emekakelvin: 3 in 1 question.

Jamb 2013.
Mathematic question4 type 1

.Evaluate 1.25*0.025/0.05, correct to 1 decimal place.
A)0.6
B)6.2
C)6.3
D)0.5

This maths question do me strong thing 4 hall.

Chemistry Jamb 2013 Type D Number 31
Which of the following compounds is a neutral oxide?
A)sulphur(VI)oxide
B)sulphur(IV)oxide
C) carbon(II)oxide
D)carbon(IV)oxide.

Physic Jamb 2014 Type B, number 4
The pair of physical quantities that are scalar only are
A) length and displacement
B)impulse and time
c)volume and area
D) moment and momentum.

Happy New Year

1.If u use either [1.25x0.025]/0.05 or 1.25x[0.025/0.05],d ansas wil be d same which is 0.63...the correct option is A
2.C.carbon(ii)oxide
3.C.volume and area

aysuccess99: This is my own questions for Physics, they are jamb based question.
1. The magnitude of the resultant of two mutually perpendicular forces, F1, and F2 is 13N. If the magnitude of F1 is 5N, what is the magnitude of F2?
A2.6N B8.0N C12.0N D18.0N
2. A vibrator of frequency 60Hz is used in generating transverse stationery waves in a long thin wire. If the average distance between successive nodes on the wire is 45cm. Find the speed of the transverse waves in the wire.
A.27m/s B 54m/s C90m/s D108m/s
*workings would be highly appreciated* Thanks a lot.

1.
usin pythagoras theorem:F2^2=13^2-5^2
F2^2=169-25
F2^2=144
F2=12
Ansa is C.12.0N

2.f=60Hz,/\=45cm=0.45m
v=/\f=0.45x60=27
the ansa is A.27m/s.

FrancisTony: Under constant tension and constant mass per unit length, the note produced by a plucked string is 500Hz when the length of the string is 0.90m. At what length is the frequency 150Hz?
A. 3m
b. 4m
c. 5m
d. 6m

i av tried my best on dis,it shd be A.3m or that the correct ansa not in the option...if its 2ruly the ansa isn't in the option,it shd be 9m.

Kunlexic: i av tried my best on dis,it shd be A.3m or that the correct ansa not in the option...if its 2ruly the ansa isn't in the option,it shd be 9m.

you're correct but show working

FrancisTony: you're correct but show working

This is the working: The frequency of a sonometer wire is inversely proportional to the length of the wire. So,

F=k/l.......where L is the length and K is the constant..
finally,it SHOULD BE
FREQ(1) * LENGTH(1)=FREQ(2) * LENGHT(2)....where * Means multiplication!

FREQ(2)=( FREQ(1) * LENGTH(1) )/( LENGTH(2) )

FREQ(2)=( 500 * 0.9) / (150)


ANS=3metre!

Dr. Laz'Devitan:


This is the working: The frequency of a sonometer wire is inversely proportional to the length of the wire. So,

F=k/l.......where L is the length and K is the constant..
finally,it SHOULD BE
FREQ(1) * LENGTH(1)=FREQ(2) * LENGHT(2)....where * Means multiplication!

FREQ(2)=( FREQ(1) * LENGTH(1) )/( LENGTH(2) )

FREQ(2)=( 500 * 0.9) / (150)


ANS=3metre!

correct sir!! thumbs up sir.....let's continue with the question and answer...
@kunlexic and ai-bi-ke, ur number1 answer is correct while number 2 is wrong.

Let's continue with the question and answer series. someone should answer this physics question:
1. The magnification of the image of an object placed in front of a convex mirror is 1/3. If the radius of curvature of the mirror is 24cm, what is the distance between the object and its image......
A8cmB16cmC24cmD32cm
2. when a yellow card is observed through a blue glass, the card would appear:
A. black B green Cred D white.
3. Three resistors with resistance 250n,500n,1kn are connected in series A 6v battery is connected to either end of the combination. Calculate the potential difference between the ends of the 250n resistor.
A.0.20vB 0.86v C1.71v D3.43v

Kunlexic:
1.
usin pythagoras theorem:F2^2=13^2-5^2
F2^2=169-25
F2^2=144
F2=12
Ansa is C.12.0N

2.f=60Hz,/\=45cm=0.45m
v=/\f=0.45x60=27
the ansa is A.27m/s.

wat of if i predicted 54 m/s......just guessing.

aysuccess99: This is my own questions for Physics, they are jamb based question.
2. A vibrator of frequency 60Hz is used in generating transverse stationery waves in a long thin wire. If the average distance between successive nodes on the wire is 45cm. Find the speed of the transverse waves in the wire.
A.27m/s B 54m/s C90m/s D108m/s
*workings would be highly appreciated* Thanks a lot.

solution
x=vt/2
x is the distance, v is the speed, t is the time.
t=1/f
therefore, x=v/2f
2f*x=v, when f=60Hz, x=45cm=0.45m
we now have: 2*60*0.45=54m/s
that's the answer my great guru......

great.david:
wat of if i predicted 54 m/s......just guessing.

that's correct sir...you're very correct....thumbs up for you.

aysuccess99:
that's correct sir...you're very correct....thumbs up for you.

I did not see that question before..dis is the working....

The length between two nodes in a string is 0.5¥=L......Where ¥ Is LAMBDA(wavelength)..
Speed=wavelenght * Frequency.....so

¥=2 * 45=90cm=0.9m

speed(V)=¥ * F
= 0.9 * 60
= 54m/s

Thanks!

Dr. Laz'Devitan:


I did not see that question before..dis is the working....

The length between two nodes in a string is 0.5¥=L......Where ¥ Is LAMBDA(wavelength)..
Speed=wavelenght * Frequency.....so

¥=2 * 45=90cm=0.9m

speed(V)=¥ * F
= 0.9 * 60
= 54m/s

Thanks!

correct sir and I love that your workings, it's great.

aysuccess99: Let's continue with the question and answer series. someone should answer this physics question:
1. The magnification of the image of an object placed in front of a convex mirror is 1/3. If the radius of curvature of the mirror is 24cm, what is the distance between the object and its image......
A8cmB16cmC24cmD32cm
2. when a yellow card is observed through a blue glass, the card would appear:
A. black B green Cred D white.
3. Three resistors with resistance 250n,500n,1kn are connected in series A 6v battery is connected to either end of the combination. Calculate the potential difference between the ends of the 250n resistor.
A.0.20vB 0.86v C1.71v D3.43v

1. B 2. D 3. B
waiting for correction

great.david:
1. B 2. D 3. B
waiting for correction

1nd2 ar correct 3

[quote

author=aysuccess99]Let's continue with the question and answer series.
someone should answer this physics question:
1. The magnification of the image of an object placed in front of a
convex mirror is 1/3. If the radius of curvature of the mirror is 24cm,
what is the distance between the object and its image......
A8cmB16cmC24cmD32cm
2. when a yellow card is observed through a blue glass, the card would
appear:
A. black B green Cred D white.
3. Three resistors with resistance 250n,500n,1kn are connected in series
A 6v battery is connected to either end of the combination. Calculate
the potential difference between the ends of the 250n resistor.
A.0.20vB 0.86v C1.71v D3.43v [/quote]

1st QUESTION from my calculation is 32cm

Mirror's formula is 1/f = (1/u) + (1/v).......if u divide both sides by v
u will get

(v/f) = M + 1......where M is magnification...
v=f(M + 1).....where f=0.5r n r=24cm.....so f=-12
v=-12(-.3333 + 1)=-8cm.......
where M=-1/3......U= -3 * -8=24cm

thus,distance between image and object is= 24-(- 8 )=24+8
=32cm

2nd QST.....The answer is GREEN....it is green because,green will be
reflected and because it is SUBRACTIVE and not ADDITIVE colours!

3rd QST....the answer is O.86v.....
When resistances are connected in series add them all...let A=500-ohms.....B=500-Ohms........C=1k-ohms=1000-ohms
Total resistance=A + B + C
=250 + 500 + 1000
=1750-Ohms......From Ohm's law.......p.d=IR....
current I=p.d/R......i.e. 6/1750
=0.00343Amps.....Current flowing through resistances in series is the same so.....
p.d across the 250-ohms= 250 * 0.00343
=0.8571
=0.86v!

you can post any jamb questions on math here https://www.nairaland.com/1147658/nairaland-mathematics-clinic/129#20655165 for quick n accurate solutions from capable hands...trust me..you will never regret it...lets help ourselves ... great work u guys r doin hia...

thanks and 1 love

Following

Dibu_Ojerinde: Following

u're writing jamb what course?
Oya take this cheap question
1) In a metre bridge experiment, two resistors 2ohm and 3ohm occupy the left and right gaps respectively. Find the balance point from the left side of the bridge.

Don't store all your eggs in one basket this again......Take IJMB and gain admission to year two without UTME.
Admission form for is now on sale...visit Apex college of advanced studies, Ilorin or call 08035812144 for more information.

[quote

author=FrancisTony] u're writing jamb what course?
Oya take this cheap question
1) In a metre bridge experiment, two resistors 2ohm and 3ohm occupy the
left and right gaps respectively. Find the balance point from the left
side of the bridge. [/quote]

Let's have it that in between the gaps is a high resistant
load(GALVANOMETER) separating the 2-ohm and 3-ohm resistors.Let's have
it that they are connected to a potentiometer of 100metre.From the
zero-mark to the gap,the left be xmetre....then the other side will be
(100-x)metre.Using the above formular may help out

Resistance(1) * Lenght(2) = Resistance(2) * Length(1)

Resistance(1)=2,Resistance(2)=3,Lenght(1)=x,Length(2)=(100-x)

2 * x=3( 100 - x)
2x=300-3x
5x=300
thus,x=60metre.........

Dibu_Ojerinde: Following

Follow wella

FrancisTony: u're writing jamb what course?

hunting

FrancisTony:
Oya take this cheap question
1) In a metre bridge experiment, two resistors 2ohm and 3ohm occupy the left and right gaps respectively. Find the balance point from the left side of the bridge.

^left/_right = ^X/_100-X

²/₃ = ^X/_100-X

200-2X = 3X

200 = 5X

X = 40cm

: . The balance point is 40cm

Dr. Laz'Devitan:


Let's have it that in between the gaps is a high resistant
load(GALVANOMETER) separating the 2-ohm and 3-ohm resistors.Let's have
it that they are connected to a potentiometer of 100metre.From the
zero-mark to the gap,the left be xmetre....then the other side will be
(100-x)metre.Using the above formular may help out

who send this one

Dr. Laz'Devitan:


Resistance(1) * Lenght(2) = Resistance(2) * Length(1)

Resistance(1)=2,Resistance(2)=3,Lenght(1)=x,Length(2)=(100-x)

2 * x=3( 100 - x)
2x=300-3x
5x=300
thus,x=60centimetre.........

this step contradicts the formula:

^left/_right = ^X/_100-X


was supposed to be 2 (100 - X) = 3X ..
.





Goodmorning boss

[quote
author=Dibu_Ojerinde]
who send this one




this step contradicts the formula:

^left/_right = ^X/_100-X


was supposed to be 2 (100 - X) = 3X ..
.





Goodmorning boss

[/quote]

Step was okay!.......the difference was that in yours you got your
values straight....but in mine....u got to have it by subtracting the
60cm from the 100cm to get the 40cm....this is because i use the other
side....not the side you used!

So i calculated knownly for the 3-ohm....i thought it was for the 2-ohm.....the subtraction...makes mine den correct!

Let me give you guys some of my tipsy questions...to grind u guys up

1)45cm^3 of a mixture of carbon( II ) and carbon ( IV ) oxide were mixed
with 45cm^3 of oxygen,and exploded.The volume of the resulting
gas,measured at the same temperature and pressure,was 70cm^3.Calculate
the volume of the carbon monoxide in the original mixture?

2)A hose directs a horizontal jet of water,moving with a velocity of 20m/s,on to a vertical wall.The cross-sectional area of the jet is 0.0005m^2.If the density of the water is 1Mg/m^3,calculate the force on the wall assuming the water is brought to rest there.

Dibu_Ojerinde:
hunting

^left/_right = ^X/_100-X

²/₃ = ^X/_100-X

200-2X = 3X

200 = 5X

X = 40cm

: . The balance point is 40cm

Uncle Dibu Ojerinde Got it.
Good one sir!

Dr. Laz'Devitan:

Let me give you guys some of my tipsy questions...to grind u guys up

1)45cm^3 of a mixture of carbon( II ) and carbon ( IV ) oxide were mixed
with 45cm^3 of oxygen,and exploded.The volume of the resulting
gas,measured at the same temperature and pressure,was 70cm^3.Calculate
the volume of the carbon monoxide in the original mixture?

2)A hose directs a horizontal jet of water,moving with a velocity of 20m/s,on to a vertical wall.The cross-sectional area of the jet is 0.0005m^2.If the density of the water is 1Mg/m^3,calculate the force on the wall assuming the water is brought to rest there.

hmmm .....these your questions get k-leg oh ... +thinking+

god bless d creator of dis thread

Dr. Laz'Devitan:

Let me give you guys some of my tipsy questions...to grind u guys up
2)A hose directs a horizontal jet of water,moving with a velocity of 20m/s,on to a vertical wall.The cross-sectional area of the jet is 0.0005m^2.If the density of the water is 1Mg/m^3,calculate the force on the wall assuming the water is brought to rest there.

F = m x a,
F = ^{m x v}/_t
V = 20ms^-1
so we have to find mass and time
density = 1mgm^-3
area = 0.0005m²
from the denstity, if mass = 1mg, volume = 1m³
so length = ^volume/_area
= 2000m
if length = 2000m,
20 x time = 2000
time = 100s
Mass = 1mg = 1 x 10⁶kg

F = (1 x 10^-6) x 20 / ₁₀₀
= 0.2 x 10^-6N

....please correct me sir

RECHECK UR QUES@LAZ

Mr Calculus: RECHECK UR QUES@LAZ

yes oh ....d questions are somehow