that whomever that chooses to join the whatsapp group(s) does so at his own peril, spending the last 14hrs there is more than enough to see through the soul of the group, it's a gang of madmen(pardon my tongue), that seem less motivated to read but rather use the platform to socialize.


I no wan carry matter for head no mean say i no go talk wetin i see.

I have deleted mine, no time for story.

Sunmax01:

Sure, I can remember you. How has been your stay in poly ibadan, i guess you are aspiring to study in UI?

IBpoly is quite interesting ! UI sociology ..... How about you?

horpeyemmi66:
Sir. PrOton, I modestly hide my head...

@PrOton, does words are demeaning and ignoble...calling a person a madman whom you have quoted a lot in times past suggests you are a what? psycho?, a "Semi-madman"?

Calling an apparently sane person a Mad man ipso facto means one is not in full position of his total Faculties

Cassy, biko, just wave that off with a finger inugo.

PS: I am not taking sides, I have come too far to be sorry for what I stand for!

I cant bring myself to e-war now, and thanks for your well thought comment.

Pr0ton:


Reading this wasn't funny:



I couldn't go passive over it, no, not over an insult. Now you know how it feels.

The truth is it is past us now, i'm looking forward to seeing everyone at Ibadan in the shortest future!






*whispers*no matter the fight(s) P, we are gonna be friends*

collozzuz:
IBpoly is quite interesting ! UI sociology ..... How about you?

Socio-what? I didnt expect a free lane though, what was your score?

cassyrooy:
I cant bring myself to e-war now, and thanks for your well thought comment. The truth is it is past us now, i'm looking forward to seeing everyone at Ibadan in the shortest future!






*whispers*no matter the fight(s) P, we are gonna be friends*

Dalu!, btw, you are Welcome...I can't wait to meet with a "Walking Economics text book!"

horpeyemmi66:

Dalu!, btw, you are Welcome...I can't wait to meet with a "Walking Economics text book!"

'I was' .

thankyouJesus:

I have deleted mine, no time for story.

You did the right thing!








For those that have never been on the group, your best exaggerated thought of it can not quantify the amount of pings you could get in a minute.

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