Damayogenius18:
....u can use three formula for elastic collision.....but I will use two here.....M1V1 = M2V2....2nd formula....
U1 + V1 = U2 + V2.....so..3× 8 + 0 = 3V1 + 2V2 for 1...for 2...8 + V1 = 0 + V2..So....8 + V1 = V2.....substitute DAT into equation 1....
3 × 8 = 3V1 + 2(8+V1)
24 = 3V1 + 16 + 2V1
24 - 16 = 5V1
V1 = 8/5 ..V1 = 1.6....
V2 = 8 + V1
V2 = 8 + 1.6
V2 = 9.6ms....V1 = 1.6ms-1, V2 =9.6

Boss, can you pls show how you derive the second formula and also show the third formula.

Orezy5:
*CHEMISTRY*


4. Calculate the mass of sodium hydroxide that must be added to 800cm³ of water to give a solution of pH 10.5

[Na = 23, O = 16, H =1, MV =22.4dm³, NA = 6.02 x 10²³ atoms]

...

Someone should pls solve this questions.

Ajet001:


Boss man!
So you also have drank the water of whining.
Nobody should whine me please...

Ajet baba. The strongest man in mbbs


You're a force to reckon with

Aceed:



Answer: u = 10cm and object distance is
5cm towards the mirror

Second one is 5.3m away

I understand you not

Nltaliban



Boss how's school?

freshsteph:
Nltaliban

Boss how's school?

School is fine.. We're trying our best.

Nltaliban:


School is fine.. We're trying our best.

okay boss....

kinda new on nairaland tho...

freshsteph:

okay boss....


kinda new on nairaland tho...

Welcome, feel free to make friends and ask any question.

Law aspirants can hit me up and ask any questions, I've got y'all covered.

freshsteph:







Velocity of each ball is 4.8m/s

4.8m/s is their ''common velocity'' which is different from ''each velocity''

Specyano:
cant move pass d upload of passport and signature...pls anyone having same issues and how will i go about it...pls quote me asap

just reduce the photo and signature size


but if the problem persists....
remove it and re-upload it

#shalom

Eben331:
Someone should pls solve this questions.

NaOH + H20 ----- Na+ + OH-

Find the hydrogen ion concentration

Ph = -log(H+)

10.5 = -(logH+)

H+ = Antilog -10.5

= Antilog -11 x Antilog +0.5
= 1 x 10^ -11 x 3.1622
= 3.16 x 10^-11 M

Volume = 0.8dm^-3

Therefore, mass conc = molar mass x molar conc

x/0.8 = 40 x 3.16 x 10^ -11

x = 1.0112 x 10^-9g

Orezy5, how correct is this?

Physics

1. A cord passing over a frictionless massless pulley has a 4kg object to one end and a 12kg object fixed to the other end. Find the acceleration and the tension in the end.
A. 4.9m/s^2, 59N
B. 6.0m/s^2, 57N
C. 2.5m/s^2, 59N
D. 9.0m/s^2, 60N
Take g as 9.81m/s^2

2. A hose lying on the ground shoots a stream of water upward at an angle of 40' to the horizontal. The speed of the water is 20m/s as it leaves the hose. How high up will it strike a wall which is 8.0m away

A. 2m high
B. 5.4m high
C. 4m high
D. 6m high

3. The ratio of linear expansivity of copper to iron is approximately 1.5. A specimen of iron and a specimen of copper expand by the same amount per unit rise in temperature. The ratio of their lengths is
A. 3
B. 1.5
C. 1.32
D. 1
E. 0.67

4. A solid object weighs 24g in air, 16g in water and 20g in oil. The ratio of the density of the solid to that of oil is??

Show your workings..

esivue007:
they havent till next week.

Have you checked.?
I said,I checked and confirmed my brother's profile as soon as I read the update here.

onlyme07:

Have you checked.?
I said,I checked and confirmed my brother's profile as soon as I read the update here.

yeah av checked its still says thus "incorrect login credentials".

esivue007:
yeah av checked its still says thus "incorrect login credentials".

Maybe yours but my brother's own which said no login credentials prior the update went through when I read the update. Make sure your Surname is in capital. Best of luck bro.

The correct answers are marked with *

Eben331:
See me see English o (1) Is the load __ heavy that you can't carry it alone? (a) much (b) so * (c) very (d) too (2) The boys look alike that one can't know __ (a) who is whom * (b) who is who (c) which is which (d) whom is whom (3) I will soon attend __ you (a) to * (b) on (c) for (d) at (4) As __ now, Bakayoko is the best player (a) of * (b) at (c) in (d) on (5) You should be __ the alert (a) at (b) on * (c) in (d) by PLS EXPLANATIONS WILL BE NEEDED!

I have confirmed the correct answers to these questions.

Ajet001:


Boss man!
So you also have drank the water of whining.
Nobody should whine me please...

Boss boss.

This is one of the biggest boss I respect in my department medicine and surgery.

No one is whining bro.

3. The ratio of linear expansivity of copper to iron is approximately 1.5. A specimen of iron and a specimen of copper expand by the same amount per unit rise in temperature. The ratio of their lengths is
A. 3
B. 1.5
C. 1.32
D. 1
E. 0.67


increase in length of copper=Lc


increase in length of iron=Li



¥c=Lc/(li@)



Lc=¥cli@



¥i=Li/(li@)


Li=¥ili@




¥c/¥1=3/2


¥c=3¥i/2


Lc=(3¥i/2)li@



Li=¥ili@



Lc/Li=(3¥ili@)/2 ÷ ¥ili@



Lc/Li=3¥ili@/2¥ili@



Lc/Li=3/2=1.5

A hose lying on the ground shoots a stream of water upward at an angle of 40' to the horizontal. The speed of the water is 20m/s as it leaves the hose. How high up will it strike a wall which is 8.0m away
A. 2m high
B. 5.4m high
C. 4m high
D. 6m high




H=maximum height reached...



H=u^2sin^2@/2g



H=20^2×sin^2 40/20


H=8.42m


R=horizontal range=8m


sin40=x/8.42


x=sin40×8.42


x=5.41m

1. A cord passing over a frictionless massless pulley has a 4kg object to one end and a 12kg object fixed to the other end. Find the acceleration and the tension in the end.
A. 4.9m/s^2, 59N
B. 6.0m/s^2, 57N
C. 2.5m/s^2, 59N
D. 9.0m/s^2, 60N
Take g as 9.81m/s^2




applying Newton's second law.




4g-T=4a


T=12a


4g-12a=4a


4g=16a



a=4g/16


a=2.5m/s^2




Tension in which end bro

freshsteph:
1. A cord passing over a frictionless massless pulley has a 4kg object to one end and a 12kg object fixed to the other end. Find the acceleration and the tension in the end.
A. 4.9m/s^2, 59N
B. 6.0m/s^2, 57N
C. 2.5m/s^2, 59N
D. 9.0m/s^2, 60N
Take g as 9.81m/s^2




applying Newton's second law.




4g-T=4a


T=12a


4g-12a=4a


4g=16a



a=4g/16


a=2.5m/s^2




Tension in which end bro

This is wrong.

freshsteph:
A hose lying on the ground shoots a stream of water upward at an angle of 40' to the horizontal. The speed of the water is 20m/s as it leaves the hose. How high up will it strike a wall which is 8.0m away
A. 2m high
B. 5.4m high
C. 4m high
D. 6m high




H=maximum height reached...



H=u^2sin^2@/2g



H=20^2×sin^2 40/20


H=8.42m


R=horizontal range=8m


sin40=x/8.42


x=sin40×8.42


x=5.41m

This one is correct, but I still can't wrap my head around how you got it..

Can you please amplify?

wait lemme post the picture...

that's it...

m2>m1



T-4g=4a


12g-T=ma


T=4g+4a



T=12g-12a


12g-12a=4a+4g


12g-4g=12a+4a


8g=16a


g=2a


9.81=2a


a=4.905m/s=4.91m/s'2


T=mg-ma

T=m(g-a)

T=12(9.81-4.91)

T=59N


A is correct....

JOIN TBOSE TUTORS TODAY AND THE ∆G of your admission this year will be < 0

(1) If 21g of alcohol of density 0.7g/cm^3 is mixed with 10g of water, what would be the density of the resulting mixture? (a) 780g/cm^3 (b) 0.78g/cm^3 (c) 30g/cm^3 (d) 10g/cm^3 (2) For a particle having an x coordinate varies in time according to the expression x=4t-2t^2, the instantaneous velocity of the particle at t=2.5secs is (a) 12m/s (b) 6m/s (c) 0m/s (d) 10m/s PLS SHOW UR WORKINGS!!

TeflonBlixx:
Physics

4. A solid object weighs 24g in air, 16g in water and 20g in oil. The ratio of the density of the solid to that of oil is??
Show your workings..

mass in air=24g, mass in water=16g, mass in oil=20g, density of water=1g/cm^3...r.d of solid=mass in air/mass of equal volume of water=24/24-16=24/8=3 >>> density of solid=r.d of solid*density of water =3*1=3g/cm^3 >>> r.d of oil=mass in air-mass in oil/mass in air-mass in water=24-20/24-16=4/8=0.5 >>> density of oil=r.d of oil*density of water=0.5*1*0.5g/cm...therefore ratio of density of solid to density of oil=density of solid/density of oil=3/0.5=6/1 >>> 6:1

Good day fellow mathematicians and lovers of mathematics. This is to create an awareness of this group created by a set of mathematicians(graduates) to bring an end to mathematical problem across all Nigerian universities.

❓Are you studying mathematics?(100 -400 level)
❓Are you having mathematics as a borrowed course? (100-400level)
❓Do you have an assignment in your mathematical courses?
❓Do you want to have answers to your tutorial questions towards your exams?
❓Do you want us to solve a whole mathematics past question for you?
❓Have u been failing your mathematical courses?

Why not join the group and see the admins putting an end to any kind of problem you have.

We have various packages on ground to attend to your questions.
1. Regular
2. VIQ
3. VVIQ
All depending on how u want us to treat your questions.

We also have packages for online personal tutorials (with chats or video calls).

With the quality of our administrators, we are still the best in what we do.

*N.B: No one knows all!*

*Monetary terms and conditions apply*.

*ELITES MATHS ABODE*

Follow this link to join the WhatsApp group: https:///88Q98c2EM1HF6qQRelePo9

Ayo199:
Good day fellow mathematicians and lovers of mathematics. This is to create an awareness of this group created by a set of mathematicians(graduates) to bring an end to mathematical problem across all Nigerian universities.

❓Are you studying mathematics?(100 -400 level)
❓Are you having mathematics as a borrowed course? (100-400level)
❓Do you have an assignment in your mathematical courses?
❓Do you want to have answers to your tutorial questions towards your exams?
❓Do you want us to solve a whole mathematics past question for you?
❓Have u been failing your mathematical courses?

Why not join the group and see the admins putting an end to any kind of problem you have.

We have various packages on ground to attend to your questions.
1. Regular
2. VIQ
3. VVIQ
All depending on how u want us to treat your questions.

We also have packages for online personal tutorials (with chats or video calls).

With the quality of our administrators, we are still the best in what we do.

*N.B: No one knows all!*

*Monetary terms and conditions apply*.

*ELITES MATHS ABODE*

Follow this link to join the WhatsApp group: https:///88Q98c2EM1HF6qQRelePo9

Whatsapp again??

Enrol now