Please help ooo

Please help oooooo[left][/left]

Thermal expansion? lemme try
I'm not a science student tho

0.0025m

ceejayx:
Thermal expansion? lemme try
I'm not a science student tho

Aw please help me

Olabagazy:
0.0025m

Which one and please explain the answer

Preciousgirl:




Aw please help me

lol d pesin I sent it to is offline

ceejayx:

lol d pesin I sent it to is offline

PPlease give me your Whatsapp line here

Simple shiit.

0.25m

I'm sending a detailed explanation in a bit

We were asked to calculate the difference in length at T2 temperature.

We were given L1 for both rods but not L2.

L1= 5m

Also, we were given the linear expansivity for the both rods.

Linear expansivity for Aluminium rod is given as 23*10^-6

Linear expansivity for Nickel rod was given as 13*10^-6

Now when dealing with length, we talk about linearity not the area. The area expansivity (B-Beta) given was just to confuse you.

The formula for linear expansivity (alpha) is;

Linear expansivity = Change in length (L2-L1)/L1(T2-T1)

I.e L.E = (L2-L1)/L1(T2-T1)

We have all parameters for all except L2, so we calculate L2

For Aluminum rod.

23*10^-6=(L2-5)/5(70-20)

23*10^-6=(L2-5)/250

23*10^-6 * 250 = L2-5

5.75*10^-3 = L2 - 5

5+5.75*10^-3=L2

L2=5.00575m (For Aluminium Rod)

For Nickel Rod

13*10^-6=(L2-5)/5(70-20)

13*10^-6=(L2-5)/250

13*10^-6 * 250 = L2-5

3.25*10^-3= L2 - 5

5+3.25*10^-3=L2

L2= 5.00325m (For nickel rod)

Now the question says the difference in length of both rods when the temp is at 70

L2 of Aluminium - L2 of Nickel = 5.00575m - 5.00325m

2.5*10^-3 or 0.0025m (Your answer)

PS: (The initial length of both rod was 5m before the temperature increased from 20 to 70, therefore you gotta find the new length first and subtract. Always understand a question first before you solve)

(I'll keep modifying until I'm done writing)

CC: Preciousgirl

Difference in length for aluminium should be 0.00575m

Preciousgirl:


PPlease give me your Whatsapp line here

register on www.quora.com then ask you question, you will get profession help from experts in physics within a day. thank me later.

chicoMX:
Difference in length for aluminium should be 0.00575m

This is so wrong. The difference in length of the both rods not the difference in length of individual rods.

Not dL of Nickel and dL of Aluminium

iamsirmichael1:


This is so wrong. The difference in length of the both rods not the difference in length of individual rods.

Not dL of Nickel and dL of Aluminium

Okay..

iamsirmichael1:
We were asked to calculate the difference in length at T2 temperature.

We were given L1 for both rods but not L2.

L1= 5m

Also, we were given the linear expansivity for the both rods.

Linear expansivity for Aluminium rod is given as 23*10^-4

Linear expansivity for Nickel rod was given as 13*10^-4

(I'll keep modifying until I'm done writing)

CC: Preciousgirl

TThanks love, waiting earnestly

Preciousgirl:



TThanks love, waiting earnestly

Can you make out my writings? Or I should go ahead with my initial post?

Lol and my title isn't 'love' but 'Sir'

The pix is from the last to the top

iamsirmichael1:


Can you make out my writings? Or I should go ahead with my initial post?

Lol and my title is 'love' but 'Sir'

Please do the normal writing

iamsirmichael1:


Can you make out my writings? Or I should go ahead with my initial post?

Lol and my title isn't 'love' but 'Sir'

The pix is from the last to the top

Also please can I get your Whatsapp number?

Not too sure

Thanks all I need sure answer please

Can you please drop Whatsapp account so I chat you on Whatsapp

That is the working

iamsirmichael1:
We were asked to calculate the difference in length at T2 temperature.

We were given L1 for both rods but not L2.

L1= 5m

Also, we were given the linear expansivity for the both rods.

Linear expansivity for Aluminium rod is given as 23*10^-6

Linear expansivity for Nickel rod was given as 13*10^-6

Now when dealing with length, we talk about linearity not the area. The area expansivity (B-Beta) given was just to confuse you.

The formula for linear expansivity (alpha) is;

Linear expansivity = Change in length (L2-L1)/L1(T2-T1)

I.e L.E = (L2-L1)/L1(T2-T1)

We have all parameters for all except L2, so we calculate L2

For Aluminum rod.

23*10^-6=(L2-5)/5(70-20)

23*10^-6=(L2-5)/250

23*10^-6 * 250 = L2-5

5.75*10^-3 = L2 - 5

5+5.75*10^-3=L2

L2=5.00575m (For Aluminium Rod)

For Nickel Rod

13*10^-6=(L2-5)/5(70-20)

13*10^-6=(L2-5)/250

13*10^-6 * 250 = L2-5

3.25*10^-3= L2 - 5

5+3.25*10^-3=L2

L2= 5.00325m (For nickel rod)

Now the question says the difference in length of both rods when the temp is at 70

L2 of Aluminium - L2 of Nickel = 5.00575m - 5.00325m

2.5*10^-3 or 0.0025m (Your answer)

PS: (The initial length of both rod was 5m before the temperature increased from 20 to 70, therefore you gotta find the new length first and subtract. Always understand a question first before you solve)

(I'll keep modifying until I'm done writing)

CC: Preciousgirl

Nairaland servers are at it again.

I'm done

If you aren't very much clear, holla at whatsapp.

iamsirmichael1:


Nairaland servers are at it again.

I'm done

Worked it on black andwhite too but servers shit as usual. Your answer would have been right if your standard form had been right , the linear expansivity values were raised to the powers of negative 6 and the answer concomitantly should be .002575m which approximately equals .0026m

Opentokwowledge:
Worked it on black andwhite too but servers shit as usual. Your answer would have been right if your standard form had been right , the linear expansivity values were raised to the powers of negative 6 and the answer concomitantly should be .002575m which approximately equals .0026m

Wait what!! You don't mean it. I thought I saw 4 ooh.

Hahahaha... Garri no good o

iamsirmichael1:


Wait what!! You don't mean it. I thought I saw 4 ooh.

Hahahaha... Garri no good o

. Garri is good oh. You just didnt add the right things with it. Nothing better than garrison doused in cold water with honey and milk playing around downed with fresh suya, baba mi

Opentokwowledge:
. Garri is good oh. You just didnt add the right things with it. Nothing better than garrison doused in cold water with honey and milk playing around downed with fresh suya, baba mi

Hahahaha, now you're making my imagination run wild.

...with those combinations you still call it garri?? Suya?? Ha! When there's money, any thing poor in diet could become wholesome

I like your moniker, what course are you studying or studied?

iamsirmichael1:


Hahahaha, now you're making my imagination run wild.

...with those combinations you still call it garri?? Suya?? Ha! When there's money, any thing poor in diet could become wholesome

I like your moniker, what course are you studying or studied?

I've always been a student of society at the University of life oh, but I work with machines and develop new ways of doing things to earn a living. What about you?

iamsirmichael1:

Wait what!! You don't mean it. I thought I saw 4 ooh.
Hahahaha... Garri no good o

I Hail Thee Oh Thou Lord Physics...

iamsirmichael1:


Hahahaha, now you're making my imagination run wild.

...with those combinations you still call it garri?? Suya?? Ha! When there's money, any thing poor in diet could become wholesome

I like your moniker, what course are you studying or studied?

Hi, I just sent you a Whatsapp message

Opentokwowledge:
I've always been a student of society at the University of life oh, but I work with machines and develop new ways of doing things to earn a living. What about you?

University of Life?? You don't say.

Anyways, I'm still a student at a university in Nigeria studying engineering. That's about as much I can say about myself.

I presume I'm conversing with my uncle's mate. I liked how you circumvent my question and gave your own answer to your own question. Hehehehe

So please Sir, answer 'my' question. We're all in the university of life